Craig’s Soil Mechanics Seventh Edition - · PDF fileCraig’s Soil Mechanics Seventh Edition Solutions Manual R.F. Craig Formerly DepartmentofCivilEngineering UniversityofDundeeUK

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Craigrsquos Soil Mechanics Seventh EditionSolutions Manual

Craigrsquos Soil Mechanics SeventhEdition Solutions Manual

RF CraigFormerly

Department of Civil Engineering

University of Dundee UK

First published 1992by E amp FN Spon an imprint of Thomson ProfessionalSecond edition 1997Third edition 200411 New Fetter Lane London EC4P 4EE

Simultaneously published in the USA and Canadaby Spon Press29 West 35th Street New York NY 10001

Spon Press is an imprint of the Taylor amp Francis Group

ordf 1992 1997 2004 RF Craig

All rights reserved No part of this book may be reprinted or reproducedor utilised in any form or by any electronic mechanical or other meansnow known or hereafter invented including photocopying and recordingor in any information storage or retrieval system without permission inwriting from the publishers

British Library Cataloguing in Publication DataA catalogue record for this book is availablefrom the British Library

Library of Congress Cataloging in Publication DataA catalog record for this book has been requested

ISBN 0ndash415ndash33294ndashX

This edition published in the Taylor amp Francis e-Library 2004

(Print edition)

ISBN 0-203-31104-3 Master e-book ISBN

ISBN 0-203-67167-8 (Adobe eReader Format)

collection of thousands of eBooks please go to wwweBookstoretandfcoukrdquoldquoTo purchase your own copy of this or any of Taylor amp Francis or Routledgersquos

Contents

1 Basic characteristics of soils 1

2 Seepage 6

3 Effective stress 14

4 Shear strength 22

5 Stresses and displacements 28

6 Lateral earth pressure 34

7 Consolidation theory 50

8 Bearing capacity 60

9 Stability of slopes 74

Authorrsquos note

In order not to short-circuit the learningprocess it is vital that the reader shouldattempt the problems before referring to thesolutions in this manual

Chapter 1

Basic characteristics of soils

11

Soil E consists of 98 coarse material (31 gravel size 67 sand size) and 2 finesIt is classified as SW well-graded gravelly SAND or in greater detail well-gradedslightly silty very gravelly SAND

Soil F consists of 63 coarse material (2 gravel size 61 sand size) and 37non-plastic fines (ie between 35 and 65 fines) therefore the soil is classified as MSsandy SILT

Soil G consists of 73 fine material (ie between 65 and 100 fines) and 27 sandsize The liquid limit is 32 and the plasticity index is 8 (ie 32 24) plotting marginallybelow the A-line in the ML zone on the plasticity chart Thus the classification is MLSILT (M-SOIL) of low plasticity (The plasticity chart is given in Figure 17)

Figure Q11

Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

12

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

191frac14 155

e frac14 055

Using Equation 113

Sr frac14 wGs

efrac14 0095 270

055frac14 0466 eth466THORN

Using Equation 119

sat frac14 Gs thorn e1thorn e w frac14 325

155 100 frac14 210Mg=m3

From Equation 114

w frac14 e

Gsfrac14 055

270frac14 0204 eth204THORN

13

Equations similar to 117ndash120 apply in the case of unit weights thus

d frac14 Gs

1thorn e w frac14272

170 98 frac14 157 kN=m3

sat frac14 Gs thorn e1thorn e w frac14 342

170 98 frac14 197 kN=m3

Using Equation 121

0 frac14 Gs 1

1thorn e w frac14 172

170 98 frac14 99 kN=m3

Using Equation 118a with Srfrac14 075

frac14 Gs thorn Sre

1thorn e w frac14 3245

170 98 frac14 187 kN=m3

2 Basic characteristics of soils

Using Equation 113

w frac14 Sre

Gsfrac14 075 070

272frac14 0193 eth193THORN

The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

14

Volume of specimenfrac14

438276 frac14 86 200mm3

Bulk density ethTHORN frac14 Mass

Volumefrac14 1680

86 200 103frac14 195Mg=m3

Water content ethwTHORN frac14 1680 1305

1305frac14 0287 eth287THORN

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

195frac14 180

e frac14 080

Using Equation 113

Sr frac14 wGs

efrac14 0287 273

080frac14 098 eth98THORN

15

Using Equation 124

d frac14

1thorn w frac14215

112frac14 192Mg=m3

From Equation 117

1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

215frac14 138

e frac14 038

Using Equation 113

Sr frac14 wGs

efrac14 012 265

038frac14 0837 eth837THORN

Basic characteristics of soils 3

Using Equation 115

Afrac14 e wGs

1thorn e frac14038 0318

138frac14 0045 eth45THORN

The zero air voids dry density is given by Equation 125

d frac14 Gs

1thorn wGsw frac14 265

1thorn eth0135 265THORN 100 frac14 195Mg=m3

ie a dry density of 200Mgm3 would not be possible

16

Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

(Mgm3) d10(Mgm3)

2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

wopt frac14 15 and dmaxfrac14 183Mg=m3

Figure Q16

4 Basic characteristics of soils

Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

d5 d10

respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

17

From Equation 120

e frac14 Gswd 1

The maximum and minimum values of void ratio are given by

emax frac14 Gsw

dmin

1

emin frac14 Gswdmax

1

From Equation 123

ID frac14 Gsweth1=dmin 1=dTHORN

Gsweth1=dmin 1=dmax

THORN

frac14 frac121 ethdmin=dTHORN1=dmin

frac121 ethdmin=dmax

THORN1=dmin

frac14 d dmin

dmax dmin

dmax

d

frac14 172 154

181 154

181

172

frac14 070 eth70THORN

Basic characteristics of soils 5

Chapter 2

Seepage

21

The coefficient of permeability is determined from the equation

k frac14 23al

At1log

h0

h1

where

a frac14

4 00052 m2 l frac14 02m

A frac14

4 012 m2 t1 frac14 3 602 s

logh0

h1frac14 log

100

035frac14 0456

k frac14 23 00052 02 0456

012 3 602frac14 49 108 m=s

22

The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

q frac14 kh Nf

Ndfrac14 106 400 37

11frac14 13 106 m3=s per m

Figure Q22

23

The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

q frac14 kh Nf

Ndfrac14 5 105 300 35

9frac14 58 105 m3=s per m

The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

Point h (m) h z (m) u frac14 w(h z)(kNm2)

1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

eg for Point 1

h1 frac14 7

9 300 frac14 233m

h1 z1 frac14 233 eth250THORN frac14 483m

Figure Q23

Seepage 7

u1 frac14 98 483 frac14 47 kN=m2

The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

24

The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

q frac14 kh Nf

Ndfrac14 40 107 550 10

11frac14 20 106 m3=s per m

25

The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

q frac14 kh Nf

Ndfrac14 20 106 500 42

9frac14 47 106 m3=s per m

Figure Q24

8 Seepage

26

The scale transformation factor in the x direction is given by Equation 221 ie

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

p 107 frac14 30 107 m=s

Thus the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 30 107 1400 325

12frac14 11 106 m3=s per m

Figure Q25

Seepage 9

27

The scale transformation factor in the x direction is

xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

pfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

p 106 frac14 45 106 m=s

The focus of the basic parabola is at point A The parabola passes through point Gsuch that

GC frac14 03HC frac14 03 30 frac14 90m

Thus the coordinates of G are

x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

480 frac14 x0 2002

4x0

Figure Q26

10 Seepage

Hence

x0 frac14 20m

Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

x frac14 20 z2

80

x 20 0 50 100 200 300z 0 400 748 980 1327 1600

The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

28

The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

Figure Q27

Seepage 11

q frac14 kh Nf

Ndfrac14 45 105 28 33

7

frac14 59 105 m3=s per m

29

The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

kx frac14 H1k1 thornH2k2

H1 thornH2frac14 106

10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

kz frac14 H1 thornH2

H1

k1thornH2

k2

frac14 10

5

eth2 106THORN thorn5

eth16 106THORNfrac14 36 106 m=s

Then the scale transformation factor is given by

xt frac14 xffiffiffiffiffikz

pffiffiffiffiffikx

p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

Figure Q28

12 Seepage

k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

qfrac14

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

p 106 frac14 57 106 m=s

Then the quantity of seepage is given by

q frac14 k0h Nf

Ndfrac14 57 106 350 56

11

frac14 10 105 m3=s per m

Figure Q29

Seepage 13

Chapter 3

Effective stress

31

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

Pore water pressure

u frac14 7 98 frac14 686 kN=m2

Effective vertical stress

0v frac14 v u frac14 1196 686 frac14 51 kN=m2

32

Buoyant unit weight

0 frac14 sat w frac14 20 98 frac14 102 kN=m3

Effective vertical stress

0v frac14 5 102 frac14 51 kN=m2 or

Total vertical stress

v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

Pore water pressure

u frac14 205 98 frac14 2009 kN=m2

Effective vertical stress

0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

33

At top of the clay

v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

u frac14 2 98 frac14 196 kN=m2

0v frac14 v u frac14 710 196 frac14 514 kN=m2

Alternatively

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

At bottom of the clay

v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

u frac14 12 98 frac14 1176 kN=m2

0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

NB The alternative method of calculation is not applicable because of the artesiancondition

Figure Q3132

Effective stress 15

34

0 frac14 20 98 frac14 102 kN=m3

At 8m depth

0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

35

0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

Figure Q33

Figure Q34

16 Effective stress

(a) Immediately after WT rise

At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

(b) Several years after WT rise

At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

At 8m depth

0v frac14 940 kN=m2 (as above)

At 12m depth

0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

Figure Q35

Effective stress 17

36

Total weight

ab frac14 210 kN

Effective weight

ac frac14 112 kN

Resultant boundary water force

be frac14 119 kN

Seepage force

ce frac14 34 kN

Resultant body force

ae frac14 99 kN eth73 to horizontalTHORN

(Refer to Figure Q36)

Figure Q36

18 Effective stress

37

Situation (1)(a)

frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 694 392 frac14 302 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

Situation (2)(a)

frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

0 frac14 u frac14 498 392 frac14 106 kN=m2

(b)

i frac14 2

4frac14 05

j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

38

The flow net is drawn in Figure Q24

Loss in total head between adjacent equipotentials

h frac14 550

Ndfrac14 550

11frac14 050m

Exit hydraulic gradient

ie frac14 h

sfrac14 050

070frac14 071

Effective stress 19

The critical hydraulic gradient is given by Equation 39

ic frac14 0

wfrac14 102

98frac14 104

Therefore factor of safety against lsquoboilingrsquo (Equation 311)

F frac14 iciefrac14 104

071frac14 15

Total head at C

hC frac14 nd

Ndh frac14 24

11 550 frac14 120m

Elevation head at C

zC frac14 250m

Pore water pressure at C

uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

Therefore effective vertical stress at C

0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

For point D

hD frac14 73

11 550 frac14 365m

zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

39

The flow net is drawn in Figure Q25

For a soil prism 150 300m adjacent to the piling

hm frac14 26

9 500 frac14 145m

20 Effective stress

Factor of safety against lsquoheavingrsquo (Equation 310)

F frac14 ic

imfrac14 0d

whmfrac14 97 300

98 145frac14 20

With a filter

F frac14 0d thorn wwhm

3 frac14 eth97 300THORN thorn w98 145

w frac14 135 kN=m2

Depth of filterfrac14 13521frac14 065m (if above water level)

Effective stress 21

Chapter 4

Shear strength

41

frac14 295 kN=m2

u frac14 120 kN=m2

0 frac14 u frac14 295 120 frac14 175 kN=m2

f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

42

03 (kNm2) 1 3 (kNm2) 01 (kNm2)

100 452 552200 908 1108400 1810 2210800 3624 4424

The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

Figure Q42

43

3 (kNm2) 1 3 (kNm2) 1 (kNm2)

200 222 422400 218 618600 220 820

The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

44

The modified shear strength parameters are

0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are

1

2eth1 2THORN frac14 1

2 170 frac14 85 kN=m2

1

2eth1 thorn 3THORN frac14 1

2eth270thorn 100THORN frac14 185 kN=m2

The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

uf frac14 36 kN=m2

Figure Q43

Figure Q44

Shear strength 23

45

3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

46

03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

Figure Q45

24 Shear strength

47

The torque required to produce shear failure is given by

T frac14 dh cud

2thorn 2

Z d=2

0

2r drcur

frac14 cud2h

2thorn 4cu

Z d=2

0

r2dr

frac14 cud2h

2thorn d

3

6

Then

35 frac14 cu52 10

2thorn 53

6

103

cu frac14 76 kN=m3

400

0 400 800 1200 1600

τ (k

Nm

2 )

σprime (kNm2)

34deg

315deg29deg

(a)

(b)

0 400

400

800 1200 1600

Failure envelope

300 500

σprime (kNm2)

τ (k

Nm

2 )

20 (kNm2)

31deg

Figure Q46

Shear strength 25

48

The relevant stress values are calculated as follows

3 frac14 600 kN=m2

1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

2(1 3) 0 40 79 107 139 159

1

2(01 thorn 03) 400 411 402 389 351 326

1

2(1 thorn 3) 600 640 679 707 739 759

The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

Af frac14 433 200

319frac14 073

49

B frac14 u33

frac14 144

350 200frac14 096

a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

10 283 65 023

Figure Q48

26 Shear strength

The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

Figure Q49

Shear strength 27

Chapter 5

Stresses and displacements

51

Vertical stress is given by

z frac14 Qz2Ip frac14 5000

52Ip

Values of Ip are obtained from Table 51

r (m) rz Ip z (kNm2)

0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

10 20 0009 2

The variation of z with radial distance (r) is plotted in Figure Q51

Figure Q51

52

Below the centre load (Figure Q52)

r

zfrac14 0 for the 7500-kN load

Ip frac14 0478

r

zfrac14 5

4frac14 125 for the 10 000- and 9000-kN loads

Ip frac14 0045

Then

z frac14X Q

z2Ip

frac14 7500 0478

42thorn 10 000 0045

42thorn 9000 0045

42

frac14 224thorn 28thorn 25 frac14 277 kN=m2

53

The vertical stress under a corner of a rectangular area is given by

z frac14 qIr

where values of Ir are obtained from Figure 510 In this case

z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

z

Figure Q52

Stresses and displacements 29

z (m) m n Ir z (kNm2)

0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

10 010 0005 5

z is plotted against z in Figure Q53

54

(a)

m frac14 125

12frac14 104

n frac14 18

12frac14 150

From Figure 510 Irfrac14 0196

z frac14 2 175 0196 frac14 68 kN=m2

Figure Q53

30 Stresses and displacements

(b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

55

Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

Px frac14 2Q

1

m2 thorn 1frac14 2 150

125frac14 76 kN=m

Equation 517 is used to obtain the pressure distribution

px frac14 4Q

h

m2n

ethm2 thorn n2THORN2 frac14150

m2n

ethm2 thorn n2THORN2 ethkN=m2THORN

Figure Q54

Stresses and displacements 31

n m2n

(m2 thorn n2)2

px(kNm2)

0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

The pressure distribution is plotted in Figure Q55

56

H

Bfrac14 10

2frac14 5

L

Bfrac14 4

2frac14 2

D

Bfrac14 1

2frac14 05

Hence from Figure 515

131 frac14 082

130 frac14 094

Figure Q55

32 Stresses and displacements

The immediate settlement is given by Equation 528

si frac14 130131qB

Eu

frac14 094 082 200 2

45frac14 7mm

Stresses and displacements 33

Chapter 6

Lateral earth pressure

61

For 0 frac14 37 the active pressure coefficient is given by

Ka frac14 1 sin 37

1thorn sin 37frac14 025

The total active thrust (Equation 66a with c0 frac14 0) is

Pa frac14 1

2KaH

2 frac14 1

2 025 17 62 frac14 765 kN=m

If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

and the thrust on the wall is

P0 frac14 1

2K0H

2 frac14 1

2 040 17 62 frac14 122 kN=m

62

The active pressure coefficients for the three soil types are as follows

Ka1 frac141 sin 35

1thorn sin 35frac14 0271

Ka2 frac141 sin 27

1thorn sin 27frac14 0375

ffiffiffiffiffiffiffiKa2

p frac14 0613

Ka3 frac141 sin 42

1thorn sin 42frac14 0198

Distribution of active pressure (plotted in Figure Q62)

Depth (m) Soil Active pressure (kNm2)

3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

Total thrust frac14 571 kNm

Point of application is (4893571) m from the top of the wall ie 857m

Force (kN) Arm (m) Moment (kN m)

(1)1

2 0271 16 32 frac14 195 20 390

(2) 0271 16 3 2 frac14 260 40 1040

(3)1

2 0271 92 22 frac14 50 433 217

(4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

(5)1

2 0375 102 32 frac14 172 70 1204

(6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

(7)1

2 0198 112 42 frac14 177 1067 1889

(8)1

2 98 92 frac14 3969 90 35721

5713 48934

Figure Q62

Lateral earth pressure 35

63

(a) For u frac14 0 Ka frac14 Kp frac14 1

Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

frac14 245

At the lower end of the piling

pa frac14 Kaqthorn Kasatz Kaccu

frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

frac14 115 kN=m2

pp frac14 Kpsatzthorn Kpccu

frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

frac14 202 kN=m2

(b) For 0 frac14 26 and frac14 1

20

Ka frac14 035

Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

pfrac14 145 ethEquation 619THORN

Kp frac14 37

Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

pfrac14 47 ethEquation 624THORN

At the lower end of the piling

pa frac14 Kaqthorn Ka0z Kacc

0

frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

frac14 187 kN=m2

pp frac14 Kp0zthorn Kpcc

0

frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

frac14 198 kN=m2

36 Lateral earth pressure

64

(a) For 0 frac14 38 Ka frac14 024

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

Force (kN) Arm (m) Moment (kN m)

(1) 024 10 66 frac14 159 33 525

(2)1

2 024 17 392 frac14 310 400 1240

(3) 024 17 39 27 frac14 430 135 580

(4)1

2 024 102 272 frac14 89 090 80

(5)1

2 98 272 frac14 357 090 321

Hfrac14 1345 MH frac14 2746

(6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

(10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

XM frac14MV MH frac14 7790 kNm

Lever arm of base resultant

M

Vfrac14 779

488frac14 160

Eccentricity of base resultant

e frac14 200 160 frac14 040m

39 m

27 m

40 m

04 m

04 m

26 m

(7)

(9)

(1)(2)

(3)

(4)

(5)

(8)(6)

(10)

WT

10 kNm2

Hydrostatic

Figure Q64

Lateral earth pressure 37

Base pressures (Equation 627)

p frac14 VB

1 6e

B

frac14 488

4eth1 060THORN

frac14 195 kN=m2 and 49 kN=m2

Factor of safety against sliding (Equation 628)

F frac14 V tan

Hfrac14 488 tan 25

1345frac14 17

(b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

Hfrac14 1633 kN

V frac14 4879 kN

MH frac14 3453 kNm

MV frac14 10536 kNm

The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

65

For 0 frac14 36 Ka frac14 026 and Kp frac14 385

Kp

Ffrac14 385

2

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

Force (kN) Arm (m) Moment (kN m)

(1)1

2 026 17 452 frac14 448 dthorn 15 448dthorn 672

(2) 026 17 45 d frac14 199d d2 995d2

(3)1

2 026 102 d2 frac14 133d2 d3 044d3

(4)1

2 385

2 17 152 frac14 368 dthorn 05 368d 184

(5)385

2 17 15 d frac14 491d d2 2455d2

(6)1

2 385

2 102 d2 frac14 982d2 d3 327d3

38 Lateral earth pressure

XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

d3 thorn 516d2 283d 1724 frac14 0

d frac14 179m

Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

Over additional 20 embedded depth

pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

66

The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

Ka frac14 sin 69=sin 105

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

26664

37775

2

frac14 050

The total active thrust (acting at 25 above the normal) is given by Equation 616

Pa frac14 1

2 050 19 7502 frac14 267 kN=m

Figure Q65

Lateral earth pressure 39

Horizontal component

Ph frac14 267 cos 40 frac14 205 kN=m

Vertical component

Pv frac14 267 sin 40 frac14 172 kN=m

Consider moments about the toe of the wall (Figure Q66) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 175 650 235 frac14 1337 258 345

(2) 050 650 235 frac14 764 175 134

(3)1

2 070 650 235 frac14 535 127 68

(4) 100 400 235 frac14 940 200 188

(5) 1

2 080 050 235 frac14 47 027 1

Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

Lever arm of base resultant

M

Vfrac14 795

525frac14 151m

Eccentricity of base resultant

e frac14 200 151 frac14 049m

Figure Q66

40 Lateral earth pressure

Base pressures (Equation 627)

p frac14 525

41 6 049

4

frac14 228 kN=m2 and 35 kN=m2

The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

67

For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

Force (kN) Arm (m) Moment (kNm)

(1)1

2 027 17 52 frac14 574 183 1050

(2) 027 17 5 3 frac14 689 500 3445

(3)1

2 027 102 32 frac14 124 550 682

(4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

(5)1

2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

(6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

(7) 1

2 267

2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

(8) 2 10ffiffiffiffiffiffiffiffiffi267p

2 d frac14 163d d2thorn 650 82d2 1060d

Tie rod force per m frac14 T 0 0

XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

d3 thorn 77d2 269d 1438 frac14 0

d frac14 467m

Depth of penetration frac14 12d frac14 560m

Lateral earth pressure 41

Algebraic sum of forces for d frac14 467m isX

F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

T frac14 905 kN=m

Force in each tie rod frac14 25T frac14 226 kN

68

(a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

0 frac14 21 98 frac14 112 kN=m3

The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

uC frac14 150

165 15 98 frac14 134 kN=m2

The average seepage pressure is

j frac14 15

165 98 frac14 09 kN=m3

Hence

0 thorn j frac14 112thorn 09 frac14 121 kN=m3

0 j frac14 112 09 frac14 103 kN=m3

Figure Q67

42 Lateral earth pressure

Consider moments about the anchor point A (per m)

Force (kN) Arm (m) Moment (kN m)

(1) 10 026 150 frac14 390 60 2340

(2)1

2 026 18 452 frac14 474 15 711

(3) 026 18 45 105 frac14 2211 825 18240

(4)1

2 026 121 1052 frac14 1734 100 17340

(5)1

2 134 15 frac14 101 40 404

(6) 134 30 frac14 402 60 2412

(7)1

2 134 60 frac14 402 95 3819

571 4527(8) Ppm

115 115PPm

XM frac14 0

Ppm frac144527

115frac14 394 kN=m

Available passive resistance

Pp frac14 1

2 385 103 62 frac14 714 kN=m

Factor of safety

Fp frac14 Pp

Ppm

frac14 714

394frac14 18

Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

Figure Q68

Lateral earth pressure 43

(b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

Consider moments (per m) about the tie point A

Force (kN) Arm (m)

(1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

(2)1

2 033 18 452 frac14 601 15

(3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

(4)1

2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

(5)1

2 134 15 frac14 101 40

(6) 134 30 frac14 402 60

(7)1

2 134 d frac14 67d d3thorn 75

(8) 1

2 30 103 d2 frac141545d2 2d3thorn 75

Moment (kN m)

(1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

d3 thorn 827d2 466d 1518 frac14 0

By trial

d frac14 544m

The minimum depth of embedment required is 544m

69

For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

0 frac14 20 98 frac14 102 kN=m3

The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

44 Lateral earth pressure

uC frac14 147

173 26 98 frac14 216 kN=m2

and the average seepage pressure around the wall is

j frac14 26

173 98 frac14 15 kN=m3

Consider moments about the prop (A) (per m)

Force (kN) Arm (m) Moment (kN m)

(1)1

2 03 17 272 frac14 186 020 37

(2) 03 17 27 53 frac14 730 335 2445

(3)1

2 03 (102thorn 15) 532 frac14 493 423 2085

(4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

(5)1

2 216 26 frac14 281 243 684

(6) 216 27 frac14 583 465 2712

(7)1

2 216 60 frac14 648 800 5184

3055(8)

1

2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

Factor of safety

Fr frac14 6885

3055frac14 225

Figure Q69

Lateral earth pressure 45

610

For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

Using the recommendations of Twine and Roscoe

p frac14 02H frac14 02 19 9 frac14 342 kN=m2

Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

611

frac14 18 kN=m3 0 frac14 34

H frac14 350m nH frac14 335m mH frac14 185m

Consider a trial value of F frac14 20 Refer to Figure 635

0m frac14 tan1tan 34

20

frac14 186

Then

frac14 45 thorn 0m2frac14 543

W frac14 1

2 18 3502 cot 543 frac14 792 kN=m

Figure Q610

46 Lateral earth pressure

P frac14 1

2 s 3352 frac14 561s kN=m

U frac14 1

2 98 1852 cosec 543 frac14 206 kN=m

Equations 630 and 631 then become

561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

ie

561s 0616N 405 frac14 0

792 0857N thorn 563 frac14 0

N frac14 848

0857frac14 989 kN=m

Then

561s 609 405 frac14 0

s frac14 649

561frac14 116 kN=m3

The calculations for trial values of F of 20 15 and 10 are summarized below

F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

Figure Q611

Lateral earth pressure 47

612

For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

45 thorn 0

2frac14 63

For the retained material between the surface and a depth of 36m

Pa frac14 1

2 030 18 362 frac14 350 kN=m

Weight of reinforced fill between the surface and a depth of 36m is

Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

Eccentricity of Rv

e frac14 263 250 frac14 013m

The average vertical stress at a depth of 36m is

z frac14 Rv

L 2efrac14 324

474frac14 68 kN=m2

(a) In the tie back wedge method K frac14 Ka and Le frac14 418m

Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

Tensile stress in the element frac14 138 103

65 3frac14 71N=mm2

Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

The weight of ABC is

W frac14 1

2 18 52 265 frac14 124 kN=m

From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

48 Lateral earth pressure

(b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

Tp frac14 032 68 120 065 frac14 170 kN

Tr frac14 213 420

418frac14 214 kN

Again the tensile failure and slipping limit states are satisfied for this element

Figure Q612

Lateral earth pressure 49

Chapter 7

Consolidation theory

71

Total change in thickness

H frac14 782 602 frac14 180mm

Average thickness frac14 1530thorn 180

2frac14 1620mm

Length of drainage path d frac14 1620

2frac14 810mm

Root time plot (Figure Q71a)

ffiffiffiffiffiffit90p frac14 33

t90 frac14 109min

cv frac14 0848d2

t90frac14 0848 8102

109 1440 365

106frac14 27m2=year

r0 frac14 782 764

782 602frac14 018

180frac14 0100

rp frac14 10eth764 645THORN9eth782 602THORN frac14

10 119

9 180frac14 0735

rs frac14 1 eth0100thorn 0735THORN frac14 0165

Log time plot (Figure Q71b)

t50 frac14 26min

cv frac14 0196d2

t50frac14 0196 8102

26 1440 365

106frac14 26m2=year

r0 frac14 782 763

782 602frac14 019

180frac14 0106

rp frac14 763 623

782 602frac14 140

180frac14 0778

rs frac14 1 eth0106thorn 0778THORN frac14 0116

Figure Q71(a)

Figure Q71(b)

Final void ratio

e1 frac14 w1Gs frac14 0232 272 frac14 0631

e

Hfrac14 1thorn e0

H0frac14 1thorn e1 thorne

H0

ie

e

180frac14 1631thorne

1710

e frac14 2936

1530frac14 0192

Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

mv frac14 1

1thorn e0 e0 e101 00

frac14 1

1823 0192

0107frac14 098m2=MN

k frac14 cvmvw frac14 265 098 98

60 1440 365 103frac14 81 1010 m=s

72

Using Equation 77 (one-dimensional method)

sc frac14 e0 e11thorn e0 H

Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

Figure Q72

52 Consolidation theory

Settlement

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

318

Notes 5 92y 460thorn 84

Heave

Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

38

73

U frac14 f ethTvTHORN frac14 f cvt

d2

Hence if cv is constant

t1

t2frac14 d

21

d22

where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

d1 frac14 95mm and d2 frac14 2500mm

for U frac14 050 t2 frac14 t1 d22

d21

frac14 20

60 24 365 25002

952frac14 263 years

for U lt 060 Tv frac14

4U2 (Equation 724(a))

t030 frac14 t050 0302

0502

frac14 263 036 frac14 095 years

Consolidation theory 53

74

The layer is open

d frac14 8

2frac14 4m

Tv frac14 cvtd2frac14 24 3

42frac14 0450

ui frac14 frac14 84 kN=m2

The excess pore water pressure is given by Equation 721

ue frac14Xmfrac141mfrac140

2ui

Msin

Mz

d

expethM2TvTHORN

In this case z frac14 d

sinMz

d

frac14 sinM

where

M frac14

23

25

2

M sin M M2Tv exp (M2Tv)

2thorn1 1110 0329

3

21 9993 457 105

ue frac14 2 84 2

1 0329 ethother terms negligibleTHORN

frac14 352 kN=m2

75

The layer is open

d frac14 6

2frac14 3m

Tv frac14 cvtd2frac14 10 3

32frac14 0333

The layer thickness will be divided into six equal parts ie m frac14 6

54 Consolidation theory

For an open layer

Tv frac14 4n

m2

n frac14 0333 62

4frac14 300

The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

i j

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

The initial and 3-year isochrones are plotted in Figure Q75

Area under initial isochrone frac14 180 units

Area under 3-year isochrone frac14 63 units

The average degree of consolidation is given by Equation 725Thus

U frac14 1 63

180frac14 065

Figure Q75

Consolidation theory 55

76

At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

0 frac14 2w frac14 2 98 frac14 196 kN=m2

The final consolidation settlement (one-dimensional method) is

sc frac14 mv0H frac14 083 196 8 frac14 130mm

Corrected time t frac14 2 1

2

40

52

frac14 1615 years

Tv frac14 cvtd2frac14 44 1615

42frac14 0444

From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

77

The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

Figure Q77

56 Consolidation theory

Point m n Ir (kNm2) sc (mm)

13020frac14 15 20

20frac14 10 0194 (4) 113 124

260

20frac14 30

20

20frac14 10 0204 (2) 59 65

360

20frac14 30

40

20frac14 20 0238 (1) 35 38

430

20frac14 15

40

20frac14 20 0224 (2) 65 72

Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

78

Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

(a) Immediate settlement

H

Bfrac14 30

35frac14 086

D

Bfrac14 2

35frac14 006

Figure Q78

Consolidation theory 57

From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

si frac14 130131qB

Eufrac14 10 032 105 35

40frac14 30mm

(b) Consolidation settlement

Layer z (m) Dz Ic (kNm2) syod (mm)

1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

3150

Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

Now

H

Bfrac14 30

35frac14 086 and A frac14 065

from Figure 712 13 frac14 079

sc frac14 13sod frac14 079 315 frac14 250mm

Total settlement

s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

79

Without sand drains

Uv frac14 025

Tv frac14 0049 ethfrom Figure 718THORN

t frac14 Tvd2

cvfrac14 0049 82

cvWith sand drains

R frac14 0564S frac14 0564 3 frac14 169m

n frac14 Rrfrac14 169

015frac14 113

Tr frac14 cht

4R2frac14 ch

4 1692 0049 82

cvethand ch frac14 cvTHORN

frac14 0275

Ur frac14 073 (from Figure 730)

58 Consolidation theory

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

U frac14 080

710

Without sand drains

Uv frac14 090

Tv frac14 0848

t frac14 Tvd2

cvfrac14 0848 102

96frac14 88 years

With sand drains

R frac14 0564S frac14 0564 4 frac14 226m

n frac14 Rrfrac14 226

015frac14 15

Tr

Tvfrac14 chcv

d2

4R2ethsame tTHORN

Tr

Tvfrac14 140

96 102

4 2262frac14 714 eth1THORN

Using Equation 740

eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

An iterative solution is required using (1) and (2) an initial value ofUv being estimated

Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

Thus

Uv frac14 0295 and Ur frac14 086

t frac14 88 00683

0848frac14 07 years

Consolidation theory 59

Chapter 8

Bearing capacity

81

(a) The ultimate bearing capacity is given by Equation 83

qf frac14 cNc thorn DNq thorn 1

2BN

For u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

The net ultimate bearing capacity is

qnf frac14 qf D frac14 540 kN=m2

The net foundation pressure is

qn frac14 q D frac14 425

2 eth21 1THORN frac14 192 kN=m2

The factor of safety (Equation 86) is

F frac14 qnfqnfrac14 540

192frac14 28

(b) For 0 frac14 28

Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

2 112 2 13

frac14 260thorn 168thorn 146 frac14 574 kN=m2

qnf frac14 574 112 frac14 563 kN=m2

F frac14 563

192frac14 29

(qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

82

For 0 frac14 38

Nq frac14 49 N frac14 67

qnf frac14 DethNq 1THORN thorn 1

2BN ethfrom Equation 83THORN

frac14 eth18 075 48THORN thorn 1

2 18 15 67

frac14 648thorn 905 frac14 1553 kN=m2

qn frac14 500

15 eth18 075THORN frac14 320 kN=m2

F frac14 qnfqnfrac14 1553

320frac14 48

0d frac14 tan1tan 38

125

frac14 32 therefore Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

2 18 15 25

frac14 15eth310thorn 337THORNfrac14 970 kN=m

Design load (action) Vd frac14 500 kN=m

The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

83

D

Bfrac14 350

225frac14 155

From Figure 85 for a square foundation

Nc frac14 81

Bearing capacity 61

For a rectangular foundation (L frac14 450m B frac14 225m)

Nc frac14 084thorn 016B

L

81 frac14 745

Using Equation 810

qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

For F frac14 3

qn frac14 1006

3frac14 335 kN=m2

q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

Design load frac14 405 450 225 frac14 4100 kN

Design undrained strength cud frac14 135

14frac14 96 kN=m2

Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

frac14 7241 kN

Design load Vd frac14 4100 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

84

For 0 frac14 40

Nq frac14 64 N frac14 95

qnf frac14 DethNq 1THORN thorn 04BN

(a) Water table 5m below ground level

qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

qn frac14 400 17 frac14 383 kN=m2

F frac14 2686

383frac14 70

(b) Water table 1m below ground level (ie at foundation level)

0 frac14 20 98 frac14 102 kN=m3

62 Bearing capacity

qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

F frac14 2040

383frac14 53

(c) Water table at ground level with upward hydraulic gradient 02

eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

F frac14 1296

392frac14 33

85

The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

Design value of 0 frac14 tan1tan 39

125

frac14 33

For 0 frac14 33 Nq frac14 26 and N frac14 29

Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

Rd gt Vd therefore the bearing resistance limit state is satisfied

86

(a) Undrained shear for u frac14 0

Nc frac14 514 Nq frac14 1 N frac14 0

qnf frac14 12cuNc

frac14 12 100 514 frac14 617 kN=m2

qn frac14 qnfFfrac14 617

3frac14 206 kN=m2

q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

Bearing capacity 63

Drained shear for 0 frac14 32

Nq frac14 23 N frac14 25

0 frac14 21 98 frac14 112 kN=m3

qnf frac14 0DethNq 1THORN thorn 040BN

frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

frac14 694 kN=m2

q frac14 694

3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

Design load frac14 42 227 frac14 3632 kN

(b) Design undrained strength cud frac14 100

14frac14 71 kNm2

Design bearing resistance Rd frac14 12cudNe area

frac14 12 71 514 42

frac14 7007 kN

For drained shear 0d frac14 tan1tan 32

125

frac14 26

Nq frac14 12 N frac14 10

Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

(c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

Layer z (m) m n Ir 0 (kNm2) sod (mm)

1 2 100 0175 0700qn 0182qn

2 6 033 0044 0176qn 0046qn

3 10 020 0017 0068qn 0018qn

0246qn

Diameter of equivalent circle B frac14 45m

H

Bfrac14 12

45frac14 27 and A frac14 042

13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

64 Bearing capacity

For sc frac14 30mm

qn frac14 30

0147frac14 204 kN=m2

q frac14 204thorn 21 frac14 225 kN=m2

Design load frac14 42 225 frac14 3600 kN

The design load is 3600 kN settlement being the limiting criterion

87

D

Bfrac14 8

4frac14 20

From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

F frac14 cuNc

Dfrac14 40 71

20 8frac14 18

88

Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

Design value of 0 frac14 tan1tan 38

125

frac14 32

Figure Q86

Bearing capacity 65

For 0 frac14 32 Nq frac14 23 and N frac14 25

Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

For B frac14 250m qn frac14 3750

2502 17 frac14 583 kN=m2

From Figure 510 m frac14 n frac14 126

6frac14 021

Ir frac14 0019

Stress increment frac14 4 0019 583 frac14 44 kN=m2

Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

The settlement is less than 20mm therefore the serviceability limit state is satisfied

89

Depth (m) N 0v (kNm2) CN N1

070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

(a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

Cw frac14 05thorn 0530

47

frac14 082

66 Bearing capacity

Thus

qa frac14 150 082 frac14 120 kN=m2

(b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

Thus

qa frac14 90 15 frac14 135 kN=m2

(c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

Ic frac14 171

1014frac14 0068

From Equation 819(a) with s frac14 25mm

q frac14 25

3507 0068frac14 150 kN=m2

810

Peak value of strain influence factor occurs at a depth of 27m and is given by

Izp frac14 05thorn 01130

16 27

05

frac14 067

Refer to Figure Q810

E frac14 25qc

Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

Ez (mm3MN)

1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

0203

C1 frac14 1 0500qnfrac14 1 05 12 16

130frac14 093

C2 frac14 1 ethsayTHORN

s frac14 C1C2qnX Iz

Ez frac14 093 1 130 0203 frac14 25mm

Bearing capacity 67

811

At pile base level

cu frac14 220 kN=m2

qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

qs frac14 00 frac14 07 1226 frac14 86 kN=m2

Then

Qf frac14 Abqb thorn Asqs

frac14

4 32 1980

thorn eth 105 139 86THORN

frac14 13 996thorn 3941 frac14 17 937 kN

0 01 02 03 04 05 06 07

0 2 4 6 8 10 12 14

1

2

3

4

5

6

7

8

(1)

(2)

(3)

(4)

(5)

qc

qc

Iz

Iz

(MNm2)

z (m)

Figure Q810

68 Bearing capacity

Allowable load

ethaTHORN Qf

2frac14 17 937

2frac14 8968 kN

ethbTHORN Abqb

3thorn Asqs frac14 13 996

3thorn 3941 frac14 8606 kN

ie allowable load frac14 8600 kN

Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

According to the limit state method

Characteristic undrained strength at base level cuk frac14 220

150kN=m2

Characteristic base resistance qbk frac14 9cuk frac14 9 220

150frac14 1320 kN=m2

Characteristic shaft resistance qsk frac14 00150

frac14 86

150frac14 57 kN=m2

Characteristic base and shaft resistances

Rbk frac14

4 32 1320 frac14 9330 kN

Rsk frac14 105 139 86

150frac14 2629 kN

For a bored pile the partial factors are b frac14 160 and s frac14 130

Design bearing resistance Rcd frac14 9330

160thorn 2629

130

frac14 5831thorn 2022

frac14 7850 kN

Adding ethDAb W) the design bearing resistance becomes 9650 kN

812

ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

qs frac14 cu frac14 040 105 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14

4 062 1305

thorn eth 06 15 42THORN

frac14 369thorn 1187 frac14 1556 kN

Bearing capacity 69

Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

(Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

(b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

qbkfrac14 9cuk frac14 9 220

150frac14 1320 kN=m2

qskfrac14cuk frac14 040 105

150frac14 28 kN=m2

Rbkfrac14

4 0602 1320 frac14 373 kN

Rskfrac14 060 15 28 frac14 791 kN

Rcdfrac14 373

160thorn 791

130frac14 233thorn 608 frac14 841 kN

Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

Rcd gt Fcd therefore the bearing resistance limit state is satisfied

(c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

q frac14 21 000

1762frac14 68 kN=m2

Immediate settlement

H

Bfrac14 15

176frac14 085

D

Bfrac14 13

176frac14 074

L

Bfrac14 1

Hence from Figure 515

130 frac14 078 and 131 frac14 041

70 Bearing capacity

Thus using Equation 528

si frac14 078 041 68 176

65frac14 6mm

Consolidation settlement

Layer z (m) Area (m2) (kNm2) mvH (mm)

1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

434 (sod)

Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

sc frac14 056 434 frac14 24mm

The total settlement is (6thorn 24) frac14 30mm

813

At base level N frac14 26 Then using Equation 830

qb frac14 40NDb

Bfrac14 40 26 2

025frac14 8320 kN=m2

ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

Figure Q812

Bearing capacity 71

Over the length embedded in sand

N frac14 21 ie18thorn 24

2

Using Equation 831

qs frac14 2N frac14 2 21 frac14 42 kN=m2

For a single pile

Qf frac14 Abqb thorn Asqs

frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

For the pile group assuming a group efficiency of 12

XQf frac14 12 9 604 frac14 6523 kN

Then the load factor is

F frac14 6523

2000thorn 1000frac14 21

(b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

Characteristic base resistance per unit area qbk frac14 8320

150frac14 5547 kNm2

Characteristic shaft resistance per unit area qsk frac14 42

150frac14 28 kNm2

Characteristic base and shaft resistances for a single pile

Rbk frac14 0252 5547 frac14 347 kN

Rsk frac14 4 025 2 28 frac14 56 kN

For a driven pile the partial factors are b frac14 s frac14 130

Design bearing resistance Rcd frac14 347

130thorn 56

130frac14 310 kN

For the pile group Rcd frac14 12 9 310 frac14 3348 kN

Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

(c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

72 Bearing capacity

N frac14 24thorn 26thorn 34

3frac14 28

Ic frac14 171

2814frac14 0016 ethEquation 818THORN

s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

The settlement is less than 20mm therefore the serviceability limit state is satisfied

814

Using Equation 841

Tf frac14 DLcu thorn

4ethD2 d2THORNcuNc

frac14 eth 02 5 06 110THORN thorn

4eth022 012THORN110 9

frac14 207thorn 23 frac14 230 kN

Figure Q813

Bearing capacity 73

Chapter 9

Stability of slopes

91

Referring to Figure Q91

W frac14 417 19 frac14 792 kN=m

Q frac14 20 28 frac14 56 kN=m

Arc lengthAB frac14

180 73 90 frac14 115m

Arc length BC frac14

180 28 90 frac14 44m

The factor of safety is given by

F frac14 rethcuLaTHORNWd1 thornQd2 frac14

90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

Depth of tension crack z0 frac14 2cu

frac14 2 20

19frac14 21m

Arc length BD frac14

180 13

1

2 90 frac14 21m

F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

Design resisting moment frac14 rXethcudLaTHORN frac14 90

14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

92

u frac14 0

Depth factor D frac14 11

9frac14 122

Using Equation 92 with F frac14 10

Ns frac14 cu

FHfrac14 30

10 19 9frac14 0175

Hence from Figure 93

frac14 50

For F frac14 12

Ns frac14 30

12 19 9frac14 0146

frac14 27

93

Refer to Figure Q93

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

74 m

214 1deg

213 1deg

39 m

WB

D

C

28 m

21 m

A

Q

Soil (1)Soil (2)

73deg

Figure Q91

Stability of slopes 75

Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

599 256 328 1372

Figure Q93

76 Stability of slopes

XW cos frac14 b

Xh cos frac14 21 2 599 frac14 2516 kN=mX

W sin frac14 bX

h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

Arc length La frac14

180 57

1

2 326 frac14 327m

The factor of safety is given by

F frac14 c0La thorn tan0ethW cos ulTHORN

W sin

frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

frac14 091

According to the limit state method

0d frac14 tan1tan 32

125

frac14 265

c0 frac14 8

160frac14 5 kN=m2

Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

Design disturbing moment frac14 1075 kN=m

The design resisting moment is less than the design disturbing moment therefore a slipwill occur

94

F frac14 1

W sin

Xfc0bthorn ethW ubTHORN tan0g sec

1thorn ethtan tan0=FTHORN

c0 frac14 8 kN=m2

0 frac14 32

c0b frac14 8 2 frac14 16 kN=m

W frac14 bh frac14 21 2 h frac14 42h kN=m

Try F frac14 100

tan0

Ffrac14 0625

Stability of slopes 77

Values of u are as obtained in Figure Q93

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

ub(kNm)

c0bthorn (W ub) tan0(kNm)

sec

1thorn (tan tan0)FProduct(kNm)

1 05 21 6 2 8 24 1078 262 13 55 31

23 33 30 1042 31

3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

224 92 72 0931 67

6 50 210 11 40 100 85 0907 777 55 231 14

12 58 112 90 0889 80

8 60 252 1812

80 114 102 0874 899 63 265 22 99 116 109 0861 94

10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

2154 88 116 0853 99

14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

1074 1091

F frac14 1091

1074frac14 102 (assumed value 100)

Thus

F frac14 101

95

F frac14 1

W sin

XfWeth1 ruTHORN tan0g sec

1thorn ethtan tan0THORN=F

0 frac14 33

ru frac14 020

W frac14 bh frac14 20 5 h frac14 100h kN=m

eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

Try F frac14 110

tan 0

Ffrac14 tan 33

110frac14 0590

78 Stability of slopes

Referring to Figure Q95

SliceNo

h(m)

W frac14 bh(kNm)

W sin(kNm)

W(1 ru) tan0(kNm)

sec

1thorn ( tan tan0)FProduct(kNm)

1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

2120 234 0892 209

4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

1185 1271

Figure Q95

Stability of slopes 79

F frac14 1271

1185frac14 107

The trial value was 110 therefore take F to be 108

96

(a) Water table at surface the factor of safety is given by Equation 912

F frac14 0

sat

tan0

tan

ptie 15 frac14 92

19

tan 36

tan

tan frac14 0234

frac14 13

Water table well below surface the factor of safety is given by Equation 911

F frac14 tan0

tan

frac14 tan 36

tan 13

frac14 31

(b) 0d frac14 tan1tan 36

125

frac14 30

Depth of potential failure surface frac14 z

Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

frac14 504z kN

Design disturbing moment per unit area Sd frac14 sat sin cos

frac14 19 z sin 13 cos 13

frac14 416z kN

Rd gtSd therefore the limit state for overall stability is satisfied

80 Stability of slopes

  • Book Cover
  • Title
  • Contents
  • Basic characteristics of soils
  • Seepage
  • Effective stress
  • Shear strength
  • Stresses and displacements
  • Lateral earth pressure
  • Consolidation theory
  • Bearing capacity
  • Stability of slopes

    Craigrsquos Soil Mechanics Seventh EditionSolutions Manual

    Craigrsquos Soil Mechanics SeventhEdition Solutions Manual

    RF CraigFormerly

    Department of Civil Engineering

    University of Dundee UK

    First published 1992by E amp FN Spon an imprint of Thomson ProfessionalSecond edition 1997Third edition 200411 New Fetter Lane London EC4P 4EE

    Simultaneously published in the USA and Canadaby Spon Press29 West 35th Street New York NY 10001

    Spon Press is an imprint of the Taylor amp Francis Group

    ordf 1992 1997 2004 RF Craig

    All rights reserved No part of this book may be reprinted or reproducedor utilised in any form or by any electronic mechanical or other meansnow known or hereafter invented including photocopying and recordingor in any information storage or retrieval system without permission inwriting from the publishers

    British Library Cataloguing in Publication DataA catalogue record for this book is availablefrom the British Library

    Library of Congress Cataloging in Publication DataA catalog record for this book has been requested

    ISBN 0ndash415ndash33294ndashX

    This edition published in the Taylor amp Francis e-Library 2004

    (Print edition)

    ISBN 0-203-31104-3 Master e-book ISBN

    ISBN 0-203-67167-8 (Adobe eReader Format)

    collection of thousands of eBooks please go to wwweBookstoretandfcoukrdquoldquoTo purchase your own copy of this or any of Taylor amp Francis or Routledgersquos

    Contents

    1 Basic characteristics of soils 1

    2 Seepage 6

    3 Effective stress 14

    4 Shear strength 22

    5 Stresses and displacements 28

    6 Lateral earth pressure 34

    7 Consolidation theory 50

    8 Bearing capacity 60

    9 Stability of slopes 74

    Authorrsquos note

    In order not to short-circuit the learningprocess it is vital that the reader shouldattempt the problems before referring to thesolutions in this manual

    Chapter 1

    Basic characteristics of soils

    11

    Soil E consists of 98 coarse material (31 gravel size 67 sand size) and 2 finesIt is classified as SW well-graded gravelly SAND or in greater detail well-gradedslightly silty very gravelly SAND

    Soil F consists of 63 coarse material (2 gravel size 61 sand size) and 37non-plastic fines (ie between 35 and 65 fines) therefore the soil is classified as MSsandy SILT

    Soil G consists of 73 fine material (ie between 65 and 100 fines) and 27 sandsize The liquid limit is 32 and the plasticity index is 8 (ie 32 24) plotting marginallybelow the A-line in the ML zone on the plasticity chart Thus the classification is MLSILT (M-SOIL) of low plasticity (The plasticity chart is given in Figure 17)

    Figure Q11

    Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

    12

    From Equation 117

    1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

    191frac14 155

    e frac14 055

    Using Equation 113

    Sr frac14 wGs

    efrac14 0095 270

    055frac14 0466 eth466THORN

    Using Equation 119

    sat frac14 Gs thorn e1thorn e w frac14 325

    155 100 frac14 210Mg=m3

    From Equation 114

    w frac14 e

    Gsfrac14 055

    270frac14 0204 eth204THORN

    13

    Equations similar to 117ndash120 apply in the case of unit weights thus

    d frac14 Gs

    1thorn e w frac14272

    170 98 frac14 157 kN=m3

    sat frac14 Gs thorn e1thorn e w frac14 342

    170 98 frac14 197 kN=m3

    Using Equation 121

    0 frac14 Gs 1

    1thorn e w frac14 172

    170 98 frac14 99 kN=m3

    Using Equation 118a with Srfrac14 075

    frac14 Gs thorn Sre

    1thorn e w frac14 3245

    170 98 frac14 187 kN=m3

    2 Basic characteristics of soils

    Using Equation 113

    w frac14 Sre

    Gsfrac14 075 070

    272frac14 0193 eth193THORN

    The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

    14

    Volume of specimenfrac14

    438276 frac14 86 200mm3

    Bulk density ethTHORN frac14 Mass

    Volumefrac14 1680

    86 200 103frac14 195Mg=m3

    Water content ethwTHORN frac14 1680 1305

    1305frac14 0287 eth287THORN

    From Equation 117

    1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

    195frac14 180

    e frac14 080

    Using Equation 113

    Sr frac14 wGs

    efrac14 0287 273

    080frac14 098 eth98THORN

    15

    Using Equation 124

    d frac14

    1thorn w frac14215

    112frac14 192Mg=m3

    From Equation 117

    1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

    215frac14 138

    e frac14 038

    Using Equation 113

    Sr frac14 wGs

    efrac14 012 265

    038frac14 0837 eth837THORN

    Basic characteristics of soils 3

    Using Equation 115

    Afrac14 e wGs

    1thorn e frac14038 0318

    138frac14 0045 eth45THORN

    The zero air voids dry density is given by Equation 125

    d frac14 Gs

    1thorn wGsw frac14 265

    1thorn eth0135 265THORN 100 frac14 195Mg=m3

    ie a dry density of 200Mgm3 would not be possible

    16

    Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

    (Mgm3) d10(Mgm3)

    2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

    In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

    wopt frac14 15 and dmaxfrac14 183Mg=m3

    Figure Q16

    4 Basic characteristics of soils

    Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

    d5 d10

    respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

    17

    From Equation 120

    e frac14 Gswd 1

    The maximum and minimum values of void ratio are given by

    emax frac14 Gsw

    dmin

    1

    emin frac14 Gswdmax

    1

    From Equation 123

    ID frac14 Gsweth1=dmin 1=dTHORN

    Gsweth1=dmin 1=dmax

    THORN

    frac14 frac121 ethdmin=dTHORN1=dmin

    frac121 ethdmin=dmax

    THORN1=dmin

    frac14 d dmin

    dmax dmin

    dmax

    d

    frac14 172 154

    181 154

    181

    172

    frac14 070 eth70THORN

    Basic characteristics of soils 5

    Chapter 2

    Seepage

    21

    The coefficient of permeability is determined from the equation

    k frac14 23al

    At1log

    h0

    h1

    where

    a frac14

    4 00052 m2 l frac14 02m

    A frac14

    4 012 m2 t1 frac14 3 602 s

    logh0

    h1frac14 log

    100

    035frac14 0456

    k frac14 23 00052 02 0456

    012 3 602frac14 49 108 m=s

    22

    The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

    q frac14 kh Nf

    Ndfrac14 106 400 37

    11frac14 13 106 m3=s per m

    Figure Q22

    23

    The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

    q frac14 kh Nf

    Ndfrac14 5 105 300 35

    9frac14 58 105 m3=s per m

    The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

    Point h (m) h z (m) u frac14 w(h z)(kNm2)

    1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

    eg for Point 1

    h1 frac14 7

    9 300 frac14 233m

    h1 z1 frac14 233 eth250THORN frac14 483m

    Figure Q23

    Seepage 7

    u1 frac14 98 483 frac14 47 kN=m2

    The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

    24

    The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

    q frac14 kh Nf

    Ndfrac14 40 107 550 10

    11frac14 20 106 m3=s per m

    25

    The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

    q frac14 kh Nf

    Ndfrac14 20 106 500 42

    9frac14 47 106 m3=s per m

    Figure Q24

    8 Seepage

    26

    The scale transformation factor in the x direction is given by Equation 221 ie

    xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

    ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

    Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

    k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

    pfrac14

    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

    p 107 frac14 30 107 m=s

    Thus the quantity of seepage is given by

    q frac14 k0h Nf

    Ndfrac14 30 107 1400 325

    12frac14 11 106 m3=s per m

    Figure Q25

    Seepage 9

    27

    The scale transformation factor in the x direction is

    xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

    ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

    Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

    k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

    pfrac14

    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

    p 106 frac14 45 106 m=s

    The focus of the basic parabola is at point A The parabola passes through point Gsuch that

    GC frac14 03HC frac14 03 30 frac14 90m

    Thus the coordinates of G are

    x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

    480 frac14 x0 2002

    4x0

    Figure Q26

    10 Seepage

    Hence

    x0 frac14 20m

    Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

    x frac14 20 z2

    80

    x 20 0 50 100 200 300z 0 400 748 980 1327 1600

    The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

    No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

    q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

    28

    The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

    Figure Q27

    Seepage 11

    q frac14 kh Nf

    Ndfrac14 45 105 28 33

    7

    frac14 59 105 m3=s per m

    29

    The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

    kx frac14 H1k1 thornH2k2

    H1 thornH2frac14 106

    10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

    kz frac14 H1 thornH2

    H1

    k1thornH2

    k2

    frac14 10

    5

    eth2 106THORN thorn5

    eth16 106THORNfrac14 36 106 m=s

    Then the scale transformation factor is given by

    xt frac14 xffiffiffiffiffikz

    pffiffiffiffiffikx

    p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

    Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

    Figure Q28

    12 Seepage

    k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

    qfrac14

    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

    p 106 frac14 57 106 m=s

    Then the quantity of seepage is given by

    q frac14 k0h Nf

    Ndfrac14 57 106 350 56

    11

    frac14 10 105 m3=s per m

    Figure Q29

    Seepage 13

    Chapter 3

    Effective stress

    31

    Buoyant unit weight

    0 frac14 sat w frac14 20 98 frac14 102 kN=m3

    Effective vertical stress

    0v frac14 5 102 frac14 51 kN=m2 or

    Total vertical stress

    v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

    Pore water pressure

    u frac14 7 98 frac14 686 kN=m2

    Effective vertical stress

    0v frac14 v u frac14 1196 686 frac14 51 kN=m2

    32

    Buoyant unit weight

    0 frac14 sat w frac14 20 98 frac14 102 kN=m3

    Effective vertical stress

    0v frac14 5 102 frac14 51 kN=m2 or

    Total vertical stress

    v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

    Pore water pressure

    u frac14 205 98 frac14 2009 kN=m2

    Effective vertical stress

    0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

    33

    At top of the clay

    v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

    u frac14 2 98 frac14 196 kN=m2

    0v frac14 v u frac14 710 196 frac14 514 kN=m2

    Alternatively

    0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

    0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

    At bottom of the clay

    v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

    u frac14 12 98 frac14 1176 kN=m2

    0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

    NB The alternative method of calculation is not applicable because of the artesiancondition

    Figure Q3132

    Effective stress 15

    34

    0 frac14 20 98 frac14 102 kN=m3

    At 8m depth

    0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

    35

    0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

    0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

    Figure Q33

    Figure Q34

    16 Effective stress

    (a) Immediately after WT rise

    At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

    0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

    At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

    0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

    (b) Several years after WT rise

    At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

    At 8m depth

    0v frac14 940 kN=m2 (as above)

    At 12m depth

    0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

    Figure Q35

    Effective stress 17

    36

    Total weight

    ab frac14 210 kN

    Effective weight

    ac frac14 112 kN

    Resultant boundary water force

    be frac14 119 kN

    Seepage force

    ce frac14 34 kN

    Resultant body force

    ae frac14 99 kN eth73 to horizontalTHORN

    (Refer to Figure Q36)

    Figure Q36

    18 Effective stress

    37

    Situation (1)(a)

    frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

    u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

    0 frac14 u frac14 694 392 frac14 302 kN=m2

    (b)

    i frac14 2

    4frac14 05

    j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

    Situation (2)(a)

    frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

    u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

    0 frac14 u frac14 498 392 frac14 106 kN=m2

    (b)

    i frac14 2

    4frac14 05

    j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

    38

    The flow net is drawn in Figure Q24

    Loss in total head between adjacent equipotentials

    h frac14 550

    Ndfrac14 550

    11frac14 050m

    Exit hydraulic gradient

    ie frac14 h

    sfrac14 050

    070frac14 071

    Effective stress 19

    The critical hydraulic gradient is given by Equation 39

    ic frac14 0

    wfrac14 102

    98frac14 104

    Therefore factor of safety against lsquoboilingrsquo (Equation 311)

    F frac14 iciefrac14 104

    071frac14 15

    Total head at C

    hC frac14 nd

    Ndh frac14 24

    11 550 frac14 120m

    Elevation head at C

    zC frac14 250m

    Pore water pressure at C

    uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

    Therefore effective vertical stress at C

    0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

    For point D

    hD frac14 73

    11 550 frac14 365m

    zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

    0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

    39

    The flow net is drawn in Figure Q25

    For a soil prism 150 300m adjacent to the piling

    hm frac14 26

    9 500 frac14 145m

    20 Effective stress

    Factor of safety against lsquoheavingrsquo (Equation 310)

    F frac14 ic

    imfrac14 0d

    whmfrac14 97 300

    98 145frac14 20

    With a filter

    F frac14 0d thorn wwhm

    3 frac14 eth97 300THORN thorn w98 145

    w frac14 135 kN=m2

    Depth of filterfrac14 13521frac14 065m (if above water level)

    Effective stress 21

    Chapter 4

    Shear strength

    41

    frac14 295 kN=m2

    u frac14 120 kN=m2

    0 frac14 u frac14 295 120 frac14 175 kN=m2

    f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

    42

    03 (kNm2) 1 3 (kNm2) 01 (kNm2)

    100 452 552200 908 1108400 1810 2210800 3624 4424

    The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

    Figure Q42

    43

    3 (kNm2) 1 3 (kNm2) 1 (kNm2)

    200 222 422400 218 618600 220 820

    The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

    44

    The modified shear strength parameters are

    0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

    a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

    The coordinates of the stress point representing failure conditions in the test are

    1

    2eth1 2THORN frac14 1

    2 170 frac14 85 kN=m2

    1

    2eth1 thorn 3THORN frac14 1

    2eth270thorn 100THORN frac14 185 kN=m2

    The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

    uf frac14 36 kN=m2

    Figure Q43

    Figure Q44

    Shear strength 23

    45

    3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

    150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

    The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

    The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

    46

    03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

    200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

    The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

    Figure Q45

    24 Shear strength

    47

    The torque required to produce shear failure is given by

    T frac14 dh cud

    2thorn 2

    Z d=2

    0

    2r drcur

    frac14 cud2h

    2thorn 4cu

    Z d=2

    0

    r2dr

    frac14 cud2h

    2thorn d

    3

    6

    Then

    35 frac14 cu52 10

    2thorn 53

    6

    103

    cu frac14 76 kN=m3

    400

    0 400 800 1200 1600

    τ (k

    Nm

    2 )

    σprime (kNm2)

    34deg

    315deg29deg

    (a)

    (b)

    0 400

    400

    800 1200 1600

    Failure envelope

    300 500

    σprime (kNm2)

    τ (k

    Nm

    2 )

    20 (kNm2)

    31deg

    Figure Q46

    Shear strength 25

    48

    The relevant stress values are calculated as follows

    3 frac14 600 kN=m2

    1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

    2(1 3) 0 40 79 107 139 159

    1

    2(01 thorn 03) 400 411 402 389 351 326

    1

    2(1 thorn 3) 600 640 679 707 739 759

    The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

    Af frac14 433 200

    319frac14 073

    49

    B frac14 u33

    frac14 144

    350 200frac14 096

    a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

    0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

    10 283 65 023

    Figure Q48

    26 Shear strength

    The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

    Figure Q49

    Shear strength 27

    Chapter 5

    Stresses and displacements

    51

    Vertical stress is given by

    z frac14 Qz2Ip frac14 5000

    52Ip

    Values of Ip are obtained from Table 51

    r (m) rz Ip z (kNm2)

    0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

    10 20 0009 2

    The variation of z with radial distance (r) is plotted in Figure Q51

    Figure Q51

    52

    Below the centre load (Figure Q52)

    r

    zfrac14 0 for the 7500-kN load

    Ip frac14 0478

    r

    zfrac14 5

    4frac14 125 for the 10 000- and 9000-kN loads

    Ip frac14 0045

    Then

    z frac14X Q

    z2Ip

    frac14 7500 0478

    42thorn 10 000 0045

    42thorn 9000 0045

    42

    frac14 224thorn 28thorn 25 frac14 277 kN=m2

    53

    The vertical stress under a corner of a rectangular area is given by

    z frac14 qIr

    where values of Ir are obtained from Figure 510 In this case

    z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

    z

    Figure Q52

    Stresses and displacements 29

    z (m) m n Ir z (kNm2)

    0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

    10 010 0005 5

    z is plotted against z in Figure Q53

    54

    (a)

    m frac14 125

    12frac14 104

    n frac14 18

    12frac14 150

    From Figure 510 Irfrac14 0196

    z frac14 2 175 0196 frac14 68 kN=m2

    Figure Q53

    30 Stresses and displacements

    (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

    z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

    55

    Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

    Px frac14 2Q

    1

    m2 thorn 1frac14 2 150

    125frac14 76 kN=m

    Equation 517 is used to obtain the pressure distribution

    px frac14 4Q

    h

    m2n

    ethm2 thorn n2THORN2 frac14150

    m2n

    ethm2 thorn n2THORN2 ethkN=m2THORN

    Figure Q54

    Stresses and displacements 31

    n m2n

    (m2 thorn n2)2

    px(kNm2)

    0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

    The pressure distribution is plotted in Figure Q55

    56

    H

    Bfrac14 10

    2frac14 5

    L

    Bfrac14 4

    2frac14 2

    D

    Bfrac14 1

    2frac14 05

    Hence from Figure 515

    131 frac14 082

    130 frac14 094

    Figure Q55

    32 Stresses and displacements

    The immediate settlement is given by Equation 528

    si frac14 130131qB

    Eu

    frac14 094 082 200 2

    45frac14 7mm

    Stresses and displacements 33

    Chapter 6

    Lateral earth pressure

    61

    For 0 frac14 37 the active pressure coefficient is given by

    Ka frac14 1 sin 37

    1thorn sin 37frac14 025

    The total active thrust (Equation 66a with c0 frac14 0) is

    Pa frac14 1

    2KaH

    2 frac14 1

    2 025 17 62 frac14 765 kN=m

    If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

    K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

    and the thrust on the wall is

    P0 frac14 1

    2K0H

    2 frac14 1

    2 040 17 62 frac14 122 kN=m

    62

    The active pressure coefficients for the three soil types are as follows

    Ka1 frac141 sin 35

    1thorn sin 35frac14 0271

    Ka2 frac141 sin 27

    1thorn sin 27frac14 0375

    ffiffiffiffiffiffiffiKa2

    p frac14 0613

    Ka3 frac141 sin 42

    1thorn sin 42frac14 0198

    Distribution of active pressure (plotted in Figure Q62)

    Depth (m) Soil Active pressure (kNm2)

    3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

    12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

    At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

    Total thrust frac14 571 kNm

    Point of application is (4893571) m from the top of the wall ie 857m

    Force (kN) Arm (m) Moment (kN m)

    (1)1

    2 0271 16 32 frac14 195 20 390

    (2) 0271 16 3 2 frac14 260 40 1040

    (3)1

    2 0271 92 22 frac14 50 433 217

    (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

    (5)1

    2 0375 102 32 frac14 172 70 1204

    (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

    (7)1

    2 0198 112 42 frac14 177 1067 1889

    (8)1

    2 98 92 frac14 3969 90 35721

    5713 48934

    Figure Q62

    Lateral earth pressure 35

    63

    (a) For u frac14 0 Ka frac14 Kp frac14 1

    Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

    frac14 245

    At the lower end of the piling

    pa frac14 Kaqthorn Kasatz Kaccu

    frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

    frac14 115 kN=m2

    pp frac14 Kpsatzthorn Kpccu

    frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

    frac14 202 kN=m2

    (b) For 0 frac14 26 and frac14 1

    20

    Ka frac14 035

    Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

    pfrac14 145 ethEquation 619THORN

    Kp frac14 37

    Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

    pfrac14 47 ethEquation 624THORN

    At the lower end of the piling

    pa frac14 Kaqthorn Ka0z Kacc

    0

    frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

    frac14 187 kN=m2

    pp frac14 Kp0zthorn Kpcc

    0

    frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

    frac14 198 kN=m2

    36 Lateral earth pressure

    64

    (a) For 0 frac14 38 Ka frac14 024

    0 frac14 20 98 frac14 102 kN=m3

    The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

    Force (kN) Arm (m) Moment (kN m)

    (1) 024 10 66 frac14 159 33 525

    (2)1

    2 024 17 392 frac14 310 400 1240

    (3) 024 17 39 27 frac14 430 135 580

    (4)1

    2 024 102 272 frac14 89 090 80

    (5)1

    2 98 272 frac14 357 090 321

    Hfrac14 1345 MH frac14 2746

    (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

    (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

    XM frac14MV MH frac14 7790 kNm

    Lever arm of base resultant

    M

    Vfrac14 779

    488frac14 160

    Eccentricity of base resultant

    e frac14 200 160 frac14 040m

    39 m

    27 m

    40 m

    04 m

    04 m

    26 m

    (7)

    (9)

    (1)(2)

    (3)

    (4)

    (5)

    (8)(6)

    (10)

    WT

    10 kNm2

    Hydrostatic

    Figure Q64

    Lateral earth pressure 37

    Base pressures (Equation 627)

    p frac14 VB

    1 6e

    B

    frac14 488

    4eth1 060THORN

    frac14 195 kN=m2 and 49 kN=m2

    Factor of safety against sliding (Equation 628)

    F frac14 V tan

    Hfrac14 488 tan 25

    1345frac14 17

    (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

    Hfrac14 1633 kN

    V frac14 4879 kN

    MH frac14 3453 kNm

    MV frac14 10536 kNm

    The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

    65

    For 0 frac14 36 Ka frac14 026 and Kp frac14 385

    Kp

    Ffrac14 385

    2

    0 frac14 20 98 frac14 102 kN=m3

    The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

    Force (kN) Arm (m) Moment (kN m)

    (1)1

    2 026 17 452 frac14 448 dthorn 15 448dthorn 672

    (2) 026 17 45 d frac14 199d d2 995d2

    (3)1

    2 026 102 d2 frac14 133d2 d3 044d3

    (4)1

    2 385

    2 17 152 frac14 368 dthorn 05 368d 184

    (5)385

    2 17 15 d frac14 491d d2 2455d2

    (6)1

    2 385

    2 102 d2 frac14 982d2 d3 327d3

    38 Lateral earth pressure

    XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

    d3 thorn 516d2 283d 1724 frac14 0

    d frac14 179m

    Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

    Over additional 20 embedded depth

    pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

    Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

    66

    The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

    Ka frac14 sin 69=sin 105

    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

    pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

    26664

    37775

    2

    frac14 050

    The total active thrust (acting at 25 above the normal) is given by Equation 616

    Pa frac14 1

    2 050 19 7502 frac14 267 kN=m

    Figure Q65

    Lateral earth pressure 39

    Horizontal component

    Ph frac14 267 cos 40 frac14 205 kN=m

    Vertical component

    Pv frac14 267 sin 40 frac14 172 kN=m

    Consider moments about the toe of the wall (Figure Q66) (per m)

    Force (kN) Arm (m) Moment (kN m)

    (1)1

    2 175 650 235 frac14 1337 258 345

    (2) 050 650 235 frac14 764 175 134

    (3)1

    2 070 650 235 frac14 535 127 68

    (4) 100 400 235 frac14 940 200 188

    (5) 1

    2 080 050 235 frac14 47 027 1

    Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

    Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

    Lever arm of base resultant

    M

    Vfrac14 795

    525frac14 151m

    Eccentricity of base resultant

    e frac14 200 151 frac14 049m

    Figure Q66

    40 Lateral earth pressure

    Base pressures (Equation 627)

    p frac14 525

    41 6 049

    4

    frac14 228 kN=m2 and 35 kN=m2

    The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

    The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

    The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

    67

    For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

    Force (kN) Arm (m) Moment (kNm)

    (1)1

    2 027 17 52 frac14 574 183 1050

    (2) 027 17 5 3 frac14 689 500 3445

    (3)1

    2 027 102 32 frac14 124 550 682

    (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

    (5)1

    2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

    (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

    (7) 1

    2 267

    2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

    (8) 2 10ffiffiffiffiffiffiffiffiffi267p

    2 d frac14 163d d2thorn 650 82d2 1060d

    Tie rod force per m frac14 T 0 0

    XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

    d3 thorn 77d2 269d 1438 frac14 0

    d frac14 467m

    Depth of penetration frac14 12d frac14 560m

    Lateral earth pressure 41

    Algebraic sum of forces for d frac14 467m isX

    F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

    T frac14 905 kN=m

    Force in each tie rod frac14 25T frac14 226 kN

    68

    (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

    0 frac14 21 98 frac14 112 kN=m3

    The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

    uC frac14 150

    165 15 98 frac14 134 kN=m2

    The average seepage pressure is

    j frac14 15

    165 98 frac14 09 kN=m3

    Hence

    0 thorn j frac14 112thorn 09 frac14 121 kN=m3

    0 j frac14 112 09 frac14 103 kN=m3

    Figure Q67

    42 Lateral earth pressure

    Consider moments about the anchor point A (per m)

    Force (kN) Arm (m) Moment (kN m)

    (1) 10 026 150 frac14 390 60 2340

    (2)1

    2 026 18 452 frac14 474 15 711

    (3) 026 18 45 105 frac14 2211 825 18240

    (4)1

    2 026 121 1052 frac14 1734 100 17340

    (5)1

    2 134 15 frac14 101 40 404

    (6) 134 30 frac14 402 60 2412

    (7)1

    2 134 60 frac14 402 95 3819

    571 4527(8) Ppm

    115 115PPm

    XM frac14 0

    Ppm frac144527

    115frac14 394 kN=m

    Available passive resistance

    Pp frac14 1

    2 385 103 62 frac14 714 kN=m

    Factor of safety

    Fp frac14 Pp

    Ppm

    frac14 714

    394frac14 18

    Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

    Figure Q68

    Lateral earth pressure 43

    (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

    Consider moments (per m) about the tie point A

    Force (kN) Arm (m)

    (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

    (2)1

    2 033 18 452 frac14 601 15

    (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

    (4)1

    2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

    (5)1

    2 134 15 frac14 101 40

    (6) 134 30 frac14 402 60

    (7)1

    2 134 d frac14 67d d3thorn 75

    (8) 1

    2 30 103 d2 frac141545d2 2d3thorn 75

    Moment (kN m)

    (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

    XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

    d3 thorn 827d2 466d 1518 frac14 0

    By trial

    d frac14 544m

    The minimum depth of embedment required is 544m

    69

    For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

    0 frac14 20 98 frac14 102 kN=m3

    The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

    44 Lateral earth pressure

    uC frac14 147

    173 26 98 frac14 216 kN=m2

    and the average seepage pressure around the wall is

    j frac14 26

    173 98 frac14 15 kN=m3

    Consider moments about the prop (A) (per m)

    Force (kN) Arm (m) Moment (kN m)

    (1)1

    2 03 17 272 frac14 186 020 37

    (2) 03 17 27 53 frac14 730 335 2445

    (3)1

    2 03 (102thorn 15) 532 frac14 493 423 2085

    (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

    (5)1

    2 216 26 frac14 281 243 684

    (6) 216 27 frac14 583 465 2712

    (7)1

    2 216 60 frac14 648 800 5184

    3055(8)

    1

    2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

    Factor of safety

    Fr frac14 6885

    3055frac14 225

    Figure Q69

    Lateral earth pressure 45

    610

    For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

    p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

    Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

    Using the recommendations of Twine and Roscoe

    p frac14 02H frac14 02 19 9 frac14 342 kN=m2

    Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

    611

    frac14 18 kN=m3 0 frac14 34

    H frac14 350m nH frac14 335m mH frac14 185m

    Consider a trial value of F frac14 20 Refer to Figure 635

    0m frac14 tan1tan 34

    20

    frac14 186

    Then

    frac14 45 thorn 0m2frac14 543

    W frac14 1

    2 18 3502 cot 543 frac14 792 kN=m

    Figure Q610

    46 Lateral earth pressure

    P frac14 1

    2 s 3352 frac14 561s kN=m

    U frac14 1

    2 98 1852 cosec 543 frac14 206 kN=m

    Equations 630 and 631 then become

    561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

    792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

    ie

    561s 0616N 405 frac14 0

    792 0857N thorn 563 frac14 0

    N frac14 848

    0857frac14 989 kN=m

    Then

    561s 609 405 frac14 0

    s frac14 649

    561frac14 116 kN=m3

    The calculations for trial values of F of 20 15 and 10 are summarized below

    F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

    20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

    s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

    Figure Q611

    Lateral earth pressure 47

    612

    For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

    45 thorn 0

    2frac14 63

    For the retained material between the surface and a depth of 36m

    Pa frac14 1

    2 030 18 362 frac14 350 kN=m

    Weight of reinforced fill between the surface and a depth of 36m is

    Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

    eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

    Eccentricity of Rv

    e frac14 263 250 frac14 013m

    The average vertical stress at a depth of 36m is

    z frac14 Rv

    L 2efrac14 324

    474frac14 68 kN=m2

    (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

    Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

    Tensile stress in the element frac14 138 103

    65 3frac14 71N=mm2

    Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

    Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

    Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

    The weight of ABC is

    W frac14 1

    2 18 52 265 frac14 124 kN=m

    From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

    48 Lateral earth pressure

    (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

    Tp frac14 032 68 120 065 frac14 170 kN

    Tr frac14 213 420

    418frac14 214 kN

    Again the tensile failure and slipping limit states are satisfied for this element

    Figure Q612

    Lateral earth pressure 49

    Chapter 7

    Consolidation theory

    71

    Total change in thickness

    H frac14 782 602 frac14 180mm

    Average thickness frac14 1530thorn 180

    2frac14 1620mm

    Length of drainage path d frac14 1620

    2frac14 810mm

    Root time plot (Figure Q71a)

    ffiffiffiffiffiffit90p frac14 33

    t90 frac14 109min

    cv frac14 0848d2

    t90frac14 0848 8102

    109 1440 365

    106frac14 27m2=year

    r0 frac14 782 764

    782 602frac14 018

    180frac14 0100

    rp frac14 10eth764 645THORN9eth782 602THORN frac14

    10 119

    9 180frac14 0735

    rs frac14 1 eth0100thorn 0735THORN frac14 0165

    Log time plot (Figure Q71b)

    t50 frac14 26min

    cv frac14 0196d2

    t50frac14 0196 8102

    26 1440 365

    106frac14 26m2=year

    r0 frac14 782 763

    782 602frac14 019

    180frac14 0106

    rp frac14 763 623

    782 602frac14 140

    180frac14 0778

    rs frac14 1 eth0106thorn 0778THORN frac14 0116

    Figure Q71(a)

    Figure Q71(b)

    Final void ratio

    e1 frac14 w1Gs frac14 0232 272 frac14 0631

    e

    Hfrac14 1thorn e0

    H0frac14 1thorn e1 thorne

    H0

    ie

    e

    180frac14 1631thorne

    1710

    e frac14 2936

    1530frac14 0192

    Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

    mv frac14 1

    1thorn e0 e0 e101 00

    frac14 1

    1823 0192

    0107frac14 098m2=MN

    k frac14 cvmvw frac14 265 098 98

    60 1440 365 103frac14 81 1010 m=s

    72

    Using Equation 77 (one-dimensional method)

    sc frac14 e0 e11thorn e0 H

    Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

    Figure Q72

    52 Consolidation theory

    Settlement

    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

    1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

    318

    Notes 5 92y 460thorn 84

    Heave

    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

    1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

    38

    73

    U frac14 f ethTvTHORN frac14 f cvt

    d2

    Hence if cv is constant

    t1

    t2frac14 d

    21

    d22

    where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

    d1 frac14 95mm and d2 frac14 2500mm

    for U frac14 050 t2 frac14 t1 d22

    d21

    frac14 20

    60 24 365 25002

    952frac14 263 years

    for U lt 060 Tv frac14

    4U2 (Equation 724(a))

    t030 frac14 t050 0302

    0502

    frac14 263 036 frac14 095 years

    Consolidation theory 53

    74

    The layer is open

    d frac14 8

    2frac14 4m

    Tv frac14 cvtd2frac14 24 3

    42frac14 0450

    ui frac14 frac14 84 kN=m2

    The excess pore water pressure is given by Equation 721

    ue frac14Xmfrac141mfrac140

    2ui

    Msin

    Mz

    d

    expethM2TvTHORN

    In this case z frac14 d

    sinMz

    d

    frac14 sinM

    where

    M frac14

    23

    25

    2

    M sin M M2Tv exp (M2Tv)

    2thorn1 1110 0329

    3

    21 9993 457 105

    ue frac14 2 84 2

    1 0329 ethother terms negligibleTHORN

    frac14 352 kN=m2

    75

    The layer is open

    d frac14 6

    2frac14 3m

    Tv frac14 cvtd2frac14 10 3

    32frac14 0333

    The layer thickness will be divided into six equal parts ie m frac14 6

    54 Consolidation theory

    For an open layer

    Tv frac14 4n

    m2

    n frac14 0333 62

    4frac14 300

    The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

    ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

    i j

    0 1 2 3 4 5 6 7 8 9 10 11 12

    0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

    The initial and 3-year isochrones are plotted in Figure Q75

    Area under initial isochrone frac14 180 units

    Area under 3-year isochrone frac14 63 units

    The average degree of consolidation is given by Equation 725Thus

    U frac14 1 63

    180frac14 065

    Figure Q75

    Consolidation theory 55

    76

    At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

    0 frac14 2w frac14 2 98 frac14 196 kN=m2

    The final consolidation settlement (one-dimensional method) is

    sc frac14 mv0H frac14 083 196 8 frac14 130mm

    Corrected time t frac14 2 1

    2

    40

    52

    frac14 1615 years

    Tv frac14 cvtd2frac14 44 1615

    42frac14 0444

    From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

    77

    The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

    Figure Q77

    56 Consolidation theory

    Point m n Ir (kNm2) sc (mm)

    13020frac14 15 20

    20frac14 10 0194 (4) 113 124

    260

    20frac14 30

    20

    20frac14 10 0204 (2) 59 65

    360

    20frac14 30

    40

    20frac14 20 0238 (1) 35 38

    430

    20frac14 15

    40

    20frac14 20 0224 (2) 65 72

    Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

    78

    Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

    (a) Immediate settlement

    H

    Bfrac14 30

    35frac14 086

    D

    Bfrac14 2

    35frac14 006

    Figure Q78

    Consolidation theory 57

    From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

    si frac14 130131qB

    Eufrac14 10 032 105 35

    40frac14 30mm

    (b) Consolidation settlement

    Layer z (m) Dz Ic (kNm2) syod (mm)

    1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

    3150

    Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

    Now

    H

    Bfrac14 30

    35frac14 086 and A frac14 065

    from Figure 712 13 frac14 079

    sc frac14 13sod frac14 079 315 frac14 250mm

    Total settlement

    s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

    79

    Without sand drains

    Uv frac14 025

    Tv frac14 0049 ethfrom Figure 718THORN

    t frac14 Tvd2

    cvfrac14 0049 82

    cvWith sand drains

    R frac14 0564S frac14 0564 3 frac14 169m

    n frac14 Rrfrac14 169

    015frac14 113

    Tr frac14 cht

    4R2frac14 ch

    4 1692 0049 82

    cvethand ch frac14 cvTHORN

    frac14 0275

    Ur frac14 073 (from Figure 730)

    58 Consolidation theory

    Using Equation 740

    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

    U frac14 080

    710

    Without sand drains

    Uv frac14 090

    Tv frac14 0848

    t frac14 Tvd2

    cvfrac14 0848 102

    96frac14 88 years

    With sand drains

    R frac14 0564S frac14 0564 4 frac14 226m

    n frac14 Rrfrac14 226

    015frac14 15

    Tr

    Tvfrac14 chcv

    d2

    4R2ethsame tTHORN

    Tr

    Tvfrac14 140

    96 102

    4 2262frac14 714 eth1THORN

    Using Equation 740

    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

    An iterative solution is required using (1) and (2) an initial value ofUv being estimated

    Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

    040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

    Thus

    Uv frac14 0295 and Ur frac14 086

    t frac14 88 00683

    0848frac14 07 years

    Consolidation theory 59

    Chapter 8

    Bearing capacity

    81

    (a) The ultimate bearing capacity is given by Equation 83

    qf frac14 cNc thorn DNq thorn 1

    2BN

    For u frac14 0

    Nc frac14 514 Nq frac14 1 N frac14 0

    qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

    The net ultimate bearing capacity is

    qnf frac14 qf D frac14 540 kN=m2

    The net foundation pressure is

    qn frac14 q D frac14 425

    2 eth21 1THORN frac14 192 kN=m2

    The factor of safety (Equation 86) is

    F frac14 qnfqnfrac14 540

    192frac14 28

    (b) For 0 frac14 28

    Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

    qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

    2 112 2 13

    frac14 260thorn 168thorn 146 frac14 574 kN=m2

    qnf frac14 574 112 frac14 563 kN=m2

    F frac14 563

    192frac14 29

    (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

    82

    For 0 frac14 38

    Nq frac14 49 N frac14 67

    qnf frac14 DethNq 1THORN thorn 1

    2BN ethfrom Equation 83THORN

    frac14 eth18 075 48THORN thorn 1

    2 18 15 67

    frac14 648thorn 905 frac14 1553 kN=m2

    qn frac14 500

    15 eth18 075THORN frac14 320 kN=m2

    F frac14 qnfqnfrac14 1553

    320frac14 48

    0d frac14 tan1tan 38

    125

    frac14 32 therefore Nq frac14 23 and N frac14 25

    Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

    2 18 15 25

    frac14 15eth310thorn 337THORNfrac14 970 kN=m

    Design load (action) Vd frac14 500 kN=m

    The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

    83

    D

    Bfrac14 350

    225frac14 155

    From Figure 85 for a square foundation

    Nc frac14 81

    Bearing capacity 61

    For a rectangular foundation (L frac14 450m B frac14 225m)

    Nc frac14 084thorn 016B

    L

    81 frac14 745

    Using Equation 810

    qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

    For F frac14 3

    qn frac14 1006

    3frac14 335 kN=m2

    q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

    Design load frac14 405 450 225 frac14 4100 kN

    Design undrained strength cud frac14 135

    14frac14 96 kN=m2

    Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

    frac14 7241 kN

    Design load Vd frac14 4100 kN

    Rd gt Vd therefore the bearing resistance limit state is satisfied

    84

    For 0 frac14 40

    Nq frac14 64 N frac14 95

    qnf frac14 DethNq 1THORN thorn 04BN

    (a) Water table 5m below ground level

    qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

    qn frac14 400 17 frac14 383 kN=m2

    F frac14 2686

    383frac14 70

    (b) Water table 1m below ground level (ie at foundation level)

    0 frac14 20 98 frac14 102 kN=m3

    62 Bearing capacity

    qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

    F frac14 2040

    383frac14 53

    (c) Water table at ground level with upward hydraulic gradient 02

    eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

    qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

    F frac14 1296

    392frac14 33

    85

    The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

    Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

    Design value of 0 frac14 tan1tan 39

    125

    frac14 33

    For 0 frac14 33 Nq frac14 26 and N frac14 29

    Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

    Rd gt Vd therefore the bearing resistance limit state is satisfied

    86

    (a) Undrained shear for u frac14 0

    Nc frac14 514 Nq frac14 1 N frac14 0

    qnf frac14 12cuNc

    frac14 12 100 514 frac14 617 kN=m2

    qn frac14 qnfFfrac14 617

    3frac14 206 kN=m2

    q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

    Bearing capacity 63

    Drained shear for 0 frac14 32

    Nq frac14 23 N frac14 25

    0 frac14 21 98 frac14 112 kN=m3

    qnf frac14 0DethNq 1THORN thorn 040BN

    frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

    frac14 694 kN=m2

    q frac14 694

    3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

    Design load frac14 42 227 frac14 3632 kN

    (b) Design undrained strength cud frac14 100

    14frac14 71 kNm2

    Design bearing resistance Rd frac14 12cudNe area

    frac14 12 71 514 42

    frac14 7007 kN

    For drained shear 0d frac14 tan1tan 32

    125

    frac14 26

    Nq frac14 12 N frac14 10

    Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

    (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

    Layer z (m) m n Ir 0 (kNm2) sod (mm)

    1 2 100 0175 0700qn 0182qn

    2 6 033 0044 0176qn 0046qn

    3 10 020 0017 0068qn 0018qn

    0246qn

    Diameter of equivalent circle B frac14 45m

    H

    Bfrac14 12

    45frac14 27 and A frac14 042

    13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

    64 Bearing capacity

    For sc frac14 30mm

    qn frac14 30

    0147frac14 204 kN=m2

    q frac14 204thorn 21 frac14 225 kN=m2

    Design load frac14 42 225 frac14 3600 kN

    The design load is 3600 kN settlement being the limiting criterion

    87

    D

    Bfrac14 8

    4frac14 20

    From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

    F frac14 cuNc

    Dfrac14 40 71

    20 8frac14 18

    88

    Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

    Design value of 0 frac14 tan1tan 38

    125

    frac14 32

    Figure Q86

    Bearing capacity 65

    For 0 frac14 32 Nq frac14 23 and N frac14 25

    Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

    The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

    Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

    For B frac14 250m qn frac14 3750

    2502 17 frac14 583 kN=m2

    From Figure 510 m frac14 n frac14 126

    6frac14 021

    Ir frac14 0019

    Stress increment frac14 4 0019 583 frac14 44 kN=m2

    Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

    The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

    The settlement is less than 20mm therefore the serviceability limit state is satisfied

    89

    Depth (m) N 0v (kNm2) CN N1

    070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

    Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

    (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

    Cw frac14 05thorn 0530

    47

    frac14 082

    66 Bearing capacity

    Thus

    qa frac14 150 082 frac14 120 kN=m2

    (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

    Thus

    qa frac14 90 15 frac14 135 kN=m2

    (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

    Ic frac14 171

    1014frac14 0068

    From Equation 819(a) with s frac14 25mm

    q frac14 25

    3507 0068frac14 150 kN=m2

    810

    Peak value of strain influence factor occurs at a depth of 27m and is given by

    Izp frac14 05thorn 01130

    16 27

    05

    frac14 067

    Refer to Figure Q810

    E frac14 25qc

    Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

    Ez (mm3MN)

    1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

    0203

    C1 frac14 1 0500qnfrac14 1 05 12 16

    130frac14 093

    C2 frac14 1 ethsayTHORN

    s frac14 C1C2qnX Iz

    Ez frac14 093 1 130 0203 frac14 25mm

    Bearing capacity 67

    811

    At pile base level

    cu frac14 220 kN=m2

    qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

    Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

    00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

    qs frac14 00 frac14 07 1226 frac14 86 kN=m2

    Then

    Qf frac14 Abqb thorn Asqs

    frac14

    4 32 1980

    thorn eth 105 139 86THORN

    frac14 13 996thorn 3941 frac14 17 937 kN

    0 01 02 03 04 05 06 07

    0 2 4 6 8 10 12 14

    1

    2

    3

    4

    5

    6

    7

    8

    (1)

    (2)

    (3)

    (4)

    (5)

    qc

    qc

    Iz

    Iz

    (MNm2)

    z (m)

    Figure Q810

    68 Bearing capacity

    Allowable load

    ethaTHORN Qf

    2frac14 17 937

    2frac14 8968 kN

    ethbTHORN Abqb

    3thorn Asqs frac14 13 996

    3thorn 3941 frac14 8606 kN

    ie allowable load frac14 8600 kN

    Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

    According to the limit state method

    Characteristic undrained strength at base level cuk frac14 220

    150kN=m2

    Characteristic base resistance qbk frac14 9cuk frac14 9 220

    150frac14 1320 kN=m2

    Characteristic shaft resistance qsk frac14 00150

    frac14 86

    150frac14 57 kN=m2

    Characteristic base and shaft resistances

    Rbk frac14

    4 32 1320 frac14 9330 kN

    Rsk frac14 105 139 86

    150frac14 2629 kN

    For a bored pile the partial factors are b frac14 160 and s frac14 130

    Design bearing resistance Rcd frac14 9330

    160thorn 2629

    130

    frac14 5831thorn 2022

    frac14 7850 kN

    Adding ethDAb W) the design bearing resistance becomes 9650 kN

    812

    ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

    qs frac14 cu frac14 040 105 frac14 42 kN=m2

    For a single pile

    Qf frac14 Abqb thorn Asqs

    frac14

    4 062 1305

    thorn eth 06 15 42THORN

    frac14 369thorn 1187 frac14 1556 kN

    Bearing capacity 69

    Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

    eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

    (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

    (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

    qbkfrac14 9cuk frac14 9 220

    150frac14 1320 kN=m2

    qskfrac14cuk frac14 040 105

    150frac14 28 kN=m2

    Rbkfrac14

    4 0602 1320 frac14 373 kN

    Rskfrac14 060 15 28 frac14 791 kN

    Rcdfrac14 373

    160thorn 791

    130frac14 233thorn 608 frac14 841 kN

    Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

    Rcd gt Fcd therefore the bearing resistance limit state is satisfied

    (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

    q frac14 21 000

    1762frac14 68 kN=m2

    Immediate settlement

    H

    Bfrac14 15

    176frac14 085

    D

    Bfrac14 13

    176frac14 074

    L

    Bfrac14 1

    Hence from Figure 515

    130 frac14 078 and 131 frac14 041

    70 Bearing capacity

    Thus using Equation 528

    si frac14 078 041 68 176

    65frac14 6mm

    Consolidation settlement

    Layer z (m) Area (m2) (kNm2) mvH (mm)

    1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

    434 (sod)

    Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

    sc frac14 056 434 frac14 24mm

    The total settlement is (6thorn 24) frac14 30mm

    813

    At base level N frac14 26 Then using Equation 830

    qb frac14 40NDb

    Bfrac14 40 26 2

    025frac14 8320 kN=m2

    ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

    Figure Q812

    Bearing capacity 71

    Over the length embedded in sand

    N frac14 21 ie18thorn 24

    2

    Using Equation 831

    qs frac14 2N frac14 2 21 frac14 42 kN=m2

    For a single pile

    Qf frac14 Abqb thorn Asqs

    frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

    For the pile group assuming a group efficiency of 12

    XQf frac14 12 9 604 frac14 6523 kN

    Then the load factor is

    F frac14 6523

    2000thorn 1000frac14 21

    (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

    Characteristic base resistance per unit area qbk frac14 8320

    150frac14 5547 kNm2

    Characteristic shaft resistance per unit area qsk frac14 42

    150frac14 28 kNm2

    Characteristic base and shaft resistances for a single pile

    Rbk frac14 0252 5547 frac14 347 kN

    Rsk frac14 4 025 2 28 frac14 56 kN

    For a driven pile the partial factors are b frac14 s frac14 130

    Design bearing resistance Rcd frac14 347

    130thorn 56

    130frac14 310 kN

    For the pile group Rcd frac14 12 9 310 frac14 3348 kN

    Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

    (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

    From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

    72 Bearing capacity

    N frac14 24thorn 26thorn 34

    3frac14 28

    Ic frac14 171

    2814frac14 0016 ethEquation 818THORN

    s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

    The settlement is less than 20mm therefore the serviceability limit state is satisfied

    814

    Using Equation 841

    Tf frac14 DLcu thorn

    4ethD2 d2THORNcuNc

    frac14 eth 02 5 06 110THORN thorn

    4eth022 012THORN110 9

    frac14 207thorn 23 frac14 230 kN

    Figure Q813

    Bearing capacity 73

    Chapter 9

    Stability of slopes

    91

    Referring to Figure Q91

    W frac14 417 19 frac14 792 kN=m

    Q frac14 20 28 frac14 56 kN=m

    Arc lengthAB frac14

    180 73 90 frac14 115m

    Arc length BC frac14

    180 28 90 frac14 44m

    The factor of safety is given by

    F frac14 rethcuLaTHORNWd1 thornQd2 frac14

    90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

    Depth of tension crack z0 frac14 2cu

    frac14 2 20

    19frac14 21m

    Arc length BD frac14

    180 13

    1

    2 90 frac14 21m

    F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

    The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

    Design resisting moment frac14 rXethcudLaTHORN frac14 90

    14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

    Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

    The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

    92

    u frac14 0

    Depth factor D frac14 11

    9frac14 122

    Using Equation 92 with F frac14 10

    Ns frac14 cu

    FHfrac14 30

    10 19 9frac14 0175

    Hence from Figure 93

    frac14 50

    For F frac14 12

    Ns frac14 30

    12 19 9frac14 0146

    frac14 27

    93

    Refer to Figure Q93

    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

    1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

    74 m

    214 1deg

    213 1deg

    39 m

    WB

    D

    C

    28 m

    21 m

    A

    Q

    Soil (1)Soil (2)

    73deg

    Figure Q91

    Stability of slopes 75

    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

    9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

    599 256 328 1372

    Figure Q93

    76 Stability of slopes

    XW cos frac14 b

    Xh cos frac14 21 2 599 frac14 2516 kN=mX

    W sin frac14 bX

    h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

    Arc length La frac14

    180 57

    1

    2 326 frac14 327m

    The factor of safety is given by

    F frac14 c0La thorn tan0ethW cos ulTHORN

    W sin

    frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

    frac14 091

    According to the limit state method

    0d frac14 tan1tan 32

    125

    frac14 265

    c0 frac14 8

    160frac14 5 kN=m2

    Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

    Design disturbing moment frac14 1075 kN=m

    The design resisting moment is less than the design disturbing moment therefore a slipwill occur

    94

    F frac14 1

    W sin

    Xfc0bthorn ethW ubTHORN tan0g sec

    1thorn ethtan tan0=FTHORN

    c0 frac14 8 kN=m2

    0 frac14 32

    c0b frac14 8 2 frac14 16 kN=m

    W frac14 bh frac14 21 2 h frac14 42h kN=m

    Try F frac14 100

    tan0

    Ffrac14 0625

    Stability of slopes 77

    Values of u are as obtained in Figure Q93

    SliceNo

    h(m)

    W frac14 bh(kNm)

    W sin(kNm)

    ub(kNm)

    c0bthorn (W ub) tan0(kNm)

    sec

    1thorn (tan tan0)FProduct(kNm)

    1 05 21 6 2 8 24 1078 262 13 55 31

    23 33 30 1042 31

    3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

    224 92 72 0931 67

    6 50 210 11 40 100 85 0907 777 55 231 14

    12 58 112 90 0889 80

    8 60 252 1812

    80 114 102 0874 899 63 265 22 99 116 109 0861 94

    10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

    2154 88 116 0853 99

    14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

    1074 1091

    F frac14 1091

    1074frac14 102 (assumed value 100)

    Thus

    F frac14 101

    95

    F frac14 1

    W sin

    XfWeth1 ruTHORN tan0g sec

    1thorn ethtan tan0THORN=F

    0 frac14 33

    ru frac14 020

    W frac14 bh frac14 20 5 h frac14 100h kN=m

    eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

    Try F frac14 110

    tan 0

    Ffrac14 tan 33

    110frac14 0590

    78 Stability of slopes

    Referring to Figure Q95

    SliceNo

    h(m)

    W frac14 bh(kNm)

    W sin(kNm)

    W(1 ru) tan0(kNm)

    sec

    1thorn ( tan tan0)FProduct(kNm)

    1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

    2120 234 0892 209

    4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

    1185 1271

    Figure Q95

    Stability of slopes 79

    F frac14 1271

    1185frac14 107

    The trial value was 110 therefore take F to be 108

    96

    (a) Water table at surface the factor of safety is given by Equation 912

    F frac14 0

    sat

    tan0

    tan

    ptie 15 frac14 92

    19

    tan 36

    tan

    tan frac14 0234

    frac14 13

    Water table well below surface the factor of safety is given by Equation 911

    F frac14 tan0

    tan

    frac14 tan 36

    tan 13

    frac14 31

    (b) 0d frac14 tan1tan 36

    125

    frac14 30

    Depth of potential failure surface frac14 z

    Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

    frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

    frac14 504z kN

    Design disturbing moment per unit area Sd frac14 sat sin cos

    frac14 19 z sin 13 cos 13

    frac14 416z kN

    Rd gtSd therefore the limit state for overall stability is satisfied

    80 Stability of slopes

    • Book Cover
    • Title
    • Contents
    • Basic characteristics of soils
    • Seepage
    • Effective stress
    • Shear strength
    • Stresses and displacements
    • Lateral earth pressure
    • Consolidation theory
    • Bearing capacity
    • Stability of slopes

      Craigrsquos Soil Mechanics SeventhEdition Solutions Manual

      RF CraigFormerly

      Department of Civil Engineering

      University of Dundee UK

      First published 1992by E amp FN Spon an imprint of Thomson ProfessionalSecond edition 1997Third edition 200411 New Fetter Lane London EC4P 4EE

      Simultaneously published in the USA and Canadaby Spon Press29 West 35th Street New York NY 10001

      Spon Press is an imprint of the Taylor amp Francis Group

      ordf 1992 1997 2004 RF Craig

      All rights reserved No part of this book may be reprinted or reproducedor utilised in any form or by any electronic mechanical or other meansnow known or hereafter invented including photocopying and recordingor in any information storage or retrieval system without permission inwriting from the publishers

      British Library Cataloguing in Publication DataA catalogue record for this book is availablefrom the British Library

      Library of Congress Cataloging in Publication DataA catalog record for this book has been requested

      ISBN 0ndash415ndash33294ndashX

      This edition published in the Taylor amp Francis e-Library 2004

      (Print edition)

      ISBN 0-203-31104-3 Master e-book ISBN

      ISBN 0-203-67167-8 (Adobe eReader Format)

      collection of thousands of eBooks please go to wwweBookstoretandfcoukrdquoldquoTo purchase your own copy of this or any of Taylor amp Francis or Routledgersquos

      Contents

      1 Basic characteristics of soils 1

      2 Seepage 6

      3 Effective stress 14

      4 Shear strength 22

      5 Stresses and displacements 28

      6 Lateral earth pressure 34

      7 Consolidation theory 50

      8 Bearing capacity 60

      9 Stability of slopes 74

      Authorrsquos note

      In order not to short-circuit the learningprocess it is vital that the reader shouldattempt the problems before referring to thesolutions in this manual

      Chapter 1

      Basic characteristics of soils

      11

      Soil E consists of 98 coarse material (31 gravel size 67 sand size) and 2 finesIt is classified as SW well-graded gravelly SAND or in greater detail well-gradedslightly silty very gravelly SAND

      Soil F consists of 63 coarse material (2 gravel size 61 sand size) and 37non-plastic fines (ie between 35 and 65 fines) therefore the soil is classified as MSsandy SILT

      Soil G consists of 73 fine material (ie between 65 and 100 fines) and 27 sandsize The liquid limit is 32 and the plasticity index is 8 (ie 32 24) plotting marginallybelow the A-line in the ML zone on the plasticity chart Thus the classification is MLSILT (M-SOIL) of low plasticity (The plasticity chart is given in Figure 17)

      Figure Q11

      Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

      12

      From Equation 117

      1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

      191frac14 155

      e frac14 055

      Using Equation 113

      Sr frac14 wGs

      efrac14 0095 270

      055frac14 0466 eth466THORN

      Using Equation 119

      sat frac14 Gs thorn e1thorn e w frac14 325

      155 100 frac14 210Mg=m3

      From Equation 114

      w frac14 e

      Gsfrac14 055

      270frac14 0204 eth204THORN

      13

      Equations similar to 117ndash120 apply in the case of unit weights thus

      d frac14 Gs

      1thorn e w frac14272

      170 98 frac14 157 kN=m3

      sat frac14 Gs thorn e1thorn e w frac14 342

      170 98 frac14 197 kN=m3

      Using Equation 121

      0 frac14 Gs 1

      1thorn e w frac14 172

      170 98 frac14 99 kN=m3

      Using Equation 118a with Srfrac14 075

      frac14 Gs thorn Sre

      1thorn e w frac14 3245

      170 98 frac14 187 kN=m3

      2 Basic characteristics of soils

      Using Equation 113

      w frac14 Sre

      Gsfrac14 075 070

      272frac14 0193 eth193THORN

      The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

      14

      Volume of specimenfrac14

      438276 frac14 86 200mm3

      Bulk density ethTHORN frac14 Mass

      Volumefrac14 1680

      86 200 103frac14 195Mg=m3

      Water content ethwTHORN frac14 1680 1305

      1305frac14 0287 eth287THORN

      From Equation 117

      1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

      195frac14 180

      e frac14 080

      Using Equation 113

      Sr frac14 wGs

      efrac14 0287 273

      080frac14 098 eth98THORN

      15

      Using Equation 124

      d frac14

      1thorn w frac14215

      112frac14 192Mg=m3

      From Equation 117

      1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

      215frac14 138

      e frac14 038

      Using Equation 113

      Sr frac14 wGs

      efrac14 012 265

      038frac14 0837 eth837THORN

      Basic characteristics of soils 3

      Using Equation 115

      Afrac14 e wGs

      1thorn e frac14038 0318

      138frac14 0045 eth45THORN

      The zero air voids dry density is given by Equation 125

      d frac14 Gs

      1thorn wGsw frac14 265

      1thorn eth0135 265THORN 100 frac14 195Mg=m3

      ie a dry density of 200Mgm3 would not be possible

      16

      Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

      (Mgm3) d10(Mgm3)

      2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

      In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

      wopt frac14 15 and dmaxfrac14 183Mg=m3

      Figure Q16

      4 Basic characteristics of soils

      Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

      d5 d10

      respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

      17

      From Equation 120

      e frac14 Gswd 1

      The maximum and minimum values of void ratio are given by

      emax frac14 Gsw

      dmin

      1

      emin frac14 Gswdmax

      1

      From Equation 123

      ID frac14 Gsweth1=dmin 1=dTHORN

      Gsweth1=dmin 1=dmax

      THORN

      frac14 frac121 ethdmin=dTHORN1=dmin

      frac121 ethdmin=dmax

      THORN1=dmin

      frac14 d dmin

      dmax dmin

      dmax

      d

      frac14 172 154

      181 154

      181

      172

      frac14 070 eth70THORN

      Basic characteristics of soils 5

      Chapter 2

      Seepage

      21

      The coefficient of permeability is determined from the equation

      k frac14 23al

      At1log

      h0

      h1

      where

      a frac14

      4 00052 m2 l frac14 02m

      A frac14

      4 012 m2 t1 frac14 3 602 s

      logh0

      h1frac14 log

      100

      035frac14 0456

      k frac14 23 00052 02 0456

      012 3 602frac14 49 108 m=s

      22

      The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

      q frac14 kh Nf

      Ndfrac14 106 400 37

      11frac14 13 106 m3=s per m

      Figure Q22

      23

      The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

      q frac14 kh Nf

      Ndfrac14 5 105 300 35

      9frac14 58 105 m3=s per m

      The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

      Point h (m) h z (m) u frac14 w(h z)(kNm2)

      1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

      eg for Point 1

      h1 frac14 7

      9 300 frac14 233m

      h1 z1 frac14 233 eth250THORN frac14 483m

      Figure Q23

      Seepage 7

      u1 frac14 98 483 frac14 47 kN=m2

      The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

      24

      The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

      q frac14 kh Nf

      Ndfrac14 40 107 550 10

      11frac14 20 106 m3=s per m

      25

      The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

      q frac14 kh Nf

      Ndfrac14 20 106 500 42

      9frac14 47 106 m3=s per m

      Figure Q24

      8 Seepage

      26

      The scale transformation factor in the x direction is given by Equation 221 ie

      xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

      ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

      Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

      k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

      pfrac14

      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

      p 107 frac14 30 107 m=s

      Thus the quantity of seepage is given by

      q frac14 k0h Nf

      Ndfrac14 30 107 1400 325

      12frac14 11 106 m3=s per m

      Figure Q25

      Seepage 9

      27

      The scale transformation factor in the x direction is

      xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

      ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

      Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

      k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

      pfrac14

      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

      p 106 frac14 45 106 m=s

      The focus of the basic parabola is at point A The parabola passes through point Gsuch that

      GC frac14 03HC frac14 03 30 frac14 90m

      Thus the coordinates of G are

      x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

      480 frac14 x0 2002

      4x0

      Figure Q26

      10 Seepage

      Hence

      x0 frac14 20m

      Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

      x frac14 20 z2

      80

      x 20 0 50 100 200 300z 0 400 748 980 1327 1600

      The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

      No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

      q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

      28

      The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

      Figure Q27

      Seepage 11

      q frac14 kh Nf

      Ndfrac14 45 105 28 33

      7

      frac14 59 105 m3=s per m

      29

      The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

      kx frac14 H1k1 thornH2k2

      H1 thornH2frac14 106

      10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

      kz frac14 H1 thornH2

      H1

      k1thornH2

      k2

      frac14 10

      5

      eth2 106THORN thorn5

      eth16 106THORNfrac14 36 106 m=s

      Then the scale transformation factor is given by

      xt frac14 xffiffiffiffiffikz

      pffiffiffiffiffikx

      p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

      Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

      Figure Q28

      12 Seepage

      k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

      qfrac14

      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

      p 106 frac14 57 106 m=s

      Then the quantity of seepage is given by

      q frac14 k0h Nf

      Ndfrac14 57 106 350 56

      11

      frac14 10 105 m3=s per m

      Figure Q29

      Seepage 13

      Chapter 3

      Effective stress

      31

      Buoyant unit weight

      0 frac14 sat w frac14 20 98 frac14 102 kN=m3

      Effective vertical stress

      0v frac14 5 102 frac14 51 kN=m2 or

      Total vertical stress

      v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

      Pore water pressure

      u frac14 7 98 frac14 686 kN=m2

      Effective vertical stress

      0v frac14 v u frac14 1196 686 frac14 51 kN=m2

      32

      Buoyant unit weight

      0 frac14 sat w frac14 20 98 frac14 102 kN=m3

      Effective vertical stress

      0v frac14 5 102 frac14 51 kN=m2 or

      Total vertical stress

      v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

      Pore water pressure

      u frac14 205 98 frac14 2009 kN=m2

      Effective vertical stress

      0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

      33

      At top of the clay

      v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

      u frac14 2 98 frac14 196 kN=m2

      0v frac14 v u frac14 710 196 frac14 514 kN=m2

      Alternatively

      0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

      0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

      At bottom of the clay

      v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

      u frac14 12 98 frac14 1176 kN=m2

      0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

      NB The alternative method of calculation is not applicable because of the artesiancondition

      Figure Q3132

      Effective stress 15

      34

      0 frac14 20 98 frac14 102 kN=m3

      At 8m depth

      0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

      35

      0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

      0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

      Figure Q33

      Figure Q34

      16 Effective stress

      (a) Immediately after WT rise

      At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

      0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

      At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

      0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

      (b) Several years after WT rise

      At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

      At 8m depth

      0v frac14 940 kN=m2 (as above)

      At 12m depth

      0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

      Figure Q35

      Effective stress 17

      36

      Total weight

      ab frac14 210 kN

      Effective weight

      ac frac14 112 kN

      Resultant boundary water force

      be frac14 119 kN

      Seepage force

      ce frac14 34 kN

      Resultant body force

      ae frac14 99 kN eth73 to horizontalTHORN

      (Refer to Figure Q36)

      Figure Q36

      18 Effective stress

      37

      Situation (1)(a)

      frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

      u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

      0 frac14 u frac14 694 392 frac14 302 kN=m2

      (b)

      i frac14 2

      4frac14 05

      j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

      Situation (2)(a)

      frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

      u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

      0 frac14 u frac14 498 392 frac14 106 kN=m2

      (b)

      i frac14 2

      4frac14 05

      j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

      38

      The flow net is drawn in Figure Q24

      Loss in total head between adjacent equipotentials

      h frac14 550

      Ndfrac14 550

      11frac14 050m

      Exit hydraulic gradient

      ie frac14 h

      sfrac14 050

      070frac14 071

      Effective stress 19

      The critical hydraulic gradient is given by Equation 39

      ic frac14 0

      wfrac14 102

      98frac14 104

      Therefore factor of safety against lsquoboilingrsquo (Equation 311)

      F frac14 iciefrac14 104

      071frac14 15

      Total head at C

      hC frac14 nd

      Ndh frac14 24

      11 550 frac14 120m

      Elevation head at C

      zC frac14 250m

      Pore water pressure at C

      uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

      Therefore effective vertical stress at C

      0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

      For point D

      hD frac14 73

      11 550 frac14 365m

      zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

      0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

      39

      The flow net is drawn in Figure Q25

      For a soil prism 150 300m adjacent to the piling

      hm frac14 26

      9 500 frac14 145m

      20 Effective stress

      Factor of safety against lsquoheavingrsquo (Equation 310)

      F frac14 ic

      imfrac14 0d

      whmfrac14 97 300

      98 145frac14 20

      With a filter

      F frac14 0d thorn wwhm

      3 frac14 eth97 300THORN thorn w98 145

      w frac14 135 kN=m2

      Depth of filterfrac14 13521frac14 065m (if above water level)

      Effective stress 21

      Chapter 4

      Shear strength

      41

      frac14 295 kN=m2

      u frac14 120 kN=m2

      0 frac14 u frac14 295 120 frac14 175 kN=m2

      f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

      42

      03 (kNm2) 1 3 (kNm2) 01 (kNm2)

      100 452 552200 908 1108400 1810 2210800 3624 4424

      The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

      Figure Q42

      43

      3 (kNm2) 1 3 (kNm2) 1 (kNm2)

      200 222 422400 218 618600 220 820

      The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

      44

      The modified shear strength parameters are

      0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

      a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

      The coordinates of the stress point representing failure conditions in the test are

      1

      2eth1 2THORN frac14 1

      2 170 frac14 85 kN=m2

      1

      2eth1 thorn 3THORN frac14 1

      2eth270thorn 100THORN frac14 185 kN=m2

      The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

      uf frac14 36 kN=m2

      Figure Q43

      Figure Q44

      Shear strength 23

      45

      3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

      150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

      The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

      The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

      46

      03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

      200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

      The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

      Figure Q45

      24 Shear strength

      47

      The torque required to produce shear failure is given by

      T frac14 dh cud

      2thorn 2

      Z d=2

      0

      2r drcur

      frac14 cud2h

      2thorn 4cu

      Z d=2

      0

      r2dr

      frac14 cud2h

      2thorn d

      3

      6

      Then

      35 frac14 cu52 10

      2thorn 53

      6

      103

      cu frac14 76 kN=m3

      400

      0 400 800 1200 1600

      τ (k

      Nm

      2 )

      σprime (kNm2)

      34deg

      315deg29deg

      (a)

      (b)

      0 400

      400

      800 1200 1600

      Failure envelope

      300 500

      σprime (kNm2)

      τ (k

      Nm

      2 )

      20 (kNm2)

      31deg

      Figure Q46

      Shear strength 25

      48

      The relevant stress values are calculated as follows

      3 frac14 600 kN=m2

      1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

      2(1 3) 0 40 79 107 139 159

      1

      2(01 thorn 03) 400 411 402 389 351 326

      1

      2(1 thorn 3) 600 640 679 707 739 759

      The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

      Af frac14 433 200

      319frac14 073

      49

      B frac14 u33

      frac14 144

      350 200frac14 096

      a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

      0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

      10 283 65 023

      Figure Q48

      26 Shear strength

      The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

      Figure Q49

      Shear strength 27

      Chapter 5

      Stresses and displacements

      51

      Vertical stress is given by

      z frac14 Qz2Ip frac14 5000

      52Ip

      Values of Ip are obtained from Table 51

      r (m) rz Ip z (kNm2)

      0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

      10 20 0009 2

      The variation of z with radial distance (r) is plotted in Figure Q51

      Figure Q51

      52

      Below the centre load (Figure Q52)

      r

      zfrac14 0 for the 7500-kN load

      Ip frac14 0478

      r

      zfrac14 5

      4frac14 125 for the 10 000- and 9000-kN loads

      Ip frac14 0045

      Then

      z frac14X Q

      z2Ip

      frac14 7500 0478

      42thorn 10 000 0045

      42thorn 9000 0045

      42

      frac14 224thorn 28thorn 25 frac14 277 kN=m2

      53

      The vertical stress under a corner of a rectangular area is given by

      z frac14 qIr

      where values of Ir are obtained from Figure 510 In this case

      z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

      z

      Figure Q52

      Stresses and displacements 29

      z (m) m n Ir z (kNm2)

      0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

      10 010 0005 5

      z is plotted against z in Figure Q53

      54

      (a)

      m frac14 125

      12frac14 104

      n frac14 18

      12frac14 150

      From Figure 510 Irfrac14 0196

      z frac14 2 175 0196 frac14 68 kN=m2

      Figure Q53

      30 Stresses and displacements

      (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

      z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

      55

      Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

      Px frac14 2Q

      1

      m2 thorn 1frac14 2 150

      125frac14 76 kN=m

      Equation 517 is used to obtain the pressure distribution

      px frac14 4Q

      h

      m2n

      ethm2 thorn n2THORN2 frac14150

      m2n

      ethm2 thorn n2THORN2 ethkN=m2THORN

      Figure Q54

      Stresses and displacements 31

      n m2n

      (m2 thorn n2)2

      px(kNm2)

      0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

      The pressure distribution is plotted in Figure Q55

      56

      H

      Bfrac14 10

      2frac14 5

      L

      Bfrac14 4

      2frac14 2

      D

      Bfrac14 1

      2frac14 05

      Hence from Figure 515

      131 frac14 082

      130 frac14 094

      Figure Q55

      32 Stresses and displacements

      The immediate settlement is given by Equation 528

      si frac14 130131qB

      Eu

      frac14 094 082 200 2

      45frac14 7mm

      Stresses and displacements 33

      Chapter 6

      Lateral earth pressure

      61

      For 0 frac14 37 the active pressure coefficient is given by

      Ka frac14 1 sin 37

      1thorn sin 37frac14 025

      The total active thrust (Equation 66a with c0 frac14 0) is

      Pa frac14 1

      2KaH

      2 frac14 1

      2 025 17 62 frac14 765 kN=m

      If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

      K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

      and the thrust on the wall is

      P0 frac14 1

      2K0H

      2 frac14 1

      2 040 17 62 frac14 122 kN=m

      62

      The active pressure coefficients for the three soil types are as follows

      Ka1 frac141 sin 35

      1thorn sin 35frac14 0271

      Ka2 frac141 sin 27

      1thorn sin 27frac14 0375

      ffiffiffiffiffiffiffiKa2

      p frac14 0613

      Ka3 frac141 sin 42

      1thorn sin 42frac14 0198

      Distribution of active pressure (plotted in Figure Q62)

      Depth (m) Soil Active pressure (kNm2)

      3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

      12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

      At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

      Total thrust frac14 571 kNm

      Point of application is (4893571) m from the top of the wall ie 857m

      Force (kN) Arm (m) Moment (kN m)

      (1)1

      2 0271 16 32 frac14 195 20 390

      (2) 0271 16 3 2 frac14 260 40 1040

      (3)1

      2 0271 92 22 frac14 50 433 217

      (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

      (5)1

      2 0375 102 32 frac14 172 70 1204

      (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

      (7)1

      2 0198 112 42 frac14 177 1067 1889

      (8)1

      2 98 92 frac14 3969 90 35721

      5713 48934

      Figure Q62

      Lateral earth pressure 35

      63

      (a) For u frac14 0 Ka frac14 Kp frac14 1

      Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

      frac14 245

      At the lower end of the piling

      pa frac14 Kaqthorn Kasatz Kaccu

      frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

      frac14 115 kN=m2

      pp frac14 Kpsatzthorn Kpccu

      frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

      frac14 202 kN=m2

      (b) For 0 frac14 26 and frac14 1

      20

      Ka frac14 035

      Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

      pfrac14 145 ethEquation 619THORN

      Kp frac14 37

      Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

      pfrac14 47 ethEquation 624THORN

      At the lower end of the piling

      pa frac14 Kaqthorn Ka0z Kacc

      0

      frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

      frac14 187 kN=m2

      pp frac14 Kp0zthorn Kpcc

      0

      frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

      frac14 198 kN=m2

      36 Lateral earth pressure

      64

      (a) For 0 frac14 38 Ka frac14 024

      0 frac14 20 98 frac14 102 kN=m3

      The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

      Force (kN) Arm (m) Moment (kN m)

      (1) 024 10 66 frac14 159 33 525

      (2)1

      2 024 17 392 frac14 310 400 1240

      (3) 024 17 39 27 frac14 430 135 580

      (4)1

      2 024 102 272 frac14 89 090 80

      (5)1

      2 98 272 frac14 357 090 321

      Hfrac14 1345 MH frac14 2746

      (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

      (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

      XM frac14MV MH frac14 7790 kNm

      Lever arm of base resultant

      M

      Vfrac14 779

      488frac14 160

      Eccentricity of base resultant

      e frac14 200 160 frac14 040m

      39 m

      27 m

      40 m

      04 m

      04 m

      26 m

      (7)

      (9)

      (1)(2)

      (3)

      (4)

      (5)

      (8)(6)

      (10)

      WT

      10 kNm2

      Hydrostatic

      Figure Q64

      Lateral earth pressure 37

      Base pressures (Equation 627)

      p frac14 VB

      1 6e

      B

      frac14 488

      4eth1 060THORN

      frac14 195 kN=m2 and 49 kN=m2

      Factor of safety against sliding (Equation 628)

      F frac14 V tan

      Hfrac14 488 tan 25

      1345frac14 17

      (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

      Hfrac14 1633 kN

      V frac14 4879 kN

      MH frac14 3453 kNm

      MV frac14 10536 kNm

      The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

      65

      For 0 frac14 36 Ka frac14 026 and Kp frac14 385

      Kp

      Ffrac14 385

      2

      0 frac14 20 98 frac14 102 kN=m3

      The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

      Force (kN) Arm (m) Moment (kN m)

      (1)1

      2 026 17 452 frac14 448 dthorn 15 448dthorn 672

      (2) 026 17 45 d frac14 199d d2 995d2

      (3)1

      2 026 102 d2 frac14 133d2 d3 044d3

      (4)1

      2 385

      2 17 152 frac14 368 dthorn 05 368d 184

      (5)385

      2 17 15 d frac14 491d d2 2455d2

      (6)1

      2 385

      2 102 d2 frac14 982d2 d3 327d3

      38 Lateral earth pressure

      XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

      d3 thorn 516d2 283d 1724 frac14 0

      d frac14 179m

      Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

      Over additional 20 embedded depth

      pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

      Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

      66

      The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

      Ka frac14 sin 69=sin 105

      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

      pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

      26664

      37775

      2

      frac14 050

      The total active thrust (acting at 25 above the normal) is given by Equation 616

      Pa frac14 1

      2 050 19 7502 frac14 267 kN=m

      Figure Q65

      Lateral earth pressure 39

      Horizontal component

      Ph frac14 267 cos 40 frac14 205 kN=m

      Vertical component

      Pv frac14 267 sin 40 frac14 172 kN=m

      Consider moments about the toe of the wall (Figure Q66) (per m)

      Force (kN) Arm (m) Moment (kN m)

      (1)1

      2 175 650 235 frac14 1337 258 345

      (2) 050 650 235 frac14 764 175 134

      (3)1

      2 070 650 235 frac14 535 127 68

      (4) 100 400 235 frac14 940 200 188

      (5) 1

      2 080 050 235 frac14 47 027 1

      Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

      Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

      Lever arm of base resultant

      M

      Vfrac14 795

      525frac14 151m

      Eccentricity of base resultant

      e frac14 200 151 frac14 049m

      Figure Q66

      40 Lateral earth pressure

      Base pressures (Equation 627)

      p frac14 525

      41 6 049

      4

      frac14 228 kN=m2 and 35 kN=m2

      The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

      The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

      The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

      67

      For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

      Force (kN) Arm (m) Moment (kNm)

      (1)1

      2 027 17 52 frac14 574 183 1050

      (2) 027 17 5 3 frac14 689 500 3445

      (3)1

      2 027 102 32 frac14 124 550 682

      (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

      (5)1

      2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

      (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

      (7) 1

      2 267

      2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

      (8) 2 10ffiffiffiffiffiffiffiffiffi267p

      2 d frac14 163d d2thorn 650 82d2 1060d

      Tie rod force per m frac14 T 0 0

      XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

      d3 thorn 77d2 269d 1438 frac14 0

      d frac14 467m

      Depth of penetration frac14 12d frac14 560m

      Lateral earth pressure 41

      Algebraic sum of forces for d frac14 467m isX

      F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

      T frac14 905 kN=m

      Force in each tie rod frac14 25T frac14 226 kN

      68

      (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

      0 frac14 21 98 frac14 112 kN=m3

      The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

      uC frac14 150

      165 15 98 frac14 134 kN=m2

      The average seepage pressure is

      j frac14 15

      165 98 frac14 09 kN=m3

      Hence

      0 thorn j frac14 112thorn 09 frac14 121 kN=m3

      0 j frac14 112 09 frac14 103 kN=m3

      Figure Q67

      42 Lateral earth pressure

      Consider moments about the anchor point A (per m)

      Force (kN) Arm (m) Moment (kN m)

      (1) 10 026 150 frac14 390 60 2340

      (2)1

      2 026 18 452 frac14 474 15 711

      (3) 026 18 45 105 frac14 2211 825 18240

      (4)1

      2 026 121 1052 frac14 1734 100 17340

      (5)1

      2 134 15 frac14 101 40 404

      (6) 134 30 frac14 402 60 2412

      (7)1

      2 134 60 frac14 402 95 3819

      571 4527(8) Ppm

      115 115PPm

      XM frac14 0

      Ppm frac144527

      115frac14 394 kN=m

      Available passive resistance

      Pp frac14 1

      2 385 103 62 frac14 714 kN=m

      Factor of safety

      Fp frac14 Pp

      Ppm

      frac14 714

      394frac14 18

      Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

      Figure Q68

      Lateral earth pressure 43

      (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

      Consider moments (per m) about the tie point A

      Force (kN) Arm (m)

      (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

      (2)1

      2 033 18 452 frac14 601 15

      (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

      (4)1

      2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

      (5)1

      2 134 15 frac14 101 40

      (6) 134 30 frac14 402 60

      (7)1

      2 134 d frac14 67d d3thorn 75

      (8) 1

      2 30 103 d2 frac141545d2 2d3thorn 75

      Moment (kN m)

      (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

      XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

      d3 thorn 827d2 466d 1518 frac14 0

      By trial

      d frac14 544m

      The minimum depth of embedment required is 544m

      69

      For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

      0 frac14 20 98 frac14 102 kN=m3

      The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

      44 Lateral earth pressure

      uC frac14 147

      173 26 98 frac14 216 kN=m2

      and the average seepage pressure around the wall is

      j frac14 26

      173 98 frac14 15 kN=m3

      Consider moments about the prop (A) (per m)

      Force (kN) Arm (m) Moment (kN m)

      (1)1

      2 03 17 272 frac14 186 020 37

      (2) 03 17 27 53 frac14 730 335 2445

      (3)1

      2 03 (102thorn 15) 532 frac14 493 423 2085

      (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

      (5)1

      2 216 26 frac14 281 243 684

      (6) 216 27 frac14 583 465 2712

      (7)1

      2 216 60 frac14 648 800 5184

      3055(8)

      1

      2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

      Factor of safety

      Fr frac14 6885

      3055frac14 225

      Figure Q69

      Lateral earth pressure 45

      610

      For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

      p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

      Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

      Using the recommendations of Twine and Roscoe

      p frac14 02H frac14 02 19 9 frac14 342 kN=m2

      Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

      611

      frac14 18 kN=m3 0 frac14 34

      H frac14 350m nH frac14 335m mH frac14 185m

      Consider a trial value of F frac14 20 Refer to Figure 635

      0m frac14 tan1tan 34

      20

      frac14 186

      Then

      frac14 45 thorn 0m2frac14 543

      W frac14 1

      2 18 3502 cot 543 frac14 792 kN=m

      Figure Q610

      46 Lateral earth pressure

      P frac14 1

      2 s 3352 frac14 561s kN=m

      U frac14 1

      2 98 1852 cosec 543 frac14 206 kN=m

      Equations 630 and 631 then become

      561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

      792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

      ie

      561s 0616N 405 frac14 0

      792 0857N thorn 563 frac14 0

      N frac14 848

      0857frac14 989 kN=m

      Then

      561s 609 405 frac14 0

      s frac14 649

      561frac14 116 kN=m3

      The calculations for trial values of F of 20 15 and 10 are summarized below

      F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

      20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

      s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

      Figure Q611

      Lateral earth pressure 47

      612

      For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

      45 thorn 0

      2frac14 63

      For the retained material between the surface and a depth of 36m

      Pa frac14 1

      2 030 18 362 frac14 350 kN=m

      Weight of reinforced fill between the surface and a depth of 36m is

      Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

      eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

      Eccentricity of Rv

      e frac14 263 250 frac14 013m

      The average vertical stress at a depth of 36m is

      z frac14 Rv

      L 2efrac14 324

      474frac14 68 kN=m2

      (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

      Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

      Tensile stress in the element frac14 138 103

      65 3frac14 71N=mm2

      Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

      Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

      Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

      The weight of ABC is

      W frac14 1

      2 18 52 265 frac14 124 kN=m

      From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

      48 Lateral earth pressure

      (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

      Tp frac14 032 68 120 065 frac14 170 kN

      Tr frac14 213 420

      418frac14 214 kN

      Again the tensile failure and slipping limit states are satisfied for this element

      Figure Q612

      Lateral earth pressure 49

      Chapter 7

      Consolidation theory

      71

      Total change in thickness

      H frac14 782 602 frac14 180mm

      Average thickness frac14 1530thorn 180

      2frac14 1620mm

      Length of drainage path d frac14 1620

      2frac14 810mm

      Root time plot (Figure Q71a)

      ffiffiffiffiffiffit90p frac14 33

      t90 frac14 109min

      cv frac14 0848d2

      t90frac14 0848 8102

      109 1440 365

      106frac14 27m2=year

      r0 frac14 782 764

      782 602frac14 018

      180frac14 0100

      rp frac14 10eth764 645THORN9eth782 602THORN frac14

      10 119

      9 180frac14 0735

      rs frac14 1 eth0100thorn 0735THORN frac14 0165

      Log time plot (Figure Q71b)

      t50 frac14 26min

      cv frac14 0196d2

      t50frac14 0196 8102

      26 1440 365

      106frac14 26m2=year

      r0 frac14 782 763

      782 602frac14 019

      180frac14 0106

      rp frac14 763 623

      782 602frac14 140

      180frac14 0778

      rs frac14 1 eth0106thorn 0778THORN frac14 0116

      Figure Q71(a)

      Figure Q71(b)

      Final void ratio

      e1 frac14 w1Gs frac14 0232 272 frac14 0631

      e

      Hfrac14 1thorn e0

      H0frac14 1thorn e1 thorne

      H0

      ie

      e

      180frac14 1631thorne

      1710

      e frac14 2936

      1530frac14 0192

      Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

      mv frac14 1

      1thorn e0 e0 e101 00

      frac14 1

      1823 0192

      0107frac14 098m2=MN

      k frac14 cvmvw frac14 265 098 98

      60 1440 365 103frac14 81 1010 m=s

      72

      Using Equation 77 (one-dimensional method)

      sc frac14 e0 e11thorn e0 H

      Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

      Figure Q72

      52 Consolidation theory

      Settlement

      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

      1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

      318

      Notes 5 92y 460thorn 84

      Heave

      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

      1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

      38

      73

      U frac14 f ethTvTHORN frac14 f cvt

      d2

      Hence if cv is constant

      t1

      t2frac14 d

      21

      d22

      where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

      d1 frac14 95mm and d2 frac14 2500mm

      for U frac14 050 t2 frac14 t1 d22

      d21

      frac14 20

      60 24 365 25002

      952frac14 263 years

      for U lt 060 Tv frac14

      4U2 (Equation 724(a))

      t030 frac14 t050 0302

      0502

      frac14 263 036 frac14 095 years

      Consolidation theory 53

      74

      The layer is open

      d frac14 8

      2frac14 4m

      Tv frac14 cvtd2frac14 24 3

      42frac14 0450

      ui frac14 frac14 84 kN=m2

      The excess pore water pressure is given by Equation 721

      ue frac14Xmfrac141mfrac140

      2ui

      Msin

      Mz

      d

      expethM2TvTHORN

      In this case z frac14 d

      sinMz

      d

      frac14 sinM

      where

      M frac14

      23

      25

      2

      M sin M M2Tv exp (M2Tv)

      2thorn1 1110 0329

      3

      21 9993 457 105

      ue frac14 2 84 2

      1 0329 ethother terms negligibleTHORN

      frac14 352 kN=m2

      75

      The layer is open

      d frac14 6

      2frac14 3m

      Tv frac14 cvtd2frac14 10 3

      32frac14 0333

      The layer thickness will be divided into six equal parts ie m frac14 6

      54 Consolidation theory

      For an open layer

      Tv frac14 4n

      m2

      n frac14 0333 62

      4frac14 300

      The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

      ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

      i j

      0 1 2 3 4 5 6 7 8 9 10 11 12

      0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

      The initial and 3-year isochrones are plotted in Figure Q75

      Area under initial isochrone frac14 180 units

      Area under 3-year isochrone frac14 63 units

      The average degree of consolidation is given by Equation 725Thus

      U frac14 1 63

      180frac14 065

      Figure Q75

      Consolidation theory 55

      76

      At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

      0 frac14 2w frac14 2 98 frac14 196 kN=m2

      The final consolidation settlement (one-dimensional method) is

      sc frac14 mv0H frac14 083 196 8 frac14 130mm

      Corrected time t frac14 2 1

      2

      40

      52

      frac14 1615 years

      Tv frac14 cvtd2frac14 44 1615

      42frac14 0444

      From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

      77

      The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

      Figure Q77

      56 Consolidation theory

      Point m n Ir (kNm2) sc (mm)

      13020frac14 15 20

      20frac14 10 0194 (4) 113 124

      260

      20frac14 30

      20

      20frac14 10 0204 (2) 59 65

      360

      20frac14 30

      40

      20frac14 20 0238 (1) 35 38

      430

      20frac14 15

      40

      20frac14 20 0224 (2) 65 72

      Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

      78

      Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

      (a) Immediate settlement

      H

      Bfrac14 30

      35frac14 086

      D

      Bfrac14 2

      35frac14 006

      Figure Q78

      Consolidation theory 57

      From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

      si frac14 130131qB

      Eufrac14 10 032 105 35

      40frac14 30mm

      (b) Consolidation settlement

      Layer z (m) Dz Ic (kNm2) syod (mm)

      1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

      3150

      Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

      Now

      H

      Bfrac14 30

      35frac14 086 and A frac14 065

      from Figure 712 13 frac14 079

      sc frac14 13sod frac14 079 315 frac14 250mm

      Total settlement

      s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

      79

      Without sand drains

      Uv frac14 025

      Tv frac14 0049 ethfrom Figure 718THORN

      t frac14 Tvd2

      cvfrac14 0049 82

      cvWith sand drains

      R frac14 0564S frac14 0564 3 frac14 169m

      n frac14 Rrfrac14 169

      015frac14 113

      Tr frac14 cht

      4R2frac14 ch

      4 1692 0049 82

      cvethand ch frac14 cvTHORN

      frac14 0275

      Ur frac14 073 (from Figure 730)

      58 Consolidation theory

      Using Equation 740

      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

      U frac14 080

      710

      Without sand drains

      Uv frac14 090

      Tv frac14 0848

      t frac14 Tvd2

      cvfrac14 0848 102

      96frac14 88 years

      With sand drains

      R frac14 0564S frac14 0564 4 frac14 226m

      n frac14 Rrfrac14 226

      015frac14 15

      Tr

      Tvfrac14 chcv

      d2

      4R2ethsame tTHORN

      Tr

      Tvfrac14 140

      96 102

      4 2262frac14 714 eth1THORN

      Using Equation 740

      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

      An iterative solution is required using (1) and (2) an initial value ofUv being estimated

      Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

      040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

      Thus

      Uv frac14 0295 and Ur frac14 086

      t frac14 88 00683

      0848frac14 07 years

      Consolidation theory 59

      Chapter 8

      Bearing capacity

      81

      (a) The ultimate bearing capacity is given by Equation 83

      qf frac14 cNc thorn DNq thorn 1

      2BN

      For u frac14 0

      Nc frac14 514 Nq frac14 1 N frac14 0

      qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

      The net ultimate bearing capacity is

      qnf frac14 qf D frac14 540 kN=m2

      The net foundation pressure is

      qn frac14 q D frac14 425

      2 eth21 1THORN frac14 192 kN=m2

      The factor of safety (Equation 86) is

      F frac14 qnfqnfrac14 540

      192frac14 28

      (b) For 0 frac14 28

      Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

      qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

      2 112 2 13

      frac14 260thorn 168thorn 146 frac14 574 kN=m2

      qnf frac14 574 112 frac14 563 kN=m2

      F frac14 563

      192frac14 29

      (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

      82

      For 0 frac14 38

      Nq frac14 49 N frac14 67

      qnf frac14 DethNq 1THORN thorn 1

      2BN ethfrom Equation 83THORN

      frac14 eth18 075 48THORN thorn 1

      2 18 15 67

      frac14 648thorn 905 frac14 1553 kN=m2

      qn frac14 500

      15 eth18 075THORN frac14 320 kN=m2

      F frac14 qnfqnfrac14 1553

      320frac14 48

      0d frac14 tan1tan 38

      125

      frac14 32 therefore Nq frac14 23 and N frac14 25

      Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

      2 18 15 25

      frac14 15eth310thorn 337THORNfrac14 970 kN=m

      Design load (action) Vd frac14 500 kN=m

      The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

      83

      D

      Bfrac14 350

      225frac14 155

      From Figure 85 for a square foundation

      Nc frac14 81

      Bearing capacity 61

      For a rectangular foundation (L frac14 450m B frac14 225m)

      Nc frac14 084thorn 016B

      L

      81 frac14 745

      Using Equation 810

      qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

      For F frac14 3

      qn frac14 1006

      3frac14 335 kN=m2

      q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

      Design load frac14 405 450 225 frac14 4100 kN

      Design undrained strength cud frac14 135

      14frac14 96 kN=m2

      Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

      frac14 7241 kN

      Design load Vd frac14 4100 kN

      Rd gt Vd therefore the bearing resistance limit state is satisfied

      84

      For 0 frac14 40

      Nq frac14 64 N frac14 95

      qnf frac14 DethNq 1THORN thorn 04BN

      (a) Water table 5m below ground level

      qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

      qn frac14 400 17 frac14 383 kN=m2

      F frac14 2686

      383frac14 70

      (b) Water table 1m below ground level (ie at foundation level)

      0 frac14 20 98 frac14 102 kN=m3

      62 Bearing capacity

      qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

      F frac14 2040

      383frac14 53

      (c) Water table at ground level with upward hydraulic gradient 02

      eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

      qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

      F frac14 1296

      392frac14 33

      85

      The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

      Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

      Design value of 0 frac14 tan1tan 39

      125

      frac14 33

      For 0 frac14 33 Nq frac14 26 and N frac14 29

      Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

      Rd gt Vd therefore the bearing resistance limit state is satisfied

      86

      (a) Undrained shear for u frac14 0

      Nc frac14 514 Nq frac14 1 N frac14 0

      qnf frac14 12cuNc

      frac14 12 100 514 frac14 617 kN=m2

      qn frac14 qnfFfrac14 617

      3frac14 206 kN=m2

      q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

      Bearing capacity 63

      Drained shear for 0 frac14 32

      Nq frac14 23 N frac14 25

      0 frac14 21 98 frac14 112 kN=m3

      qnf frac14 0DethNq 1THORN thorn 040BN

      frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

      frac14 694 kN=m2

      q frac14 694

      3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

      Design load frac14 42 227 frac14 3632 kN

      (b) Design undrained strength cud frac14 100

      14frac14 71 kNm2

      Design bearing resistance Rd frac14 12cudNe area

      frac14 12 71 514 42

      frac14 7007 kN

      For drained shear 0d frac14 tan1tan 32

      125

      frac14 26

      Nq frac14 12 N frac14 10

      Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

      (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

      Layer z (m) m n Ir 0 (kNm2) sod (mm)

      1 2 100 0175 0700qn 0182qn

      2 6 033 0044 0176qn 0046qn

      3 10 020 0017 0068qn 0018qn

      0246qn

      Diameter of equivalent circle B frac14 45m

      H

      Bfrac14 12

      45frac14 27 and A frac14 042

      13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

      64 Bearing capacity

      For sc frac14 30mm

      qn frac14 30

      0147frac14 204 kN=m2

      q frac14 204thorn 21 frac14 225 kN=m2

      Design load frac14 42 225 frac14 3600 kN

      The design load is 3600 kN settlement being the limiting criterion

      87

      D

      Bfrac14 8

      4frac14 20

      From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

      F frac14 cuNc

      Dfrac14 40 71

      20 8frac14 18

      88

      Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

      Design value of 0 frac14 tan1tan 38

      125

      frac14 32

      Figure Q86

      Bearing capacity 65

      For 0 frac14 32 Nq frac14 23 and N frac14 25

      Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

      The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

      Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

      For B frac14 250m qn frac14 3750

      2502 17 frac14 583 kN=m2

      From Figure 510 m frac14 n frac14 126

      6frac14 021

      Ir frac14 0019

      Stress increment frac14 4 0019 583 frac14 44 kN=m2

      Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

      The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

      The settlement is less than 20mm therefore the serviceability limit state is satisfied

      89

      Depth (m) N 0v (kNm2) CN N1

      070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

      Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

      (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

      Cw frac14 05thorn 0530

      47

      frac14 082

      66 Bearing capacity

      Thus

      qa frac14 150 082 frac14 120 kN=m2

      (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

      Thus

      qa frac14 90 15 frac14 135 kN=m2

      (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

      Ic frac14 171

      1014frac14 0068

      From Equation 819(a) with s frac14 25mm

      q frac14 25

      3507 0068frac14 150 kN=m2

      810

      Peak value of strain influence factor occurs at a depth of 27m and is given by

      Izp frac14 05thorn 01130

      16 27

      05

      frac14 067

      Refer to Figure Q810

      E frac14 25qc

      Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

      Ez (mm3MN)

      1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

      0203

      C1 frac14 1 0500qnfrac14 1 05 12 16

      130frac14 093

      C2 frac14 1 ethsayTHORN

      s frac14 C1C2qnX Iz

      Ez frac14 093 1 130 0203 frac14 25mm

      Bearing capacity 67

      811

      At pile base level

      cu frac14 220 kN=m2

      qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

      Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

      00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

      qs frac14 00 frac14 07 1226 frac14 86 kN=m2

      Then

      Qf frac14 Abqb thorn Asqs

      frac14

      4 32 1980

      thorn eth 105 139 86THORN

      frac14 13 996thorn 3941 frac14 17 937 kN

      0 01 02 03 04 05 06 07

      0 2 4 6 8 10 12 14

      1

      2

      3

      4

      5

      6

      7

      8

      (1)

      (2)

      (3)

      (4)

      (5)

      qc

      qc

      Iz

      Iz

      (MNm2)

      z (m)

      Figure Q810

      68 Bearing capacity

      Allowable load

      ethaTHORN Qf

      2frac14 17 937

      2frac14 8968 kN

      ethbTHORN Abqb

      3thorn Asqs frac14 13 996

      3thorn 3941 frac14 8606 kN

      ie allowable load frac14 8600 kN

      Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

      According to the limit state method

      Characteristic undrained strength at base level cuk frac14 220

      150kN=m2

      Characteristic base resistance qbk frac14 9cuk frac14 9 220

      150frac14 1320 kN=m2

      Characteristic shaft resistance qsk frac14 00150

      frac14 86

      150frac14 57 kN=m2

      Characteristic base and shaft resistances

      Rbk frac14

      4 32 1320 frac14 9330 kN

      Rsk frac14 105 139 86

      150frac14 2629 kN

      For a bored pile the partial factors are b frac14 160 and s frac14 130

      Design bearing resistance Rcd frac14 9330

      160thorn 2629

      130

      frac14 5831thorn 2022

      frac14 7850 kN

      Adding ethDAb W) the design bearing resistance becomes 9650 kN

      812

      ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

      qs frac14 cu frac14 040 105 frac14 42 kN=m2

      For a single pile

      Qf frac14 Abqb thorn Asqs

      frac14

      4 062 1305

      thorn eth 06 15 42THORN

      frac14 369thorn 1187 frac14 1556 kN

      Bearing capacity 69

      Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

      eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

      (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

      (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

      qbkfrac14 9cuk frac14 9 220

      150frac14 1320 kN=m2

      qskfrac14cuk frac14 040 105

      150frac14 28 kN=m2

      Rbkfrac14

      4 0602 1320 frac14 373 kN

      Rskfrac14 060 15 28 frac14 791 kN

      Rcdfrac14 373

      160thorn 791

      130frac14 233thorn 608 frac14 841 kN

      Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

      Rcd gt Fcd therefore the bearing resistance limit state is satisfied

      (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

      q frac14 21 000

      1762frac14 68 kN=m2

      Immediate settlement

      H

      Bfrac14 15

      176frac14 085

      D

      Bfrac14 13

      176frac14 074

      L

      Bfrac14 1

      Hence from Figure 515

      130 frac14 078 and 131 frac14 041

      70 Bearing capacity

      Thus using Equation 528

      si frac14 078 041 68 176

      65frac14 6mm

      Consolidation settlement

      Layer z (m) Area (m2) (kNm2) mvH (mm)

      1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

      434 (sod)

      Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

      sc frac14 056 434 frac14 24mm

      The total settlement is (6thorn 24) frac14 30mm

      813

      At base level N frac14 26 Then using Equation 830

      qb frac14 40NDb

      Bfrac14 40 26 2

      025frac14 8320 kN=m2

      ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

      Figure Q812

      Bearing capacity 71

      Over the length embedded in sand

      N frac14 21 ie18thorn 24

      2

      Using Equation 831

      qs frac14 2N frac14 2 21 frac14 42 kN=m2

      For a single pile

      Qf frac14 Abqb thorn Asqs

      frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

      For the pile group assuming a group efficiency of 12

      XQf frac14 12 9 604 frac14 6523 kN

      Then the load factor is

      F frac14 6523

      2000thorn 1000frac14 21

      (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

      Characteristic base resistance per unit area qbk frac14 8320

      150frac14 5547 kNm2

      Characteristic shaft resistance per unit area qsk frac14 42

      150frac14 28 kNm2

      Characteristic base and shaft resistances for a single pile

      Rbk frac14 0252 5547 frac14 347 kN

      Rsk frac14 4 025 2 28 frac14 56 kN

      For a driven pile the partial factors are b frac14 s frac14 130

      Design bearing resistance Rcd frac14 347

      130thorn 56

      130frac14 310 kN

      For the pile group Rcd frac14 12 9 310 frac14 3348 kN

      Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

      (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

      From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

      72 Bearing capacity

      N frac14 24thorn 26thorn 34

      3frac14 28

      Ic frac14 171

      2814frac14 0016 ethEquation 818THORN

      s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

      The settlement is less than 20mm therefore the serviceability limit state is satisfied

      814

      Using Equation 841

      Tf frac14 DLcu thorn

      4ethD2 d2THORNcuNc

      frac14 eth 02 5 06 110THORN thorn

      4eth022 012THORN110 9

      frac14 207thorn 23 frac14 230 kN

      Figure Q813

      Bearing capacity 73

      Chapter 9

      Stability of slopes

      91

      Referring to Figure Q91

      W frac14 417 19 frac14 792 kN=m

      Q frac14 20 28 frac14 56 kN=m

      Arc lengthAB frac14

      180 73 90 frac14 115m

      Arc length BC frac14

      180 28 90 frac14 44m

      The factor of safety is given by

      F frac14 rethcuLaTHORNWd1 thornQd2 frac14

      90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

      Depth of tension crack z0 frac14 2cu

      frac14 2 20

      19frac14 21m

      Arc length BD frac14

      180 13

      1

      2 90 frac14 21m

      F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

      The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

      Design resisting moment frac14 rXethcudLaTHORN frac14 90

      14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

      Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

      The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

      92

      u frac14 0

      Depth factor D frac14 11

      9frac14 122

      Using Equation 92 with F frac14 10

      Ns frac14 cu

      FHfrac14 30

      10 19 9frac14 0175

      Hence from Figure 93

      frac14 50

      For F frac14 12

      Ns frac14 30

      12 19 9frac14 0146

      frac14 27

      93

      Refer to Figure Q93

      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

      1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

      74 m

      214 1deg

      213 1deg

      39 m

      WB

      D

      C

      28 m

      21 m

      A

      Q

      Soil (1)Soil (2)

      73deg

      Figure Q91

      Stability of slopes 75

      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

      9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

      599 256 328 1372

      Figure Q93

      76 Stability of slopes

      XW cos frac14 b

      Xh cos frac14 21 2 599 frac14 2516 kN=mX

      W sin frac14 bX

      h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

      Arc length La frac14

      180 57

      1

      2 326 frac14 327m

      The factor of safety is given by

      F frac14 c0La thorn tan0ethW cos ulTHORN

      W sin

      frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

      frac14 091

      According to the limit state method

      0d frac14 tan1tan 32

      125

      frac14 265

      c0 frac14 8

      160frac14 5 kN=m2

      Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

      Design disturbing moment frac14 1075 kN=m

      The design resisting moment is less than the design disturbing moment therefore a slipwill occur

      94

      F frac14 1

      W sin

      Xfc0bthorn ethW ubTHORN tan0g sec

      1thorn ethtan tan0=FTHORN

      c0 frac14 8 kN=m2

      0 frac14 32

      c0b frac14 8 2 frac14 16 kN=m

      W frac14 bh frac14 21 2 h frac14 42h kN=m

      Try F frac14 100

      tan0

      Ffrac14 0625

      Stability of slopes 77

      Values of u are as obtained in Figure Q93

      SliceNo

      h(m)

      W frac14 bh(kNm)

      W sin(kNm)

      ub(kNm)

      c0bthorn (W ub) tan0(kNm)

      sec

      1thorn (tan tan0)FProduct(kNm)

      1 05 21 6 2 8 24 1078 262 13 55 31

      23 33 30 1042 31

      3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

      224 92 72 0931 67

      6 50 210 11 40 100 85 0907 777 55 231 14

      12 58 112 90 0889 80

      8 60 252 1812

      80 114 102 0874 899 63 265 22 99 116 109 0861 94

      10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

      2154 88 116 0853 99

      14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

      1074 1091

      F frac14 1091

      1074frac14 102 (assumed value 100)

      Thus

      F frac14 101

      95

      F frac14 1

      W sin

      XfWeth1 ruTHORN tan0g sec

      1thorn ethtan tan0THORN=F

      0 frac14 33

      ru frac14 020

      W frac14 bh frac14 20 5 h frac14 100h kN=m

      eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

      Try F frac14 110

      tan 0

      Ffrac14 tan 33

      110frac14 0590

      78 Stability of slopes

      Referring to Figure Q95

      SliceNo

      h(m)

      W frac14 bh(kNm)

      W sin(kNm)

      W(1 ru) tan0(kNm)

      sec

      1thorn ( tan tan0)FProduct(kNm)

      1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

      2120 234 0892 209

      4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

      1185 1271

      Figure Q95

      Stability of slopes 79

      F frac14 1271

      1185frac14 107

      The trial value was 110 therefore take F to be 108

      96

      (a) Water table at surface the factor of safety is given by Equation 912

      F frac14 0

      sat

      tan0

      tan

      ptie 15 frac14 92

      19

      tan 36

      tan

      tan frac14 0234

      frac14 13

      Water table well below surface the factor of safety is given by Equation 911

      F frac14 tan0

      tan

      frac14 tan 36

      tan 13

      frac14 31

      (b) 0d frac14 tan1tan 36

      125

      frac14 30

      Depth of potential failure surface frac14 z

      Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

      frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

      frac14 504z kN

      Design disturbing moment per unit area Sd frac14 sat sin cos

      frac14 19 z sin 13 cos 13

      frac14 416z kN

      Rd gtSd therefore the limit state for overall stability is satisfied

      80 Stability of slopes

      • Book Cover
      • Title
      • Contents
      • Basic characteristics of soils
      • Seepage
      • Effective stress
      • Shear strength
      • Stresses and displacements
      • Lateral earth pressure
      • Consolidation theory
      • Bearing capacity
      • Stability of slopes

        First published 1992by E amp FN Spon an imprint of Thomson ProfessionalSecond edition 1997Third edition 200411 New Fetter Lane London EC4P 4EE

        Simultaneously published in the USA and Canadaby Spon Press29 West 35th Street New York NY 10001

        Spon Press is an imprint of the Taylor amp Francis Group

        ordf 1992 1997 2004 RF Craig

        All rights reserved No part of this book may be reprinted or reproducedor utilised in any form or by any electronic mechanical or other meansnow known or hereafter invented including photocopying and recordingor in any information storage or retrieval system without permission inwriting from the publishers

        British Library Cataloguing in Publication DataA catalogue record for this book is availablefrom the British Library

        Library of Congress Cataloging in Publication DataA catalog record for this book has been requested

        ISBN 0ndash415ndash33294ndashX

        This edition published in the Taylor amp Francis e-Library 2004

        (Print edition)

        ISBN 0-203-31104-3 Master e-book ISBN

        ISBN 0-203-67167-8 (Adobe eReader Format)

        collection of thousands of eBooks please go to wwweBookstoretandfcoukrdquoldquoTo purchase your own copy of this or any of Taylor amp Francis or Routledgersquos

        Contents

        1 Basic characteristics of soils 1

        2 Seepage 6

        3 Effective stress 14

        4 Shear strength 22

        5 Stresses and displacements 28

        6 Lateral earth pressure 34

        7 Consolidation theory 50

        8 Bearing capacity 60

        9 Stability of slopes 74

        Authorrsquos note

        In order not to short-circuit the learningprocess it is vital that the reader shouldattempt the problems before referring to thesolutions in this manual

        Chapter 1

        Basic characteristics of soils

        11

        Soil E consists of 98 coarse material (31 gravel size 67 sand size) and 2 finesIt is classified as SW well-graded gravelly SAND or in greater detail well-gradedslightly silty very gravelly SAND

        Soil F consists of 63 coarse material (2 gravel size 61 sand size) and 37non-plastic fines (ie between 35 and 65 fines) therefore the soil is classified as MSsandy SILT

        Soil G consists of 73 fine material (ie between 65 and 100 fines) and 27 sandsize The liquid limit is 32 and the plasticity index is 8 (ie 32 24) plotting marginallybelow the A-line in the ML zone on the plasticity chart Thus the classification is MLSILT (M-SOIL) of low plasticity (The plasticity chart is given in Figure 17)

        Figure Q11

        Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

        12

        From Equation 117

        1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

        191frac14 155

        e frac14 055

        Using Equation 113

        Sr frac14 wGs

        efrac14 0095 270

        055frac14 0466 eth466THORN

        Using Equation 119

        sat frac14 Gs thorn e1thorn e w frac14 325

        155 100 frac14 210Mg=m3

        From Equation 114

        w frac14 e

        Gsfrac14 055

        270frac14 0204 eth204THORN

        13

        Equations similar to 117ndash120 apply in the case of unit weights thus

        d frac14 Gs

        1thorn e w frac14272

        170 98 frac14 157 kN=m3

        sat frac14 Gs thorn e1thorn e w frac14 342

        170 98 frac14 197 kN=m3

        Using Equation 121

        0 frac14 Gs 1

        1thorn e w frac14 172

        170 98 frac14 99 kN=m3

        Using Equation 118a with Srfrac14 075

        frac14 Gs thorn Sre

        1thorn e w frac14 3245

        170 98 frac14 187 kN=m3

        2 Basic characteristics of soils

        Using Equation 113

        w frac14 Sre

        Gsfrac14 075 070

        272frac14 0193 eth193THORN

        The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

        14

        Volume of specimenfrac14

        438276 frac14 86 200mm3

        Bulk density ethTHORN frac14 Mass

        Volumefrac14 1680

        86 200 103frac14 195Mg=m3

        Water content ethwTHORN frac14 1680 1305

        1305frac14 0287 eth287THORN

        From Equation 117

        1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

        195frac14 180

        e frac14 080

        Using Equation 113

        Sr frac14 wGs

        efrac14 0287 273

        080frac14 098 eth98THORN

        15

        Using Equation 124

        d frac14

        1thorn w frac14215

        112frac14 192Mg=m3

        From Equation 117

        1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

        215frac14 138

        e frac14 038

        Using Equation 113

        Sr frac14 wGs

        efrac14 012 265

        038frac14 0837 eth837THORN

        Basic characteristics of soils 3

        Using Equation 115

        Afrac14 e wGs

        1thorn e frac14038 0318

        138frac14 0045 eth45THORN

        The zero air voids dry density is given by Equation 125

        d frac14 Gs

        1thorn wGsw frac14 265

        1thorn eth0135 265THORN 100 frac14 195Mg=m3

        ie a dry density of 200Mgm3 would not be possible

        16

        Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

        (Mgm3) d10(Mgm3)

        2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

        In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

        wopt frac14 15 and dmaxfrac14 183Mg=m3

        Figure Q16

        4 Basic characteristics of soils

        Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

        d5 d10

        respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

        17

        From Equation 120

        e frac14 Gswd 1

        The maximum and minimum values of void ratio are given by

        emax frac14 Gsw

        dmin

        1

        emin frac14 Gswdmax

        1

        From Equation 123

        ID frac14 Gsweth1=dmin 1=dTHORN

        Gsweth1=dmin 1=dmax

        THORN

        frac14 frac121 ethdmin=dTHORN1=dmin

        frac121 ethdmin=dmax

        THORN1=dmin

        frac14 d dmin

        dmax dmin

        dmax

        d

        frac14 172 154

        181 154

        181

        172

        frac14 070 eth70THORN

        Basic characteristics of soils 5

        Chapter 2

        Seepage

        21

        The coefficient of permeability is determined from the equation

        k frac14 23al

        At1log

        h0

        h1

        where

        a frac14

        4 00052 m2 l frac14 02m

        A frac14

        4 012 m2 t1 frac14 3 602 s

        logh0

        h1frac14 log

        100

        035frac14 0456

        k frac14 23 00052 02 0456

        012 3 602frac14 49 108 m=s

        22

        The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

        q frac14 kh Nf

        Ndfrac14 106 400 37

        11frac14 13 106 m3=s per m

        Figure Q22

        23

        The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

        q frac14 kh Nf

        Ndfrac14 5 105 300 35

        9frac14 58 105 m3=s per m

        The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

        Point h (m) h z (m) u frac14 w(h z)(kNm2)

        1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

        eg for Point 1

        h1 frac14 7

        9 300 frac14 233m

        h1 z1 frac14 233 eth250THORN frac14 483m

        Figure Q23

        Seepage 7

        u1 frac14 98 483 frac14 47 kN=m2

        The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

        24

        The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

        q frac14 kh Nf

        Ndfrac14 40 107 550 10

        11frac14 20 106 m3=s per m

        25

        The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

        q frac14 kh Nf

        Ndfrac14 20 106 500 42

        9frac14 47 106 m3=s per m

        Figure Q24

        8 Seepage

        26

        The scale transformation factor in the x direction is given by Equation 221 ie

        xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

        ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

        Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

        k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

        pfrac14

        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

        p 107 frac14 30 107 m=s

        Thus the quantity of seepage is given by

        q frac14 k0h Nf

        Ndfrac14 30 107 1400 325

        12frac14 11 106 m3=s per m

        Figure Q25

        Seepage 9

        27

        The scale transformation factor in the x direction is

        xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

        ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

        Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

        k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

        pfrac14

        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

        p 106 frac14 45 106 m=s

        The focus of the basic parabola is at point A The parabola passes through point Gsuch that

        GC frac14 03HC frac14 03 30 frac14 90m

        Thus the coordinates of G are

        x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

        480 frac14 x0 2002

        4x0

        Figure Q26

        10 Seepage

        Hence

        x0 frac14 20m

        Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

        x frac14 20 z2

        80

        x 20 0 50 100 200 300z 0 400 748 980 1327 1600

        The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

        No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

        q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

        28

        The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

        Figure Q27

        Seepage 11

        q frac14 kh Nf

        Ndfrac14 45 105 28 33

        7

        frac14 59 105 m3=s per m

        29

        The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

        kx frac14 H1k1 thornH2k2

        H1 thornH2frac14 106

        10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

        kz frac14 H1 thornH2

        H1

        k1thornH2

        k2

        frac14 10

        5

        eth2 106THORN thorn5

        eth16 106THORNfrac14 36 106 m=s

        Then the scale transformation factor is given by

        xt frac14 xffiffiffiffiffikz

        pffiffiffiffiffikx

        p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

        Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

        Figure Q28

        12 Seepage

        k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

        qfrac14

        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

        p 106 frac14 57 106 m=s

        Then the quantity of seepage is given by

        q frac14 k0h Nf

        Ndfrac14 57 106 350 56

        11

        frac14 10 105 m3=s per m

        Figure Q29

        Seepage 13

        Chapter 3

        Effective stress

        31

        Buoyant unit weight

        0 frac14 sat w frac14 20 98 frac14 102 kN=m3

        Effective vertical stress

        0v frac14 5 102 frac14 51 kN=m2 or

        Total vertical stress

        v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

        Pore water pressure

        u frac14 7 98 frac14 686 kN=m2

        Effective vertical stress

        0v frac14 v u frac14 1196 686 frac14 51 kN=m2

        32

        Buoyant unit weight

        0 frac14 sat w frac14 20 98 frac14 102 kN=m3

        Effective vertical stress

        0v frac14 5 102 frac14 51 kN=m2 or

        Total vertical stress

        v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

        Pore water pressure

        u frac14 205 98 frac14 2009 kN=m2

        Effective vertical stress

        0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

        33

        At top of the clay

        v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

        u frac14 2 98 frac14 196 kN=m2

        0v frac14 v u frac14 710 196 frac14 514 kN=m2

        Alternatively

        0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

        0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

        At bottom of the clay

        v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

        u frac14 12 98 frac14 1176 kN=m2

        0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

        NB The alternative method of calculation is not applicable because of the artesiancondition

        Figure Q3132

        Effective stress 15

        34

        0 frac14 20 98 frac14 102 kN=m3

        At 8m depth

        0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

        35

        0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

        0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

        Figure Q33

        Figure Q34

        16 Effective stress

        (a) Immediately after WT rise

        At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

        0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

        At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

        0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

        (b) Several years after WT rise

        At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

        At 8m depth

        0v frac14 940 kN=m2 (as above)

        At 12m depth

        0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

        Figure Q35

        Effective stress 17

        36

        Total weight

        ab frac14 210 kN

        Effective weight

        ac frac14 112 kN

        Resultant boundary water force

        be frac14 119 kN

        Seepage force

        ce frac14 34 kN

        Resultant body force

        ae frac14 99 kN eth73 to horizontalTHORN

        (Refer to Figure Q36)

        Figure Q36

        18 Effective stress

        37

        Situation (1)(a)

        frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

        u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

        0 frac14 u frac14 694 392 frac14 302 kN=m2

        (b)

        i frac14 2

        4frac14 05

        j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

        Situation (2)(a)

        frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

        u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

        0 frac14 u frac14 498 392 frac14 106 kN=m2

        (b)

        i frac14 2

        4frac14 05

        j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

        38

        The flow net is drawn in Figure Q24

        Loss in total head between adjacent equipotentials

        h frac14 550

        Ndfrac14 550

        11frac14 050m

        Exit hydraulic gradient

        ie frac14 h

        sfrac14 050

        070frac14 071

        Effective stress 19

        The critical hydraulic gradient is given by Equation 39

        ic frac14 0

        wfrac14 102

        98frac14 104

        Therefore factor of safety against lsquoboilingrsquo (Equation 311)

        F frac14 iciefrac14 104

        071frac14 15

        Total head at C

        hC frac14 nd

        Ndh frac14 24

        11 550 frac14 120m

        Elevation head at C

        zC frac14 250m

        Pore water pressure at C

        uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

        Therefore effective vertical stress at C

        0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

        For point D

        hD frac14 73

        11 550 frac14 365m

        zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

        0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

        39

        The flow net is drawn in Figure Q25

        For a soil prism 150 300m adjacent to the piling

        hm frac14 26

        9 500 frac14 145m

        20 Effective stress

        Factor of safety against lsquoheavingrsquo (Equation 310)

        F frac14 ic

        imfrac14 0d

        whmfrac14 97 300

        98 145frac14 20

        With a filter

        F frac14 0d thorn wwhm

        3 frac14 eth97 300THORN thorn w98 145

        w frac14 135 kN=m2

        Depth of filterfrac14 13521frac14 065m (if above water level)

        Effective stress 21

        Chapter 4

        Shear strength

        41

        frac14 295 kN=m2

        u frac14 120 kN=m2

        0 frac14 u frac14 295 120 frac14 175 kN=m2

        f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

        42

        03 (kNm2) 1 3 (kNm2) 01 (kNm2)

        100 452 552200 908 1108400 1810 2210800 3624 4424

        The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

        Figure Q42

        43

        3 (kNm2) 1 3 (kNm2) 1 (kNm2)

        200 222 422400 218 618600 220 820

        The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

        44

        The modified shear strength parameters are

        0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

        a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

        The coordinates of the stress point representing failure conditions in the test are

        1

        2eth1 2THORN frac14 1

        2 170 frac14 85 kN=m2

        1

        2eth1 thorn 3THORN frac14 1

        2eth270thorn 100THORN frac14 185 kN=m2

        The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

        uf frac14 36 kN=m2

        Figure Q43

        Figure Q44

        Shear strength 23

        45

        3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

        150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

        The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

        The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

        46

        03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

        200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

        The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

        Figure Q45

        24 Shear strength

        47

        The torque required to produce shear failure is given by

        T frac14 dh cud

        2thorn 2

        Z d=2

        0

        2r drcur

        frac14 cud2h

        2thorn 4cu

        Z d=2

        0

        r2dr

        frac14 cud2h

        2thorn d

        3

        6

        Then

        35 frac14 cu52 10

        2thorn 53

        6

        103

        cu frac14 76 kN=m3

        400

        0 400 800 1200 1600

        τ (k

        Nm

        2 )

        σprime (kNm2)

        34deg

        315deg29deg

        (a)

        (b)

        0 400

        400

        800 1200 1600

        Failure envelope

        300 500

        σprime (kNm2)

        τ (k

        Nm

        2 )

        20 (kNm2)

        31deg

        Figure Q46

        Shear strength 25

        48

        The relevant stress values are calculated as follows

        3 frac14 600 kN=m2

        1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

        2(1 3) 0 40 79 107 139 159

        1

        2(01 thorn 03) 400 411 402 389 351 326

        1

        2(1 thorn 3) 600 640 679 707 739 759

        The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

        Af frac14 433 200

        319frac14 073

        49

        B frac14 u33

        frac14 144

        350 200frac14 096

        a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

        0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

        10 283 65 023

        Figure Q48

        26 Shear strength

        The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

        Figure Q49

        Shear strength 27

        Chapter 5

        Stresses and displacements

        51

        Vertical stress is given by

        z frac14 Qz2Ip frac14 5000

        52Ip

        Values of Ip are obtained from Table 51

        r (m) rz Ip z (kNm2)

        0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

        10 20 0009 2

        The variation of z with radial distance (r) is plotted in Figure Q51

        Figure Q51

        52

        Below the centre load (Figure Q52)

        r

        zfrac14 0 for the 7500-kN load

        Ip frac14 0478

        r

        zfrac14 5

        4frac14 125 for the 10 000- and 9000-kN loads

        Ip frac14 0045

        Then

        z frac14X Q

        z2Ip

        frac14 7500 0478

        42thorn 10 000 0045

        42thorn 9000 0045

        42

        frac14 224thorn 28thorn 25 frac14 277 kN=m2

        53

        The vertical stress under a corner of a rectangular area is given by

        z frac14 qIr

        where values of Ir are obtained from Figure 510 In this case

        z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

        z

        Figure Q52

        Stresses and displacements 29

        z (m) m n Ir z (kNm2)

        0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

        10 010 0005 5

        z is plotted against z in Figure Q53

        54

        (a)

        m frac14 125

        12frac14 104

        n frac14 18

        12frac14 150

        From Figure 510 Irfrac14 0196

        z frac14 2 175 0196 frac14 68 kN=m2

        Figure Q53

        30 Stresses and displacements

        (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

        z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

        55

        Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

        Px frac14 2Q

        1

        m2 thorn 1frac14 2 150

        125frac14 76 kN=m

        Equation 517 is used to obtain the pressure distribution

        px frac14 4Q

        h

        m2n

        ethm2 thorn n2THORN2 frac14150

        m2n

        ethm2 thorn n2THORN2 ethkN=m2THORN

        Figure Q54

        Stresses and displacements 31

        n m2n

        (m2 thorn n2)2

        px(kNm2)

        0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

        The pressure distribution is plotted in Figure Q55

        56

        H

        Bfrac14 10

        2frac14 5

        L

        Bfrac14 4

        2frac14 2

        D

        Bfrac14 1

        2frac14 05

        Hence from Figure 515

        131 frac14 082

        130 frac14 094

        Figure Q55

        32 Stresses and displacements

        The immediate settlement is given by Equation 528

        si frac14 130131qB

        Eu

        frac14 094 082 200 2

        45frac14 7mm

        Stresses and displacements 33

        Chapter 6

        Lateral earth pressure

        61

        For 0 frac14 37 the active pressure coefficient is given by

        Ka frac14 1 sin 37

        1thorn sin 37frac14 025

        The total active thrust (Equation 66a with c0 frac14 0) is

        Pa frac14 1

        2KaH

        2 frac14 1

        2 025 17 62 frac14 765 kN=m

        If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

        K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

        and the thrust on the wall is

        P0 frac14 1

        2K0H

        2 frac14 1

        2 040 17 62 frac14 122 kN=m

        62

        The active pressure coefficients for the three soil types are as follows

        Ka1 frac141 sin 35

        1thorn sin 35frac14 0271

        Ka2 frac141 sin 27

        1thorn sin 27frac14 0375

        ffiffiffiffiffiffiffiKa2

        p frac14 0613

        Ka3 frac141 sin 42

        1thorn sin 42frac14 0198

        Distribution of active pressure (plotted in Figure Q62)

        Depth (m) Soil Active pressure (kNm2)

        3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

        12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

        At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

        Total thrust frac14 571 kNm

        Point of application is (4893571) m from the top of the wall ie 857m

        Force (kN) Arm (m) Moment (kN m)

        (1)1

        2 0271 16 32 frac14 195 20 390

        (2) 0271 16 3 2 frac14 260 40 1040

        (3)1

        2 0271 92 22 frac14 50 433 217

        (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

        (5)1

        2 0375 102 32 frac14 172 70 1204

        (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

        (7)1

        2 0198 112 42 frac14 177 1067 1889

        (8)1

        2 98 92 frac14 3969 90 35721

        5713 48934

        Figure Q62

        Lateral earth pressure 35

        63

        (a) For u frac14 0 Ka frac14 Kp frac14 1

        Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

        frac14 245

        At the lower end of the piling

        pa frac14 Kaqthorn Kasatz Kaccu

        frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

        frac14 115 kN=m2

        pp frac14 Kpsatzthorn Kpccu

        frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

        frac14 202 kN=m2

        (b) For 0 frac14 26 and frac14 1

        20

        Ka frac14 035

        Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

        pfrac14 145 ethEquation 619THORN

        Kp frac14 37

        Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

        pfrac14 47 ethEquation 624THORN

        At the lower end of the piling

        pa frac14 Kaqthorn Ka0z Kacc

        0

        frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

        frac14 187 kN=m2

        pp frac14 Kp0zthorn Kpcc

        0

        frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

        frac14 198 kN=m2

        36 Lateral earth pressure

        64

        (a) For 0 frac14 38 Ka frac14 024

        0 frac14 20 98 frac14 102 kN=m3

        The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

        Force (kN) Arm (m) Moment (kN m)

        (1) 024 10 66 frac14 159 33 525

        (2)1

        2 024 17 392 frac14 310 400 1240

        (3) 024 17 39 27 frac14 430 135 580

        (4)1

        2 024 102 272 frac14 89 090 80

        (5)1

        2 98 272 frac14 357 090 321

        Hfrac14 1345 MH frac14 2746

        (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

        (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

        XM frac14MV MH frac14 7790 kNm

        Lever arm of base resultant

        M

        Vfrac14 779

        488frac14 160

        Eccentricity of base resultant

        e frac14 200 160 frac14 040m

        39 m

        27 m

        40 m

        04 m

        04 m

        26 m

        (7)

        (9)

        (1)(2)

        (3)

        (4)

        (5)

        (8)(6)

        (10)

        WT

        10 kNm2

        Hydrostatic

        Figure Q64

        Lateral earth pressure 37

        Base pressures (Equation 627)

        p frac14 VB

        1 6e

        B

        frac14 488

        4eth1 060THORN

        frac14 195 kN=m2 and 49 kN=m2

        Factor of safety against sliding (Equation 628)

        F frac14 V tan

        Hfrac14 488 tan 25

        1345frac14 17

        (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

        Hfrac14 1633 kN

        V frac14 4879 kN

        MH frac14 3453 kNm

        MV frac14 10536 kNm

        The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

        65

        For 0 frac14 36 Ka frac14 026 and Kp frac14 385

        Kp

        Ffrac14 385

        2

        0 frac14 20 98 frac14 102 kN=m3

        The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

        Force (kN) Arm (m) Moment (kN m)

        (1)1

        2 026 17 452 frac14 448 dthorn 15 448dthorn 672

        (2) 026 17 45 d frac14 199d d2 995d2

        (3)1

        2 026 102 d2 frac14 133d2 d3 044d3

        (4)1

        2 385

        2 17 152 frac14 368 dthorn 05 368d 184

        (5)385

        2 17 15 d frac14 491d d2 2455d2

        (6)1

        2 385

        2 102 d2 frac14 982d2 d3 327d3

        38 Lateral earth pressure

        XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

        d3 thorn 516d2 283d 1724 frac14 0

        d frac14 179m

        Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

        Over additional 20 embedded depth

        pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

        Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

        66

        The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

        Ka frac14 sin 69=sin 105

        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

        pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

        26664

        37775

        2

        frac14 050

        The total active thrust (acting at 25 above the normal) is given by Equation 616

        Pa frac14 1

        2 050 19 7502 frac14 267 kN=m

        Figure Q65

        Lateral earth pressure 39

        Horizontal component

        Ph frac14 267 cos 40 frac14 205 kN=m

        Vertical component

        Pv frac14 267 sin 40 frac14 172 kN=m

        Consider moments about the toe of the wall (Figure Q66) (per m)

        Force (kN) Arm (m) Moment (kN m)

        (1)1

        2 175 650 235 frac14 1337 258 345

        (2) 050 650 235 frac14 764 175 134

        (3)1

        2 070 650 235 frac14 535 127 68

        (4) 100 400 235 frac14 940 200 188

        (5) 1

        2 080 050 235 frac14 47 027 1

        Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

        Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

        Lever arm of base resultant

        M

        Vfrac14 795

        525frac14 151m

        Eccentricity of base resultant

        e frac14 200 151 frac14 049m

        Figure Q66

        40 Lateral earth pressure

        Base pressures (Equation 627)

        p frac14 525

        41 6 049

        4

        frac14 228 kN=m2 and 35 kN=m2

        The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

        The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

        The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

        67

        For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

        Force (kN) Arm (m) Moment (kNm)

        (1)1

        2 027 17 52 frac14 574 183 1050

        (2) 027 17 5 3 frac14 689 500 3445

        (3)1

        2 027 102 32 frac14 124 550 682

        (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

        (5)1

        2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

        (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

        (7) 1

        2 267

        2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

        (8) 2 10ffiffiffiffiffiffiffiffiffi267p

        2 d frac14 163d d2thorn 650 82d2 1060d

        Tie rod force per m frac14 T 0 0

        XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

        d3 thorn 77d2 269d 1438 frac14 0

        d frac14 467m

        Depth of penetration frac14 12d frac14 560m

        Lateral earth pressure 41

        Algebraic sum of forces for d frac14 467m isX

        F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

        T frac14 905 kN=m

        Force in each tie rod frac14 25T frac14 226 kN

        68

        (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

        0 frac14 21 98 frac14 112 kN=m3

        The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

        uC frac14 150

        165 15 98 frac14 134 kN=m2

        The average seepage pressure is

        j frac14 15

        165 98 frac14 09 kN=m3

        Hence

        0 thorn j frac14 112thorn 09 frac14 121 kN=m3

        0 j frac14 112 09 frac14 103 kN=m3

        Figure Q67

        42 Lateral earth pressure

        Consider moments about the anchor point A (per m)

        Force (kN) Arm (m) Moment (kN m)

        (1) 10 026 150 frac14 390 60 2340

        (2)1

        2 026 18 452 frac14 474 15 711

        (3) 026 18 45 105 frac14 2211 825 18240

        (4)1

        2 026 121 1052 frac14 1734 100 17340

        (5)1

        2 134 15 frac14 101 40 404

        (6) 134 30 frac14 402 60 2412

        (7)1

        2 134 60 frac14 402 95 3819

        571 4527(8) Ppm

        115 115PPm

        XM frac14 0

        Ppm frac144527

        115frac14 394 kN=m

        Available passive resistance

        Pp frac14 1

        2 385 103 62 frac14 714 kN=m

        Factor of safety

        Fp frac14 Pp

        Ppm

        frac14 714

        394frac14 18

        Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

        Figure Q68

        Lateral earth pressure 43

        (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

        Consider moments (per m) about the tie point A

        Force (kN) Arm (m)

        (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

        (2)1

        2 033 18 452 frac14 601 15

        (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

        (4)1

        2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

        (5)1

        2 134 15 frac14 101 40

        (6) 134 30 frac14 402 60

        (7)1

        2 134 d frac14 67d d3thorn 75

        (8) 1

        2 30 103 d2 frac141545d2 2d3thorn 75

        Moment (kN m)

        (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

        XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

        d3 thorn 827d2 466d 1518 frac14 0

        By trial

        d frac14 544m

        The minimum depth of embedment required is 544m

        69

        For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

        0 frac14 20 98 frac14 102 kN=m3

        The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

        44 Lateral earth pressure

        uC frac14 147

        173 26 98 frac14 216 kN=m2

        and the average seepage pressure around the wall is

        j frac14 26

        173 98 frac14 15 kN=m3

        Consider moments about the prop (A) (per m)

        Force (kN) Arm (m) Moment (kN m)

        (1)1

        2 03 17 272 frac14 186 020 37

        (2) 03 17 27 53 frac14 730 335 2445

        (3)1

        2 03 (102thorn 15) 532 frac14 493 423 2085

        (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

        (5)1

        2 216 26 frac14 281 243 684

        (6) 216 27 frac14 583 465 2712

        (7)1

        2 216 60 frac14 648 800 5184

        3055(8)

        1

        2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

        Factor of safety

        Fr frac14 6885

        3055frac14 225

        Figure Q69

        Lateral earth pressure 45

        610

        For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

        p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

        Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

        Using the recommendations of Twine and Roscoe

        p frac14 02H frac14 02 19 9 frac14 342 kN=m2

        Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

        611

        frac14 18 kN=m3 0 frac14 34

        H frac14 350m nH frac14 335m mH frac14 185m

        Consider a trial value of F frac14 20 Refer to Figure 635

        0m frac14 tan1tan 34

        20

        frac14 186

        Then

        frac14 45 thorn 0m2frac14 543

        W frac14 1

        2 18 3502 cot 543 frac14 792 kN=m

        Figure Q610

        46 Lateral earth pressure

        P frac14 1

        2 s 3352 frac14 561s kN=m

        U frac14 1

        2 98 1852 cosec 543 frac14 206 kN=m

        Equations 630 and 631 then become

        561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

        792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

        ie

        561s 0616N 405 frac14 0

        792 0857N thorn 563 frac14 0

        N frac14 848

        0857frac14 989 kN=m

        Then

        561s 609 405 frac14 0

        s frac14 649

        561frac14 116 kN=m3

        The calculations for trial values of F of 20 15 and 10 are summarized below

        F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

        20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

        s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

        Figure Q611

        Lateral earth pressure 47

        612

        For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

        45 thorn 0

        2frac14 63

        For the retained material between the surface and a depth of 36m

        Pa frac14 1

        2 030 18 362 frac14 350 kN=m

        Weight of reinforced fill between the surface and a depth of 36m is

        Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

        eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

        Eccentricity of Rv

        e frac14 263 250 frac14 013m

        The average vertical stress at a depth of 36m is

        z frac14 Rv

        L 2efrac14 324

        474frac14 68 kN=m2

        (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

        Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

        Tensile stress in the element frac14 138 103

        65 3frac14 71N=mm2

        Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

        Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

        Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

        The weight of ABC is

        W frac14 1

        2 18 52 265 frac14 124 kN=m

        From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

        48 Lateral earth pressure

        (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

        Tp frac14 032 68 120 065 frac14 170 kN

        Tr frac14 213 420

        418frac14 214 kN

        Again the tensile failure and slipping limit states are satisfied for this element

        Figure Q612

        Lateral earth pressure 49

        Chapter 7

        Consolidation theory

        71

        Total change in thickness

        H frac14 782 602 frac14 180mm

        Average thickness frac14 1530thorn 180

        2frac14 1620mm

        Length of drainage path d frac14 1620

        2frac14 810mm

        Root time plot (Figure Q71a)

        ffiffiffiffiffiffit90p frac14 33

        t90 frac14 109min

        cv frac14 0848d2

        t90frac14 0848 8102

        109 1440 365

        106frac14 27m2=year

        r0 frac14 782 764

        782 602frac14 018

        180frac14 0100

        rp frac14 10eth764 645THORN9eth782 602THORN frac14

        10 119

        9 180frac14 0735

        rs frac14 1 eth0100thorn 0735THORN frac14 0165

        Log time plot (Figure Q71b)

        t50 frac14 26min

        cv frac14 0196d2

        t50frac14 0196 8102

        26 1440 365

        106frac14 26m2=year

        r0 frac14 782 763

        782 602frac14 019

        180frac14 0106

        rp frac14 763 623

        782 602frac14 140

        180frac14 0778

        rs frac14 1 eth0106thorn 0778THORN frac14 0116

        Figure Q71(a)

        Figure Q71(b)

        Final void ratio

        e1 frac14 w1Gs frac14 0232 272 frac14 0631

        e

        Hfrac14 1thorn e0

        H0frac14 1thorn e1 thorne

        H0

        ie

        e

        180frac14 1631thorne

        1710

        e frac14 2936

        1530frac14 0192

        Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

        mv frac14 1

        1thorn e0 e0 e101 00

        frac14 1

        1823 0192

        0107frac14 098m2=MN

        k frac14 cvmvw frac14 265 098 98

        60 1440 365 103frac14 81 1010 m=s

        72

        Using Equation 77 (one-dimensional method)

        sc frac14 e0 e11thorn e0 H

        Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

        Figure Q72

        52 Consolidation theory

        Settlement

        Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

        1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

        318

        Notes 5 92y 460thorn 84

        Heave

        Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

        1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

        38

        73

        U frac14 f ethTvTHORN frac14 f cvt

        d2

        Hence if cv is constant

        t1

        t2frac14 d

        21

        d22

        where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

        d1 frac14 95mm and d2 frac14 2500mm

        for U frac14 050 t2 frac14 t1 d22

        d21

        frac14 20

        60 24 365 25002

        952frac14 263 years

        for U lt 060 Tv frac14

        4U2 (Equation 724(a))

        t030 frac14 t050 0302

        0502

        frac14 263 036 frac14 095 years

        Consolidation theory 53

        74

        The layer is open

        d frac14 8

        2frac14 4m

        Tv frac14 cvtd2frac14 24 3

        42frac14 0450

        ui frac14 frac14 84 kN=m2

        The excess pore water pressure is given by Equation 721

        ue frac14Xmfrac141mfrac140

        2ui

        Msin

        Mz

        d

        expethM2TvTHORN

        In this case z frac14 d

        sinMz

        d

        frac14 sinM

        where

        M frac14

        23

        25

        2

        M sin M M2Tv exp (M2Tv)

        2thorn1 1110 0329

        3

        21 9993 457 105

        ue frac14 2 84 2

        1 0329 ethother terms negligibleTHORN

        frac14 352 kN=m2

        75

        The layer is open

        d frac14 6

        2frac14 3m

        Tv frac14 cvtd2frac14 10 3

        32frac14 0333

        The layer thickness will be divided into six equal parts ie m frac14 6

        54 Consolidation theory

        For an open layer

        Tv frac14 4n

        m2

        n frac14 0333 62

        4frac14 300

        The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

        ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

        i j

        0 1 2 3 4 5 6 7 8 9 10 11 12

        0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

        The initial and 3-year isochrones are plotted in Figure Q75

        Area under initial isochrone frac14 180 units

        Area under 3-year isochrone frac14 63 units

        The average degree of consolidation is given by Equation 725Thus

        U frac14 1 63

        180frac14 065

        Figure Q75

        Consolidation theory 55

        76

        At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

        0 frac14 2w frac14 2 98 frac14 196 kN=m2

        The final consolidation settlement (one-dimensional method) is

        sc frac14 mv0H frac14 083 196 8 frac14 130mm

        Corrected time t frac14 2 1

        2

        40

        52

        frac14 1615 years

        Tv frac14 cvtd2frac14 44 1615

        42frac14 0444

        From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

        77

        The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

        Figure Q77

        56 Consolidation theory

        Point m n Ir (kNm2) sc (mm)

        13020frac14 15 20

        20frac14 10 0194 (4) 113 124

        260

        20frac14 30

        20

        20frac14 10 0204 (2) 59 65

        360

        20frac14 30

        40

        20frac14 20 0238 (1) 35 38

        430

        20frac14 15

        40

        20frac14 20 0224 (2) 65 72

        Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

        78

        Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

        (a) Immediate settlement

        H

        Bfrac14 30

        35frac14 086

        D

        Bfrac14 2

        35frac14 006

        Figure Q78

        Consolidation theory 57

        From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

        si frac14 130131qB

        Eufrac14 10 032 105 35

        40frac14 30mm

        (b) Consolidation settlement

        Layer z (m) Dz Ic (kNm2) syod (mm)

        1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

        3150

        Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

        Now

        H

        Bfrac14 30

        35frac14 086 and A frac14 065

        from Figure 712 13 frac14 079

        sc frac14 13sod frac14 079 315 frac14 250mm

        Total settlement

        s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

        79

        Without sand drains

        Uv frac14 025

        Tv frac14 0049 ethfrom Figure 718THORN

        t frac14 Tvd2

        cvfrac14 0049 82

        cvWith sand drains

        R frac14 0564S frac14 0564 3 frac14 169m

        n frac14 Rrfrac14 169

        015frac14 113

        Tr frac14 cht

        4R2frac14 ch

        4 1692 0049 82

        cvethand ch frac14 cvTHORN

        frac14 0275

        Ur frac14 073 (from Figure 730)

        58 Consolidation theory

        Using Equation 740

        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

        U frac14 080

        710

        Without sand drains

        Uv frac14 090

        Tv frac14 0848

        t frac14 Tvd2

        cvfrac14 0848 102

        96frac14 88 years

        With sand drains

        R frac14 0564S frac14 0564 4 frac14 226m

        n frac14 Rrfrac14 226

        015frac14 15

        Tr

        Tvfrac14 chcv

        d2

        4R2ethsame tTHORN

        Tr

        Tvfrac14 140

        96 102

        4 2262frac14 714 eth1THORN

        Using Equation 740

        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

        An iterative solution is required using (1) and (2) an initial value ofUv being estimated

        Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

        040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

        Thus

        Uv frac14 0295 and Ur frac14 086

        t frac14 88 00683

        0848frac14 07 years

        Consolidation theory 59

        Chapter 8

        Bearing capacity

        81

        (a) The ultimate bearing capacity is given by Equation 83

        qf frac14 cNc thorn DNq thorn 1

        2BN

        For u frac14 0

        Nc frac14 514 Nq frac14 1 N frac14 0

        qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

        The net ultimate bearing capacity is

        qnf frac14 qf D frac14 540 kN=m2

        The net foundation pressure is

        qn frac14 q D frac14 425

        2 eth21 1THORN frac14 192 kN=m2

        The factor of safety (Equation 86) is

        F frac14 qnfqnfrac14 540

        192frac14 28

        (b) For 0 frac14 28

        Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

        qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

        2 112 2 13

        frac14 260thorn 168thorn 146 frac14 574 kN=m2

        qnf frac14 574 112 frac14 563 kN=m2

        F frac14 563

        192frac14 29

        (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

        82

        For 0 frac14 38

        Nq frac14 49 N frac14 67

        qnf frac14 DethNq 1THORN thorn 1

        2BN ethfrom Equation 83THORN

        frac14 eth18 075 48THORN thorn 1

        2 18 15 67

        frac14 648thorn 905 frac14 1553 kN=m2

        qn frac14 500

        15 eth18 075THORN frac14 320 kN=m2

        F frac14 qnfqnfrac14 1553

        320frac14 48

        0d frac14 tan1tan 38

        125

        frac14 32 therefore Nq frac14 23 and N frac14 25

        Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

        2 18 15 25

        frac14 15eth310thorn 337THORNfrac14 970 kN=m

        Design load (action) Vd frac14 500 kN=m

        The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

        83

        D

        Bfrac14 350

        225frac14 155

        From Figure 85 for a square foundation

        Nc frac14 81

        Bearing capacity 61

        For a rectangular foundation (L frac14 450m B frac14 225m)

        Nc frac14 084thorn 016B

        L

        81 frac14 745

        Using Equation 810

        qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

        For F frac14 3

        qn frac14 1006

        3frac14 335 kN=m2

        q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

        Design load frac14 405 450 225 frac14 4100 kN

        Design undrained strength cud frac14 135

        14frac14 96 kN=m2

        Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

        frac14 7241 kN

        Design load Vd frac14 4100 kN

        Rd gt Vd therefore the bearing resistance limit state is satisfied

        84

        For 0 frac14 40

        Nq frac14 64 N frac14 95

        qnf frac14 DethNq 1THORN thorn 04BN

        (a) Water table 5m below ground level

        qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

        qn frac14 400 17 frac14 383 kN=m2

        F frac14 2686

        383frac14 70

        (b) Water table 1m below ground level (ie at foundation level)

        0 frac14 20 98 frac14 102 kN=m3

        62 Bearing capacity

        qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

        F frac14 2040

        383frac14 53

        (c) Water table at ground level with upward hydraulic gradient 02

        eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

        qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

        F frac14 1296

        392frac14 33

        85

        The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

        Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

        Design value of 0 frac14 tan1tan 39

        125

        frac14 33

        For 0 frac14 33 Nq frac14 26 and N frac14 29

        Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

        Rd gt Vd therefore the bearing resistance limit state is satisfied

        86

        (a) Undrained shear for u frac14 0

        Nc frac14 514 Nq frac14 1 N frac14 0

        qnf frac14 12cuNc

        frac14 12 100 514 frac14 617 kN=m2

        qn frac14 qnfFfrac14 617

        3frac14 206 kN=m2

        q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

        Bearing capacity 63

        Drained shear for 0 frac14 32

        Nq frac14 23 N frac14 25

        0 frac14 21 98 frac14 112 kN=m3

        qnf frac14 0DethNq 1THORN thorn 040BN

        frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

        frac14 694 kN=m2

        q frac14 694

        3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

        Design load frac14 42 227 frac14 3632 kN

        (b) Design undrained strength cud frac14 100

        14frac14 71 kNm2

        Design bearing resistance Rd frac14 12cudNe area

        frac14 12 71 514 42

        frac14 7007 kN

        For drained shear 0d frac14 tan1tan 32

        125

        frac14 26

        Nq frac14 12 N frac14 10

        Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

        (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

        Layer z (m) m n Ir 0 (kNm2) sod (mm)

        1 2 100 0175 0700qn 0182qn

        2 6 033 0044 0176qn 0046qn

        3 10 020 0017 0068qn 0018qn

        0246qn

        Diameter of equivalent circle B frac14 45m

        H

        Bfrac14 12

        45frac14 27 and A frac14 042

        13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

        64 Bearing capacity

        For sc frac14 30mm

        qn frac14 30

        0147frac14 204 kN=m2

        q frac14 204thorn 21 frac14 225 kN=m2

        Design load frac14 42 225 frac14 3600 kN

        The design load is 3600 kN settlement being the limiting criterion

        87

        D

        Bfrac14 8

        4frac14 20

        From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

        F frac14 cuNc

        Dfrac14 40 71

        20 8frac14 18

        88

        Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

        Design value of 0 frac14 tan1tan 38

        125

        frac14 32

        Figure Q86

        Bearing capacity 65

        For 0 frac14 32 Nq frac14 23 and N frac14 25

        Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

        The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

        Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

        For B frac14 250m qn frac14 3750

        2502 17 frac14 583 kN=m2

        From Figure 510 m frac14 n frac14 126

        6frac14 021

        Ir frac14 0019

        Stress increment frac14 4 0019 583 frac14 44 kN=m2

        Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

        The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

        The settlement is less than 20mm therefore the serviceability limit state is satisfied

        89

        Depth (m) N 0v (kNm2) CN N1

        070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

        Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

        (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

        Cw frac14 05thorn 0530

        47

        frac14 082

        66 Bearing capacity

        Thus

        qa frac14 150 082 frac14 120 kN=m2

        (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

        Thus

        qa frac14 90 15 frac14 135 kN=m2

        (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

        Ic frac14 171

        1014frac14 0068

        From Equation 819(a) with s frac14 25mm

        q frac14 25

        3507 0068frac14 150 kN=m2

        810

        Peak value of strain influence factor occurs at a depth of 27m and is given by

        Izp frac14 05thorn 01130

        16 27

        05

        frac14 067

        Refer to Figure Q810

        E frac14 25qc

        Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

        Ez (mm3MN)

        1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

        0203

        C1 frac14 1 0500qnfrac14 1 05 12 16

        130frac14 093

        C2 frac14 1 ethsayTHORN

        s frac14 C1C2qnX Iz

        Ez frac14 093 1 130 0203 frac14 25mm

        Bearing capacity 67

        811

        At pile base level

        cu frac14 220 kN=m2

        qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

        Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

        00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

        qs frac14 00 frac14 07 1226 frac14 86 kN=m2

        Then

        Qf frac14 Abqb thorn Asqs

        frac14

        4 32 1980

        thorn eth 105 139 86THORN

        frac14 13 996thorn 3941 frac14 17 937 kN

        0 01 02 03 04 05 06 07

        0 2 4 6 8 10 12 14

        1

        2

        3

        4

        5

        6

        7

        8

        (1)

        (2)

        (3)

        (4)

        (5)

        qc

        qc

        Iz

        Iz

        (MNm2)

        z (m)

        Figure Q810

        68 Bearing capacity

        Allowable load

        ethaTHORN Qf

        2frac14 17 937

        2frac14 8968 kN

        ethbTHORN Abqb

        3thorn Asqs frac14 13 996

        3thorn 3941 frac14 8606 kN

        ie allowable load frac14 8600 kN

        Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

        According to the limit state method

        Characteristic undrained strength at base level cuk frac14 220

        150kN=m2

        Characteristic base resistance qbk frac14 9cuk frac14 9 220

        150frac14 1320 kN=m2

        Characteristic shaft resistance qsk frac14 00150

        frac14 86

        150frac14 57 kN=m2

        Characteristic base and shaft resistances

        Rbk frac14

        4 32 1320 frac14 9330 kN

        Rsk frac14 105 139 86

        150frac14 2629 kN

        For a bored pile the partial factors are b frac14 160 and s frac14 130

        Design bearing resistance Rcd frac14 9330

        160thorn 2629

        130

        frac14 5831thorn 2022

        frac14 7850 kN

        Adding ethDAb W) the design bearing resistance becomes 9650 kN

        812

        ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

        qs frac14 cu frac14 040 105 frac14 42 kN=m2

        For a single pile

        Qf frac14 Abqb thorn Asqs

        frac14

        4 062 1305

        thorn eth 06 15 42THORN

        frac14 369thorn 1187 frac14 1556 kN

        Bearing capacity 69

        Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

        eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

        (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

        (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

        qbkfrac14 9cuk frac14 9 220

        150frac14 1320 kN=m2

        qskfrac14cuk frac14 040 105

        150frac14 28 kN=m2

        Rbkfrac14

        4 0602 1320 frac14 373 kN

        Rskfrac14 060 15 28 frac14 791 kN

        Rcdfrac14 373

        160thorn 791

        130frac14 233thorn 608 frac14 841 kN

        Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

        Rcd gt Fcd therefore the bearing resistance limit state is satisfied

        (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

        q frac14 21 000

        1762frac14 68 kN=m2

        Immediate settlement

        H

        Bfrac14 15

        176frac14 085

        D

        Bfrac14 13

        176frac14 074

        L

        Bfrac14 1

        Hence from Figure 515

        130 frac14 078 and 131 frac14 041

        70 Bearing capacity

        Thus using Equation 528

        si frac14 078 041 68 176

        65frac14 6mm

        Consolidation settlement

        Layer z (m) Area (m2) (kNm2) mvH (mm)

        1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

        434 (sod)

        Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

        sc frac14 056 434 frac14 24mm

        The total settlement is (6thorn 24) frac14 30mm

        813

        At base level N frac14 26 Then using Equation 830

        qb frac14 40NDb

        Bfrac14 40 26 2

        025frac14 8320 kN=m2

        ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

        Figure Q812

        Bearing capacity 71

        Over the length embedded in sand

        N frac14 21 ie18thorn 24

        2

        Using Equation 831

        qs frac14 2N frac14 2 21 frac14 42 kN=m2

        For a single pile

        Qf frac14 Abqb thorn Asqs

        frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

        For the pile group assuming a group efficiency of 12

        XQf frac14 12 9 604 frac14 6523 kN

        Then the load factor is

        F frac14 6523

        2000thorn 1000frac14 21

        (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

        Characteristic base resistance per unit area qbk frac14 8320

        150frac14 5547 kNm2

        Characteristic shaft resistance per unit area qsk frac14 42

        150frac14 28 kNm2

        Characteristic base and shaft resistances for a single pile

        Rbk frac14 0252 5547 frac14 347 kN

        Rsk frac14 4 025 2 28 frac14 56 kN

        For a driven pile the partial factors are b frac14 s frac14 130

        Design bearing resistance Rcd frac14 347

        130thorn 56

        130frac14 310 kN

        For the pile group Rcd frac14 12 9 310 frac14 3348 kN

        Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

        (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

        From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

        72 Bearing capacity

        N frac14 24thorn 26thorn 34

        3frac14 28

        Ic frac14 171

        2814frac14 0016 ethEquation 818THORN

        s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

        The settlement is less than 20mm therefore the serviceability limit state is satisfied

        814

        Using Equation 841

        Tf frac14 DLcu thorn

        4ethD2 d2THORNcuNc

        frac14 eth 02 5 06 110THORN thorn

        4eth022 012THORN110 9

        frac14 207thorn 23 frac14 230 kN

        Figure Q813

        Bearing capacity 73

        Chapter 9

        Stability of slopes

        91

        Referring to Figure Q91

        W frac14 417 19 frac14 792 kN=m

        Q frac14 20 28 frac14 56 kN=m

        Arc lengthAB frac14

        180 73 90 frac14 115m

        Arc length BC frac14

        180 28 90 frac14 44m

        The factor of safety is given by

        F frac14 rethcuLaTHORNWd1 thornQd2 frac14

        90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

        Depth of tension crack z0 frac14 2cu

        frac14 2 20

        19frac14 21m

        Arc length BD frac14

        180 13

        1

        2 90 frac14 21m

        F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

        The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

        Design resisting moment frac14 rXethcudLaTHORN frac14 90

        14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

        Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

        The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

        92

        u frac14 0

        Depth factor D frac14 11

        9frac14 122

        Using Equation 92 with F frac14 10

        Ns frac14 cu

        FHfrac14 30

        10 19 9frac14 0175

        Hence from Figure 93

        frac14 50

        For F frac14 12

        Ns frac14 30

        12 19 9frac14 0146

        frac14 27

        93

        Refer to Figure Q93

        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

        1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

        74 m

        214 1deg

        213 1deg

        39 m

        WB

        D

        C

        28 m

        21 m

        A

        Q

        Soil (1)Soil (2)

        73deg

        Figure Q91

        Stability of slopes 75

        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

        9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

        599 256 328 1372

        Figure Q93

        76 Stability of slopes

        XW cos frac14 b

        Xh cos frac14 21 2 599 frac14 2516 kN=mX

        W sin frac14 bX

        h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

        Arc length La frac14

        180 57

        1

        2 326 frac14 327m

        The factor of safety is given by

        F frac14 c0La thorn tan0ethW cos ulTHORN

        W sin

        frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

        frac14 091

        According to the limit state method

        0d frac14 tan1tan 32

        125

        frac14 265

        c0 frac14 8

        160frac14 5 kN=m2

        Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

        Design disturbing moment frac14 1075 kN=m

        The design resisting moment is less than the design disturbing moment therefore a slipwill occur

        94

        F frac14 1

        W sin

        Xfc0bthorn ethW ubTHORN tan0g sec

        1thorn ethtan tan0=FTHORN

        c0 frac14 8 kN=m2

        0 frac14 32

        c0b frac14 8 2 frac14 16 kN=m

        W frac14 bh frac14 21 2 h frac14 42h kN=m

        Try F frac14 100

        tan0

        Ffrac14 0625

        Stability of slopes 77

        Values of u are as obtained in Figure Q93

        SliceNo

        h(m)

        W frac14 bh(kNm)

        W sin(kNm)

        ub(kNm)

        c0bthorn (W ub) tan0(kNm)

        sec

        1thorn (tan tan0)FProduct(kNm)

        1 05 21 6 2 8 24 1078 262 13 55 31

        23 33 30 1042 31

        3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

        224 92 72 0931 67

        6 50 210 11 40 100 85 0907 777 55 231 14

        12 58 112 90 0889 80

        8 60 252 1812

        80 114 102 0874 899 63 265 22 99 116 109 0861 94

        10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

        2154 88 116 0853 99

        14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

        1074 1091

        F frac14 1091

        1074frac14 102 (assumed value 100)

        Thus

        F frac14 101

        95

        F frac14 1

        W sin

        XfWeth1 ruTHORN tan0g sec

        1thorn ethtan tan0THORN=F

        0 frac14 33

        ru frac14 020

        W frac14 bh frac14 20 5 h frac14 100h kN=m

        eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

        Try F frac14 110

        tan 0

        Ffrac14 tan 33

        110frac14 0590

        78 Stability of slopes

        Referring to Figure Q95

        SliceNo

        h(m)

        W frac14 bh(kNm)

        W sin(kNm)

        W(1 ru) tan0(kNm)

        sec

        1thorn ( tan tan0)FProduct(kNm)

        1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

        2120 234 0892 209

        4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

        1185 1271

        Figure Q95

        Stability of slopes 79

        F frac14 1271

        1185frac14 107

        The trial value was 110 therefore take F to be 108

        96

        (a) Water table at surface the factor of safety is given by Equation 912

        F frac14 0

        sat

        tan0

        tan

        ptie 15 frac14 92

        19

        tan 36

        tan

        tan frac14 0234

        frac14 13

        Water table well below surface the factor of safety is given by Equation 911

        F frac14 tan0

        tan

        frac14 tan 36

        tan 13

        frac14 31

        (b) 0d frac14 tan1tan 36

        125

        frac14 30

        Depth of potential failure surface frac14 z

        Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

        frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

        frac14 504z kN

        Design disturbing moment per unit area Sd frac14 sat sin cos

        frac14 19 z sin 13 cos 13

        frac14 416z kN

        Rd gtSd therefore the limit state for overall stability is satisfied

        80 Stability of slopes

        • Book Cover
        • Title
        • Contents
        • Basic characteristics of soils
        • Seepage
        • Effective stress
        • Shear strength
        • Stresses and displacements
        • Lateral earth pressure
        • Consolidation theory
        • Bearing capacity
        • Stability of slopes

          Contents

          1 Basic characteristics of soils 1

          2 Seepage 6

          3 Effective stress 14

          4 Shear strength 22

          5 Stresses and displacements 28

          6 Lateral earth pressure 34

          7 Consolidation theory 50

          8 Bearing capacity 60

          9 Stability of slopes 74

          Authorrsquos note

          In order not to short-circuit the learningprocess it is vital that the reader shouldattempt the problems before referring to thesolutions in this manual

          Chapter 1

          Basic characteristics of soils

          11

          Soil E consists of 98 coarse material (31 gravel size 67 sand size) and 2 finesIt is classified as SW well-graded gravelly SAND or in greater detail well-gradedslightly silty very gravelly SAND

          Soil F consists of 63 coarse material (2 gravel size 61 sand size) and 37non-plastic fines (ie between 35 and 65 fines) therefore the soil is classified as MSsandy SILT

          Soil G consists of 73 fine material (ie between 65 and 100 fines) and 27 sandsize The liquid limit is 32 and the plasticity index is 8 (ie 32 24) plotting marginallybelow the A-line in the ML zone on the plasticity chart Thus the classification is MLSILT (M-SOIL) of low plasticity (The plasticity chart is given in Figure 17)

          Figure Q11

          Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

          12

          From Equation 117

          1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

          191frac14 155

          e frac14 055

          Using Equation 113

          Sr frac14 wGs

          efrac14 0095 270

          055frac14 0466 eth466THORN

          Using Equation 119

          sat frac14 Gs thorn e1thorn e w frac14 325

          155 100 frac14 210Mg=m3

          From Equation 114

          w frac14 e

          Gsfrac14 055

          270frac14 0204 eth204THORN

          13

          Equations similar to 117ndash120 apply in the case of unit weights thus

          d frac14 Gs

          1thorn e w frac14272

          170 98 frac14 157 kN=m3

          sat frac14 Gs thorn e1thorn e w frac14 342

          170 98 frac14 197 kN=m3

          Using Equation 121

          0 frac14 Gs 1

          1thorn e w frac14 172

          170 98 frac14 99 kN=m3

          Using Equation 118a with Srfrac14 075

          frac14 Gs thorn Sre

          1thorn e w frac14 3245

          170 98 frac14 187 kN=m3

          2 Basic characteristics of soils

          Using Equation 113

          w frac14 Sre

          Gsfrac14 075 070

          272frac14 0193 eth193THORN

          The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

          14

          Volume of specimenfrac14

          438276 frac14 86 200mm3

          Bulk density ethTHORN frac14 Mass

          Volumefrac14 1680

          86 200 103frac14 195Mg=m3

          Water content ethwTHORN frac14 1680 1305

          1305frac14 0287 eth287THORN

          From Equation 117

          1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

          195frac14 180

          e frac14 080

          Using Equation 113

          Sr frac14 wGs

          efrac14 0287 273

          080frac14 098 eth98THORN

          15

          Using Equation 124

          d frac14

          1thorn w frac14215

          112frac14 192Mg=m3

          From Equation 117

          1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

          215frac14 138

          e frac14 038

          Using Equation 113

          Sr frac14 wGs

          efrac14 012 265

          038frac14 0837 eth837THORN

          Basic characteristics of soils 3

          Using Equation 115

          Afrac14 e wGs

          1thorn e frac14038 0318

          138frac14 0045 eth45THORN

          The zero air voids dry density is given by Equation 125

          d frac14 Gs

          1thorn wGsw frac14 265

          1thorn eth0135 265THORN 100 frac14 195Mg=m3

          ie a dry density of 200Mgm3 would not be possible

          16

          Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

          (Mgm3) d10(Mgm3)

          2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

          In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

          wopt frac14 15 and dmaxfrac14 183Mg=m3

          Figure Q16

          4 Basic characteristics of soils

          Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

          d5 d10

          respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

          17

          From Equation 120

          e frac14 Gswd 1

          The maximum and minimum values of void ratio are given by

          emax frac14 Gsw

          dmin

          1

          emin frac14 Gswdmax

          1

          From Equation 123

          ID frac14 Gsweth1=dmin 1=dTHORN

          Gsweth1=dmin 1=dmax

          THORN

          frac14 frac121 ethdmin=dTHORN1=dmin

          frac121 ethdmin=dmax

          THORN1=dmin

          frac14 d dmin

          dmax dmin

          dmax

          d

          frac14 172 154

          181 154

          181

          172

          frac14 070 eth70THORN

          Basic characteristics of soils 5

          Chapter 2

          Seepage

          21

          The coefficient of permeability is determined from the equation

          k frac14 23al

          At1log

          h0

          h1

          where

          a frac14

          4 00052 m2 l frac14 02m

          A frac14

          4 012 m2 t1 frac14 3 602 s

          logh0

          h1frac14 log

          100

          035frac14 0456

          k frac14 23 00052 02 0456

          012 3 602frac14 49 108 m=s

          22

          The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

          q frac14 kh Nf

          Ndfrac14 106 400 37

          11frac14 13 106 m3=s per m

          Figure Q22

          23

          The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

          q frac14 kh Nf

          Ndfrac14 5 105 300 35

          9frac14 58 105 m3=s per m

          The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

          Point h (m) h z (m) u frac14 w(h z)(kNm2)

          1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

          eg for Point 1

          h1 frac14 7

          9 300 frac14 233m

          h1 z1 frac14 233 eth250THORN frac14 483m

          Figure Q23

          Seepage 7

          u1 frac14 98 483 frac14 47 kN=m2

          The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

          24

          The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

          q frac14 kh Nf

          Ndfrac14 40 107 550 10

          11frac14 20 106 m3=s per m

          25

          The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

          q frac14 kh Nf

          Ndfrac14 20 106 500 42

          9frac14 47 106 m3=s per m

          Figure Q24

          8 Seepage

          26

          The scale transformation factor in the x direction is given by Equation 221 ie

          xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

          ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

          Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

          k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

          pfrac14

          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

          p 107 frac14 30 107 m=s

          Thus the quantity of seepage is given by

          q frac14 k0h Nf

          Ndfrac14 30 107 1400 325

          12frac14 11 106 m3=s per m

          Figure Q25

          Seepage 9

          27

          The scale transformation factor in the x direction is

          xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

          ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

          Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

          k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

          pfrac14

          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

          p 106 frac14 45 106 m=s

          The focus of the basic parabola is at point A The parabola passes through point Gsuch that

          GC frac14 03HC frac14 03 30 frac14 90m

          Thus the coordinates of G are

          x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

          480 frac14 x0 2002

          4x0

          Figure Q26

          10 Seepage

          Hence

          x0 frac14 20m

          Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

          x frac14 20 z2

          80

          x 20 0 50 100 200 300z 0 400 748 980 1327 1600

          The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

          No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

          q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

          28

          The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

          Figure Q27

          Seepage 11

          q frac14 kh Nf

          Ndfrac14 45 105 28 33

          7

          frac14 59 105 m3=s per m

          29

          The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

          kx frac14 H1k1 thornH2k2

          H1 thornH2frac14 106

          10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

          kz frac14 H1 thornH2

          H1

          k1thornH2

          k2

          frac14 10

          5

          eth2 106THORN thorn5

          eth16 106THORNfrac14 36 106 m=s

          Then the scale transformation factor is given by

          xt frac14 xffiffiffiffiffikz

          pffiffiffiffiffikx

          p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

          Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

          Figure Q28

          12 Seepage

          k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

          qfrac14

          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

          p 106 frac14 57 106 m=s

          Then the quantity of seepage is given by

          q frac14 k0h Nf

          Ndfrac14 57 106 350 56

          11

          frac14 10 105 m3=s per m

          Figure Q29

          Seepage 13

          Chapter 3

          Effective stress

          31

          Buoyant unit weight

          0 frac14 sat w frac14 20 98 frac14 102 kN=m3

          Effective vertical stress

          0v frac14 5 102 frac14 51 kN=m2 or

          Total vertical stress

          v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

          Pore water pressure

          u frac14 7 98 frac14 686 kN=m2

          Effective vertical stress

          0v frac14 v u frac14 1196 686 frac14 51 kN=m2

          32

          Buoyant unit weight

          0 frac14 sat w frac14 20 98 frac14 102 kN=m3

          Effective vertical stress

          0v frac14 5 102 frac14 51 kN=m2 or

          Total vertical stress

          v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

          Pore water pressure

          u frac14 205 98 frac14 2009 kN=m2

          Effective vertical stress

          0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

          33

          At top of the clay

          v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

          u frac14 2 98 frac14 196 kN=m2

          0v frac14 v u frac14 710 196 frac14 514 kN=m2

          Alternatively

          0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

          0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

          At bottom of the clay

          v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

          u frac14 12 98 frac14 1176 kN=m2

          0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

          NB The alternative method of calculation is not applicable because of the artesiancondition

          Figure Q3132

          Effective stress 15

          34

          0 frac14 20 98 frac14 102 kN=m3

          At 8m depth

          0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

          35

          0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

          0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

          Figure Q33

          Figure Q34

          16 Effective stress

          (a) Immediately after WT rise

          At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

          0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

          At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

          0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

          (b) Several years after WT rise

          At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

          At 8m depth

          0v frac14 940 kN=m2 (as above)

          At 12m depth

          0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

          Figure Q35

          Effective stress 17

          36

          Total weight

          ab frac14 210 kN

          Effective weight

          ac frac14 112 kN

          Resultant boundary water force

          be frac14 119 kN

          Seepage force

          ce frac14 34 kN

          Resultant body force

          ae frac14 99 kN eth73 to horizontalTHORN

          (Refer to Figure Q36)

          Figure Q36

          18 Effective stress

          37

          Situation (1)(a)

          frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

          u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

          0 frac14 u frac14 694 392 frac14 302 kN=m2

          (b)

          i frac14 2

          4frac14 05

          j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

          Situation (2)(a)

          frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

          u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

          0 frac14 u frac14 498 392 frac14 106 kN=m2

          (b)

          i frac14 2

          4frac14 05

          j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

          38

          The flow net is drawn in Figure Q24

          Loss in total head between adjacent equipotentials

          h frac14 550

          Ndfrac14 550

          11frac14 050m

          Exit hydraulic gradient

          ie frac14 h

          sfrac14 050

          070frac14 071

          Effective stress 19

          The critical hydraulic gradient is given by Equation 39

          ic frac14 0

          wfrac14 102

          98frac14 104

          Therefore factor of safety against lsquoboilingrsquo (Equation 311)

          F frac14 iciefrac14 104

          071frac14 15

          Total head at C

          hC frac14 nd

          Ndh frac14 24

          11 550 frac14 120m

          Elevation head at C

          zC frac14 250m

          Pore water pressure at C

          uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

          Therefore effective vertical stress at C

          0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

          For point D

          hD frac14 73

          11 550 frac14 365m

          zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

          0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

          39

          The flow net is drawn in Figure Q25

          For a soil prism 150 300m adjacent to the piling

          hm frac14 26

          9 500 frac14 145m

          20 Effective stress

          Factor of safety against lsquoheavingrsquo (Equation 310)

          F frac14 ic

          imfrac14 0d

          whmfrac14 97 300

          98 145frac14 20

          With a filter

          F frac14 0d thorn wwhm

          3 frac14 eth97 300THORN thorn w98 145

          w frac14 135 kN=m2

          Depth of filterfrac14 13521frac14 065m (if above water level)

          Effective stress 21

          Chapter 4

          Shear strength

          41

          frac14 295 kN=m2

          u frac14 120 kN=m2

          0 frac14 u frac14 295 120 frac14 175 kN=m2

          f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

          42

          03 (kNm2) 1 3 (kNm2) 01 (kNm2)

          100 452 552200 908 1108400 1810 2210800 3624 4424

          The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

          Figure Q42

          43

          3 (kNm2) 1 3 (kNm2) 1 (kNm2)

          200 222 422400 218 618600 220 820

          The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

          44

          The modified shear strength parameters are

          0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

          a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

          The coordinates of the stress point representing failure conditions in the test are

          1

          2eth1 2THORN frac14 1

          2 170 frac14 85 kN=m2

          1

          2eth1 thorn 3THORN frac14 1

          2eth270thorn 100THORN frac14 185 kN=m2

          The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

          uf frac14 36 kN=m2

          Figure Q43

          Figure Q44

          Shear strength 23

          45

          3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

          150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

          The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

          The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

          46

          03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

          200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

          The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

          Figure Q45

          24 Shear strength

          47

          The torque required to produce shear failure is given by

          T frac14 dh cud

          2thorn 2

          Z d=2

          0

          2r drcur

          frac14 cud2h

          2thorn 4cu

          Z d=2

          0

          r2dr

          frac14 cud2h

          2thorn d

          3

          6

          Then

          35 frac14 cu52 10

          2thorn 53

          6

          103

          cu frac14 76 kN=m3

          400

          0 400 800 1200 1600

          τ (k

          Nm

          2 )

          σprime (kNm2)

          34deg

          315deg29deg

          (a)

          (b)

          0 400

          400

          800 1200 1600

          Failure envelope

          300 500

          σprime (kNm2)

          τ (k

          Nm

          2 )

          20 (kNm2)

          31deg

          Figure Q46

          Shear strength 25

          48

          The relevant stress values are calculated as follows

          3 frac14 600 kN=m2

          1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

          2(1 3) 0 40 79 107 139 159

          1

          2(01 thorn 03) 400 411 402 389 351 326

          1

          2(1 thorn 3) 600 640 679 707 739 759

          The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

          Af frac14 433 200

          319frac14 073

          49

          B frac14 u33

          frac14 144

          350 200frac14 096

          a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

          0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

          10 283 65 023

          Figure Q48

          26 Shear strength

          The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

          Figure Q49

          Shear strength 27

          Chapter 5

          Stresses and displacements

          51

          Vertical stress is given by

          z frac14 Qz2Ip frac14 5000

          52Ip

          Values of Ip are obtained from Table 51

          r (m) rz Ip z (kNm2)

          0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

          10 20 0009 2

          The variation of z with radial distance (r) is plotted in Figure Q51

          Figure Q51

          52

          Below the centre load (Figure Q52)

          r

          zfrac14 0 for the 7500-kN load

          Ip frac14 0478

          r

          zfrac14 5

          4frac14 125 for the 10 000- and 9000-kN loads

          Ip frac14 0045

          Then

          z frac14X Q

          z2Ip

          frac14 7500 0478

          42thorn 10 000 0045

          42thorn 9000 0045

          42

          frac14 224thorn 28thorn 25 frac14 277 kN=m2

          53

          The vertical stress under a corner of a rectangular area is given by

          z frac14 qIr

          where values of Ir are obtained from Figure 510 In this case

          z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

          z

          Figure Q52

          Stresses and displacements 29

          z (m) m n Ir z (kNm2)

          0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

          10 010 0005 5

          z is plotted against z in Figure Q53

          54

          (a)

          m frac14 125

          12frac14 104

          n frac14 18

          12frac14 150

          From Figure 510 Irfrac14 0196

          z frac14 2 175 0196 frac14 68 kN=m2

          Figure Q53

          30 Stresses and displacements

          (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

          z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

          55

          Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

          Px frac14 2Q

          1

          m2 thorn 1frac14 2 150

          125frac14 76 kN=m

          Equation 517 is used to obtain the pressure distribution

          px frac14 4Q

          h

          m2n

          ethm2 thorn n2THORN2 frac14150

          m2n

          ethm2 thorn n2THORN2 ethkN=m2THORN

          Figure Q54

          Stresses and displacements 31

          n m2n

          (m2 thorn n2)2

          px(kNm2)

          0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

          The pressure distribution is plotted in Figure Q55

          56

          H

          Bfrac14 10

          2frac14 5

          L

          Bfrac14 4

          2frac14 2

          D

          Bfrac14 1

          2frac14 05

          Hence from Figure 515

          131 frac14 082

          130 frac14 094

          Figure Q55

          32 Stresses and displacements

          The immediate settlement is given by Equation 528

          si frac14 130131qB

          Eu

          frac14 094 082 200 2

          45frac14 7mm

          Stresses and displacements 33

          Chapter 6

          Lateral earth pressure

          61

          For 0 frac14 37 the active pressure coefficient is given by

          Ka frac14 1 sin 37

          1thorn sin 37frac14 025

          The total active thrust (Equation 66a with c0 frac14 0) is

          Pa frac14 1

          2KaH

          2 frac14 1

          2 025 17 62 frac14 765 kN=m

          If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

          K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

          and the thrust on the wall is

          P0 frac14 1

          2K0H

          2 frac14 1

          2 040 17 62 frac14 122 kN=m

          62

          The active pressure coefficients for the three soil types are as follows

          Ka1 frac141 sin 35

          1thorn sin 35frac14 0271

          Ka2 frac141 sin 27

          1thorn sin 27frac14 0375

          ffiffiffiffiffiffiffiKa2

          p frac14 0613

          Ka3 frac141 sin 42

          1thorn sin 42frac14 0198

          Distribution of active pressure (plotted in Figure Q62)

          Depth (m) Soil Active pressure (kNm2)

          3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

          12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

          At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

          Total thrust frac14 571 kNm

          Point of application is (4893571) m from the top of the wall ie 857m

          Force (kN) Arm (m) Moment (kN m)

          (1)1

          2 0271 16 32 frac14 195 20 390

          (2) 0271 16 3 2 frac14 260 40 1040

          (3)1

          2 0271 92 22 frac14 50 433 217

          (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

          (5)1

          2 0375 102 32 frac14 172 70 1204

          (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

          (7)1

          2 0198 112 42 frac14 177 1067 1889

          (8)1

          2 98 92 frac14 3969 90 35721

          5713 48934

          Figure Q62

          Lateral earth pressure 35

          63

          (a) For u frac14 0 Ka frac14 Kp frac14 1

          Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

          frac14 245

          At the lower end of the piling

          pa frac14 Kaqthorn Kasatz Kaccu

          frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

          frac14 115 kN=m2

          pp frac14 Kpsatzthorn Kpccu

          frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

          frac14 202 kN=m2

          (b) For 0 frac14 26 and frac14 1

          20

          Ka frac14 035

          Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

          pfrac14 145 ethEquation 619THORN

          Kp frac14 37

          Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

          pfrac14 47 ethEquation 624THORN

          At the lower end of the piling

          pa frac14 Kaqthorn Ka0z Kacc

          0

          frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

          frac14 187 kN=m2

          pp frac14 Kp0zthorn Kpcc

          0

          frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

          frac14 198 kN=m2

          36 Lateral earth pressure

          64

          (a) For 0 frac14 38 Ka frac14 024

          0 frac14 20 98 frac14 102 kN=m3

          The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

          Force (kN) Arm (m) Moment (kN m)

          (1) 024 10 66 frac14 159 33 525

          (2)1

          2 024 17 392 frac14 310 400 1240

          (3) 024 17 39 27 frac14 430 135 580

          (4)1

          2 024 102 272 frac14 89 090 80

          (5)1

          2 98 272 frac14 357 090 321

          Hfrac14 1345 MH frac14 2746

          (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

          (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

          XM frac14MV MH frac14 7790 kNm

          Lever arm of base resultant

          M

          Vfrac14 779

          488frac14 160

          Eccentricity of base resultant

          e frac14 200 160 frac14 040m

          39 m

          27 m

          40 m

          04 m

          04 m

          26 m

          (7)

          (9)

          (1)(2)

          (3)

          (4)

          (5)

          (8)(6)

          (10)

          WT

          10 kNm2

          Hydrostatic

          Figure Q64

          Lateral earth pressure 37

          Base pressures (Equation 627)

          p frac14 VB

          1 6e

          B

          frac14 488

          4eth1 060THORN

          frac14 195 kN=m2 and 49 kN=m2

          Factor of safety against sliding (Equation 628)

          F frac14 V tan

          Hfrac14 488 tan 25

          1345frac14 17

          (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

          Hfrac14 1633 kN

          V frac14 4879 kN

          MH frac14 3453 kNm

          MV frac14 10536 kNm

          The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

          65

          For 0 frac14 36 Ka frac14 026 and Kp frac14 385

          Kp

          Ffrac14 385

          2

          0 frac14 20 98 frac14 102 kN=m3

          The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

          Force (kN) Arm (m) Moment (kN m)

          (1)1

          2 026 17 452 frac14 448 dthorn 15 448dthorn 672

          (2) 026 17 45 d frac14 199d d2 995d2

          (3)1

          2 026 102 d2 frac14 133d2 d3 044d3

          (4)1

          2 385

          2 17 152 frac14 368 dthorn 05 368d 184

          (5)385

          2 17 15 d frac14 491d d2 2455d2

          (6)1

          2 385

          2 102 d2 frac14 982d2 d3 327d3

          38 Lateral earth pressure

          XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

          d3 thorn 516d2 283d 1724 frac14 0

          d frac14 179m

          Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

          Over additional 20 embedded depth

          pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

          Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

          66

          The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

          Ka frac14 sin 69=sin 105

          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

          pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

          26664

          37775

          2

          frac14 050

          The total active thrust (acting at 25 above the normal) is given by Equation 616

          Pa frac14 1

          2 050 19 7502 frac14 267 kN=m

          Figure Q65

          Lateral earth pressure 39

          Horizontal component

          Ph frac14 267 cos 40 frac14 205 kN=m

          Vertical component

          Pv frac14 267 sin 40 frac14 172 kN=m

          Consider moments about the toe of the wall (Figure Q66) (per m)

          Force (kN) Arm (m) Moment (kN m)

          (1)1

          2 175 650 235 frac14 1337 258 345

          (2) 050 650 235 frac14 764 175 134

          (3)1

          2 070 650 235 frac14 535 127 68

          (4) 100 400 235 frac14 940 200 188

          (5) 1

          2 080 050 235 frac14 47 027 1

          Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

          Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

          Lever arm of base resultant

          M

          Vfrac14 795

          525frac14 151m

          Eccentricity of base resultant

          e frac14 200 151 frac14 049m

          Figure Q66

          40 Lateral earth pressure

          Base pressures (Equation 627)

          p frac14 525

          41 6 049

          4

          frac14 228 kN=m2 and 35 kN=m2

          The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

          The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

          The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

          67

          For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

          Force (kN) Arm (m) Moment (kNm)

          (1)1

          2 027 17 52 frac14 574 183 1050

          (2) 027 17 5 3 frac14 689 500 3445

          (3)1

          2 027 102 32 frac14 124 550 682

          (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

          (5)1

          2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

          (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

          (7) 1

          2 267

          2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

          (8) 2 10ffiffiffiffiffiffiffiffiffi267p

          2 d frac14 163d d2thorn 650 82d2 1060d

          Tie rod force per m frac14 T 0 0

          XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

          d3 thorn 77d2 269d 1438 frac14 0

          d frac14 467m

          Depth of penetration frac14 12d frac14 560m

          Lateral earth pressure 41

          Algebraic sum of forces for d frac14 467m isX

          F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

          T frac14 905 kN=m

          Force in each tie rod frac14 25T frac14 226 kN

          68

          (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

          0 frac14 21 98 frac14 112 kN=m3

          The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

          uC frac14 150

          165 15 98 frac14 134 kN=m2

          The average seepage pressure is

          j frac14 15

          165 98 frac14 09 kN=m3

          Hence

          0 thorn j frac14 112thorn 09 frac14 121 kN=m3

          0 j frac14 112 09 frac14 103 kN=m3

          Figure Q67

          42 Lateral earth pressure

          Consider moments about the anchor point A (per m)

          Force (kN) Arm (m) Moment (kN m)

          (1) 10 026 150 frac14 390 60 2340

          (2)1

          2 026 18 452 frac14 474 15 711

          (3) 026 18 45 105 frac14 2211 825 18240

          (4)1

          2 026 121 1052 frac14 1734 100 17340

          (5)1

          2 134 15 frac14 101 40 404

          (6) 134 30 frac14 402 60 2412

          (7)1

          2 134 60 frac14 402 95 3819

          571 4527(8) Ppm

          115 115PPm

          XM frac14 0

          Ppm frac144527

          115frac14 394 kN=m

          Available passive resistance

          Pp frac14 1

          2 385 103 62 frac14 714 kN=m

          Factor of safety

          Fp frac14 Pp

          Ppm

          frac14 714

          394frac14 18

          Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

          Figure Q68

          Lateral earth pressure 43

          (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

          Consider moments (per m) about the tie point A

          Force (kN) Arm (m)

          (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

          (2)1

          2 033 18 452 frac14 601 15

          (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

          (4)1

          2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

          (5)1

          2 134 15 frac14 101 40

          (6) 134 30 frac14 402 60

          (7)1

          2 134 d frac14 67d d3thorn 75

          (8) 1

          2 30 103 d2 frac141545d2 2d3thorn 75

          Moment (kN m)

          (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

          XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

          d3 thorn 827d2 466d 1518 frac14 0

          By trial

          d frac14 544m

          The minimum depth of embedment required is 544m

          69

          For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

          0 frac14 20 98 frac14 102 kN=m3

          The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

          44 Lateral earth pressure

          uC frac14 147

          173 26 98 frac14 216 kN=m2

          and the average seepage pressure around the wall is

          j frac14 26

          173 98 frac14 15 kN=m3

          Consider moments about the prop (A) (per m)

          Force (kN) Arm (m) Moment (kN m)

          (1)1

          2 03 17 272 frac14 186 020 37

          (2) 03 17 27 53 frac14 730 335 2445

          (3)1

          2 03 (102thorn 15) 532 frac14 493 423 2085

          (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

          (5)1

          2 216 26 frac14 281 243 684

          (6) 216 27 frac14 583 465 2712

          (7)1

          2 216 60 frac14 648 800 5184

          3055(8)

          1

          2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

          Factor of safety

          Fr frac14 6885

          3055frac14 225

          Figure Q69

          Lateral earth pressure 45

          610

          For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

          p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

          Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

          Using the recommendations of Twine and Roscoe

          p frac14 02H frac14 02 19 9 frac14 342 kN=m2

          Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

          611

          frac14 18 kN=m3 0 frac14 34

          H frac14 350m nH frac14 335m mH frac14 185m

          Consider a trial value of F frac14 20 Refer to Figure 635

          0m frac14 tan1tan 34

          20

          frac14 186

          Then

          frac14 45 thorn 0m2frac14 543

          W frac14 1

          2 18 3502 cot 543 frac14 792 kN=m

          Figure Q610

          46 Lateral earth pressure

          P frac14 1

          2 s 3352 frac14 561s kN=m

          U frac14 1

          2 98 1852 cosec 543 frac14 206 kN=m

          Equations 630 and 631 then become

          561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

          792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

          ie

          561s 0616N 405 frac14 0

          792 0857N thorn 563 frac14 0

          N frac14 848

          0857frac14 989 kN=m

          Then

          561s 609 405 frac14 0

          s frac14 649

          561frac14 116 kN=m3

          The calculations for trial values of F of 20 15 and 10 are summarized below

          F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

          20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

          s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

          Figure Q611

          Lateral earth pressure 47

          612

          For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

          45 thorn 0

          2frac14 63

          For the retained material between the surface and a depth of 36m

          Pa frac14 1

          2 030 18 362 frac14 350 kN=m

          Weight of reinforced fill between the surface and a depth of 36m is

          Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

          eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

          Eccentricity of Rv

          e frac14 263 250 frac14 013m

          The average vertical stress at a depth of 36m is

          z frac14 Rv

          L 2efrac14 324

          474frac14 68 kN=m2

          (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

          Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

          Tensile stress in the element frac14 138 103

          65 3frac14 71N=mm2

          Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

          Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

          Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

          The weight of ABC is

          W frac14 1

          2 18 52 265 frac14 124 kN=m

          From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

          48 Lateral earth pressure

          (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

          Tp frac14 032 68 120 065 frac14 170 kN

          Tr frac14 213 420

          418frac14 214 kN

          Again the tensile failure and slipping limit states are satisfied for this element

          Figure Q612

          Lateral earth pressure 49

          Chapter 7

          Consolidation theory

          71

          Total change in thickness

          H frac14 782 602 frac14 180mm

          Average thickness frac14 1530thorn 180

          2frac14 1620mm

          Length of drainage path d frac14 1620

          2frac14 810mm

          Root time plot (Figure Q71a)

          ffiffiffiffiffiffit90p frac14 33

          t90 frac14 109min

          cv frac14 0848d2

          t90frac14 0848 8102

          109 1440 365

          106frac14 27m2=year

          r0 frac14 782 764

          782 602frac14 018

          180frac14 0100

          rp frac14 10eth764 645THORN9eth782 602THORN frac14

          10 119

          9 180frac14 0735

          rs frac14 1 eth0100thorn 0735THORN frac14 0165

          Log time plot (Figure Q71b)

          t50 frac14 26min

          cv frac14 0196d2

          t50frac14 0196 8102

          26 1440 365

          106frac14 26m2=year

          r0 frac14 782 763

          782 602frac14 019

          180frac14 0106

          rp frac14 763 623

          782 602frac14 140

          180frac14 0778

          rs frac14 1 eth0106thorn 0778THORN frac14 0116

          Figure Q71(a)

          Figure Q71(b)

          Final void ratio

          e1 frac14 w1Gs frac14 0232 272 frac14 0631

          e

          Hfrac14 1thorn e0

          H0frac14 1thorn e1 thorne

          H0

          ie

          e

          180frac14 1631thorne

          1710

          e frac14 2936

          1530frac14 0192

          Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

          mv frac14 1

          1thorn e0 e0 e101 00

          frac14 1

          1823 0192

          0107frac14 098m2=MN

          k frac14 cvmvw frac14 265 098 98

          60 1440 365 103frac14 81 1010 m=s

          72

          Using Equation 77 (one-dimensional method)

          sc frac14 e0 e11thorn e0 H

          Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

          Figure Q72

          52 Consolidation theory

          Settlement

          Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

          1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

          318

          Notes 5 92y 460thorn 84

          Heave

          Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

          1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

          38

          73

          U frac14 f ethTvTHORN frac14 f cvt

          d2

          Hence if cv is constant

          t1

          t2frac14 d

          21

          d22

          where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

          d1 frac14 95mm and d2 frac14 2500mm

          for U frac14 050 t2 frac14 t1 d22

          d21

          frac14 20

          60 24 365 25002

          952frac14 263 years

          for U lt 060 Tv frac14

          4U2 (Equation 724(a))

          t030 frac14 t050 0302

          0502

          frac14 263 036 frac14 095 years

          Consolidation theory 53

          74

          The layer is open

          d frac14 8

          2frac14 4m

          Tv frac14 cvtd2frac14 24 3

          42frac14 0450

          ui frac14 frac14 84 kN=m2

          The excess pore water pressure is given by Equation 721

          ue frac14Xmfrac141mfrac140

          2ui

          Msin

          Mz

          d

          expethM2TvTHORN

          In this case z frac14 d

          sinMz

          d

          frac14 sinM

          where

          M frac14

          23

          25

          2

          M sin M M2Tv exp (M2Tv)

          2thorn1 1110 0329

          3

          21 9993 457 105

          ue frac14 2 84 2

          1 0329 ethother terms negligibleTHORN

          frac14 352 kN=m2

          75

          The layer is open

          d frac14 6

          2frac14 3m

          Tv frac14 cvtd2frac14 10 3

          32frac14 0333

          The layer thickness will be divided into six equal parts ie m frac14 6

          54 Consolidation theory

          For an open layer

          Tv frac14 4n

          m2

          n frac14 0333 62

          4frac14 300

          The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

          ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

          i j

          0 1 2 3 4 5 6 7 8 9 10 11 12

          0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

          The initial and 3-year isochrones are plotted in Figure Q75

          Area under initial isochrone frac14 180 units

          Area under 3-year isochrone frac14 63 units

          The average degree of consolidation is given by Equation 725Thus

          U frac14 1 63

          180frac14 065

          Figure Q75

          Consolidation theory 55

          76

          At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

          0 frac14 2w frac14 2 98 frac14 196 kN=m2

          The final consolidation settlement (one-dimensional method) is

          sc frac14 mv0H frac14 083 196 8 frac14 130mm

          Corrected time t frac14 2 1

          2

          40

          52

          frac14 1615 years

          Tv frac14 cvtd2frac14 44 1615

          42frac14 0444

          From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

          77

          The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

          Figure Q77

          56 Consolidation theory

          Point m n Ir (kNm2) sc (mm)

          13020frac14 15 20

          20frac14 10 0194 (4) 113 124

          260

          20frac14 30

          20

          20frac14 10 0204 (2) 59 65

          360

          20frac14 30

          40

          20frac14 20 0238 (1) 35 38

          430

          20frac14 15

          40

          20frac14 20 0224 (2) 65 72

          Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

          78

          Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

          (a) Immediate settlement

          H

          Bfrac14 30

          35frac14 086

          D

          Bfrac14 2

          35frac14 006

          Figure Q78

          Consolidation theory 57

          From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

          si frac14 130131qB

          Eufrac14 10 032 105 35

          40frac14 30mm

          (b) Consolidation settlement

          Layer z (m) Dz Ic (kNm2) syod (mm)

          1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

          3150

          Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

          Now

          H

          Bfrac14 30

          35frac14 086 and A frac14 065

          from Figure 712 13 frac14 079

          sc frac14 13sod frac14 079 315 frac14 250mm

          Total settlement

          s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

          79

          Without sand drains

          Uv frac14 025

          Tv frac14 0049 ethfrom Figure 718THORN

          t frac14 Tvd2

          cvfrac14 0049 82

          cvWith sand drains

          R frac14 0564S frac14 0564 3 frac14 169m

          n frac14 Rrfrac14 169

          015frac14 113

          Tr frac14 cht

          4R2frac14 ch

          4 1692 0049 82

          cvethand ch frac14 cvTHORN

          frac14 0275

          Ur frac14 073 (from Figure 730)

          58 Consolidation theory

          Using Equation 740

          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

          U frac14 080

          710

          Without sand drains

          Uv frac14 090

          Tv frac14 0848

          t frac14 Tvd2

          cvfrac14 0848 102

          96frac14 88 years

          With sand drains

          R frac14 0564S frac14 0564 4 frac14 226m

          n frac14 Rrfrac14 226

          015frac14 15

          Tr

          Tvfrac14 chcv

          d2

          4R2ethsame tTHORN

          Tr

          Tvfrac14 140

          96 102

          4 2262frac14 714 eth1THORN

          Using Equation 740

          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

          An iterative solution is required using (1) and (2) an initial value ofUv being estimated

          Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

          040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

          Thus

          Uv frac14 0295 and Ur frac14 086

          t frac14 88 00683

          0848frac14 07 years

          Consolidation theory 59

          Chapter 8

          Bearing capacity

          81

          (a) The ultimate bearing capacity is given by Equation 83

          qf frac14 cNc thorn DNq thorn 1

          2BN

          For u frac14 0

          Nc frac14 514 Nq frac14 1 N frac14 0

          qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

          The net ultimate bearing capacity is

          qnf frac14 qf D frac14 540 kN=m2

          The net foundation pressure is

          qn frac14 q D frac14 425

          2 eth21 1THORN frac14 192 kN=m2

          The factor of safety (Equation 86) is

          F frac14 qnfqnfrac14 540

          192frac14 28

          (b) For 0 frac14 28

          Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

          qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

          2 112 2 13

          frac14 260thorn 168thorn 146 frac14 574 kN=m2

          qnf frac14 574 112 frac14 563 kN=m2

          F frac14 563

          192frac14 29

          (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

          82

          For 0 frac14 38

          Nq frac14 49 N frac14 67

          qnf frac14 DethNq 1THORN thorn 1

          2BN ethfrom Equation 83THORN

          frac14 eth18 075 48THORN thorn 1

          2 18 15 67

          frac14 648thorn 905 frac14 1553 kN=m2

          qn frac14 500

          15 eth18 075THORN frac14 320 kN=m2

          F frac14 qnfqnfrac14 1553

          320frac14 48

          0d frac14 tan1tan 38

          125

          frac14 32 therefore Nq frac14 23 and N frac14 25

          Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

          2 18 15 25

          frac14 15eth310thorn 337THORNfrac14 970 kN=m

          Design load (action) Vd frac14 500 kN=m

          The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

          83

          D

          Bfrac14 350

          225frac14 155

          From Figure 85 for a square foundation

          Nc frac14 81

          Bearing capacity 61

          For a rectangular foundation (L frac14 450m B frac14 225m)

          Nc frac14 084thorn 016B

          L

          81 frac14 745

          Using Equation 810

          qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

          For F frac14 3

          qn frac14 1006

          3frac14 335 kN=m2

          q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

          Design load frac14 405 450 225 frac14 4100 kN

          Design undrained strength cud frac14 135

          14frac14 96 kN=m2

          Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

          frac14 7241 kN

          Design load Vd frac14 4100 kN

          Rd gt Vd therefore the bearing resistance limit state is satisfied

          84

          For 0 frac14 40

          Nq frac14 64 N frac14 95

          qnf frac14 DethNq 1THORN thorn 04BN

          (a) Water table 5m below ground level

          qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

          qn frac14 400 17 frac14 383 kN=m2

          F frac14 2686

          383frac14 70

          (b) Water table 1m below ground level (ie at foundation level)

          0 frac14 20 98 frac14 102 kN=m3

          62 Bearing capacity

          qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

          F frac14 2040

          383frac14 53

          (c) Water table at ground level with upward hydraulic gradient 02

          eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

          qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

          F frac14 1296

          392frac14 33

          85

          The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

          Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

          Design value of 0 frac14 tan1tan 39

          125

          frac14 33

          For 0 frac14 33 Nq frac14 26 and N frac14 29

          Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

          Rd gt Vd therefore the bearing resistance limit state is satisfied

          86

          (a) Undrained shear for u frac14 0

          Nc frac14 514 Nq frac14 1 N frac14 0

          qnf frac14 12cuNc

          frac14 12 100 514 frac14 617 kN=m2

          qn frac14 qnfFfrac14 617

          3frac14 206 kN=m2

          q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

          Bearing capacity 63

          Drained shear for 0 frac14 32

          Nq frac14 23 N frac14 25

          0 frac14 21 98 frac14 112 kN=m3

          qnf frac14 0DethNq 1THORN thorn 040BN

          frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

          frac14 694 kN=m2

          q frac14 694

          3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

          Design load frac14 42 227 frac14 3632 kN

          (b) Design undrained strength cud frac14 100

          14frac14 71 kNm2

          Design bearing resistance Rd frac14 12cudNe area

          frac14 12 71 514 42

          frac14 7007 kN

          For drained shear 0d frac14 tan1tan 32

          125

          frac14 26

          Nq frac14 12 N frac14 10

          Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

          (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

          Layer z (m) m n Ir 0 (kNm2) sod (mm)

          1 2 100 0175 0700qn 0182qn

          2 6 033 0044 0176qn 0046qn

          3 10 020 0017 0068qn 0018qn

          0246qn

          Diameter of equivalent circle B frac14 45m

          H

          Bfrac14 12

          45frac14 27 and A frac14 042

          13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

          64 Bearing capacity

          For sc frac14 30mm

          qn frac14 30

          0147frac14 204 kN=m2

          q frac14 204thorn 21 frac14 225 kN=m2

          Design load frac14 42 225 frac14 3600 kN

          The design load is 3600 kN settlement being the limiting criterion

          87

          D

          Bfrac14 8

          4frac14 20

          From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

          F frac14 cuNc

          Dfrac14 40 71

          20 8frac14 18

          88

          Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

          Design value of 0 frac14 tan1tan 38

          125

          frac14 32

          Figure Q86

          Bearing capacity 65

          For 0 frac14 32 Nq frac14 23 and N frac14 25

          Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

          The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

          Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

          For B frac14 250m qn frac14 3750

          2502 17 frac14 583 kN=m2

          From Figure 510 m frac14 n frac14 126

          6frac14 021

          Ir frac14 0019

          Stress increment frac14 4 0019 583 frac14 44 kN=m2

          Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

          The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

          The settlement is less than 20mm therefore the serviceability limit state is satisfied

          89

          Depth (m) N 0v (kNm2) CN N1

          070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

          Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

          (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

          Cw frac14 05thorn 0530

          47

          frac14 082

          66 Bearing capacity

          Thus

          qa frac14 150 082 frac14 120 kN=m2

          (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

          Thus

          qa frac14 90 15 frac14 135 kN=m2

          (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

          Ic frac14 171

          1014frac14 0068

          From Equation 819(a) with s frac14 25mm

          q frac14 25

          3507 0068frac14 150 kN=m2

          810

          Peak value of strain influence factor occurs at a depth of 27m and is given by

          Izp frac14 05thorn 01130

          16 27

          05

          frac14 067

          Refer to Figure Q810

          E frac14 25qc

          Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

          Ez (mm3MN)

          1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

          0203

          C1 frac14 1 0500qnfrac14 1 05 12 16

          130frac14 093

          C2 frac14 1 ethsayTHORN

          s frac14 C1C2qnX Iz

          Ez frac14 093 1 130 0203 frac14 25mm

          Bearing capacity 67

          811

          At pile base level

          cu frac14 220 kN=m2

          qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

          Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

          00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

          qs frac14 00 frac14 07 1226 frac14 86 kN=m2

          Then

          Qf frac14 Abqb thorn Asqs

          frac14

          4 32 1980

          thorn eth 105 139 86THORN

          frac14 13 996thorn 3941 frac14 17 937 kN

          0 01 02 03 04 05 06 07

          0 2 4 6 8 10 12 14

          1

          2

          3

          4

          5

          6

          7

          8

          (1)

          (2)

          (3)

          (4)

          (5)

          qc

          qc

          Iz

          Iz

          (MNm2)

          z (m)

          Figure Q810

          68 Bearing capacity

          Allowable load

          ethaTHORN Qf

          2frac14 17 937

          2frac14 8968 kN

          ethbTHORN Abqb

          3thorn Asqs frac14 13 996

          3thorn 3941 frac14 8606 kN

          ie allowable load frac14 8600 kN

          Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

          According to the limit state method

          Characteristic undrained strength at base level cuk frac14 220

          150kN=m2

          Characteristic base resistance qbk frac14 9cuk frac14 9 220

          150frac14 1320 kN=m2

          Characteristic shaft resistance qsk frac14 00150

          frac14 86

          150frac14 57 kN=m2

          Characteristic base and shaft resistances

          Rbk frac14

          4 32 1320 frac14 9330 kN

          Rsk frac14 105 139 86

          150frac14 2629 kN

          For a bored pile the partial factors are b frac14 160 and s frac14 130

          Design bearing resistance Rcd frac14 9330

          160thorn 2629

          130

          frac14 5831thorn 2022

          frac14 7850 kN

          Adding ethDAb W) the design bearing resistance becomes 9650 kN

          812

          ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

          qs frac14 cu frac14 040 105 frac14 42 kN=m2

          For a single pile

          Qf frac14 Abqb thorn Asqs

          frac14

          4 062 1305

          thorn eth 06 15 42THORN

          frac14 369thorn 1187 frac14 1556 kN

          Bearing capacity 69

          Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

          eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

          (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

          (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

          qbkfrac14 9cuk frac14 9 220

          150frac14 1320 kN=m2

          qskfrac14cuk frac14 040 105

          150frac14 28 kN=m2

          Rbkfrac14

          4 0602 1320 frac14 373 kN

          Rskfrac14 060 15 28 frac14 791 kN

          Rcdfrac14 373

          160thorn 791

          130frac14 233thorn 608 frac14 841 kN

          Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

          Rcd gt Fcd therefore the bearing resistance limit state is satisfied

          (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

          q frac14 21 000

          1762frac14 68 kN=m2

          Immediate settlement

          H

          Bfrac14 15

          176frac14 085

          D

          Bfrac14 13

          176frac14 074

          L

          Bfrac14 1

          Hence from Figure 515

          130 frac14 078 and 131 frac14 041

          70 Bearing capacity

          Thus using Equation 528

          si frac14 078 041 68 176

          65frac14 6mm

          Consolidation settlement

          Layer z (m) Area (m2) (kNm2) mvH (mm)

          1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

          434 (sod)

          Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

          sc frac14 056 434 frac14 24mm

          The total settlement is (6thorn 24) frac14 30mm

          813

          At base level N frac14 26 Then using Equation 830

          qb frac14 40NDb

          Bfrac14 40 26 2

          025frac14 8320 kN=m2

          ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

          Figure Q812

          Bearing capacity 71

          Over the length embedded in sand

          N frac14 21 ie18thorn 24

          2

          Using Equation 831

          qs frac14 2N frac14 2 21 frac14 42 kN=m2

          For a single pile

          Qf frac14 Abqb thorn Asqs

          frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

          For the pile group assuming a group efficiency of 12

          XQf frac14 12 9 604 frac14 6523 kN

          Then the load factor is

          F frac14 6523

          2000thorn 1000frac14 21

          (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

          Characteristic base resistance per unit area qbk frac14 8320

          150frac14 5547 kNm2

          Characteristic shaft resistance per unit area qsk frac14 42

          150frac14 28 kNm2

          Characteristic base and shaft resistances for a single pile

          Rbk frac14 0252 5547 frac14 347 kN

          Rsk frac14 4 025 2 28 frac14 56 kN

          For a driven pile the partial factors are b frac14 s frac14 130

          Design bearing resistance Rcd frac14 347

          130thorn 56

          130frac14 310 kN

          For the pile group Rcd frac14 12 9 310 frac14 3348 kN

          Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

          (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

          From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

          72 Bearing capacity

          N frac14 24thorn 26thorn 34

          3frac14 28

          Ic frac14 171

          2814frac14 0016 ethEquation 818THORN

          s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

          The settlement is less than 20mm therefore the serviceability limit state is satisfied

          814

          Using Equation 841

          Tf frac14 DLcu thorn

          4ethD2 d2THORNcuNc

          frac14 eth 02 5 06 110THORN thorn

          4eth022 012THORN110 9

          frac14 207thorn 23 frac14 230 kN

          Figure Q813

          Bearing capacity 73

          Chapter 9

          Stability of slopes

          91

          Referring to Figure Q91

          W frac14 417 19 frac14 792 kN=m

          Q frac14 20 28 frac14 56 kN=m

          Arc lengthAB frac14

          180 73 90 frac14 115m

          Arc length BC frac14

          180 28 90 frac14 44m

          The factor of safety is given by

          F frac14 rethcuLaTHORNWd1 thornQd2 frac14

          90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

          Depth of tension crack z0 frac14 2cu

          frac14 2 20

          19frac14 21m

          Arc length BD frac14

          180 13

          1

          2 90 frac14 21m

          F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

          The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

          Design resisting moment frac14 rXethcudLaTHORN frac14 90

          14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

          Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

          The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

          92

          u frac14 0

          Depth factor D frac14 11

          9frac14 122

          Using Equation 92 with F frac14 10

          Ns frac14 cu

          FHfrac14 30

          10 19 9frac14 0175

          Hence from Figure 93

          frac14 50

          For F frac14 12

          Ns frac14 30

          12 19 9frac14 0146

          frac14 27

          93

          Refer to Figure Q93

          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

          1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

          74 m

          214 1deg

          213 1deg

          39 m

          WB

          D

          C

          28 m

          21 m

          A

          Q

          Soil (1)Soil (2)

          73deg

          Figure Q91

          Stability of slopes 75

          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

          9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

          599 256 328 1372

          Figure Q93

          76 Stability of slopes

          XW cos frac14 b

          Xh cos frac14 21 2 599 frac14 2516 kN=mX

          W sin frac14 bX

          h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

          Arc length La frac14

          180 57

          1

          2 326 frac14 327m

          The factor of safety is given by

          F frac14 c0La thorn tan0ethW cos ulTHORN

          W sin

          frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

          frac14 091

          According to the limit state method

          0d frac14 tan1tan 32

          125

          frac14 265

          c0 frac14 8

          160frac14 5 kN=m2

          Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

          Design disturbing moment frac14 1075 kN=m

          The design resisting moment is less than the design disturbing moment therefore a slipwill occur

          94

          F frac14 1

          W sin

          Xfc0bthorn ethW ubTHORN tan0g sec

          1thorn ethtan tan0=FTHORN

          c0 frac14 8 kN=m2

          0 frac14 32

          c0b frac14 8 2 frac14 16 kN=m

          W frac14 bh frac14 21 2 h frac14 42h kN=m

          Try F frac14 100

          tan0

          Ffrac14 0625

          Stability of slopes 77

          Values of u are as obtained in Figure Q93

          SliceNo

          h(m)

          W frac14 bh(kNm)

          W sin(kNm)

          ub(kNm)

          c0bthorn (W ub) tan0(kNm)

          sec

          1thorn (tan tan0)FProduct(kNm)

          1 05 21 6 2 8 24 1078 262 13 55 31

          23 33 30 1042 31

          3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

          224 92 72 0931 67

          6 50 210 11 40 100 85 0907 777 55 231 14

          12 58 112 90 0889 80

          8 60 252 1812

          80 114 102 0874 899 63 265 22 99 116 109 0861 94

          10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

          2154 88 116 0853 99

          14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

          1074 1091

          F frac14 1091

          1074frac14 102 (assumed value 100)

          Thus

          F frac14 101

          95

          F frac14 1

          W sin

          XfWeth1 ruTHORN tan0g sec

          1thorn ethtan tan0THORN=F

          0 frac14 33

          ru frac14 020

          W frac14 bh frac14 20 5 h frac14 100h kN=m

          eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

          Try F frac14 110

          tan 0

          Ffrac14 tan 33

          110frac14 0590

          78 Stability of slopes

          Referring to Figure Q95

          SliceNo

          h(m)

          W frac14 bh(kNm)

          W sin(kNm)

          W(1 ru) tan0(kNm)

          sec

          1thorn ( tan tan0)FProduct(kNm)

          1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

          2120 234 0892 209

          4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

          1185 1271

          Figure Q95

          Stability of slopes 79

          F frac14 1271

          1185frac14 107

          The trial value was 110 therefore take F to be 108

          96

          (a) Water table at surface the factor of safety is given by Equation 912

          F frac14 0

          sat

          tan0

          tan

          ptie 15 frac14 92

          19

          tan 36

          tan

          tan frac14 0234

          frac14 13

          Water table well below surface the factor of safety is given by Equation 911

          F frac14 tan0

          tan

          frac14 tan 36

          tan 13

          frac14 31

          (b) 0d frac14 tan1tan 36

          125

          frac14 30

          Depth of potential failure surface frac14 z

          Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

          frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

          frac14 504z kN

          Design disturbing moment per unit area Sd frac14 sat sin cos

          frac14 19 z sin 13 cos 13

          frac14 416z kN

          Rd gtSd therefore the limit state for overall stability is satisfied

          80 Stability of slopes

          • Book Cover
          • Title
          • Contents
          • Basic characteristics of soils
          • Seepage
          • Effective stress
          • Shear strength
          • Stresses and displacements
          • Lateral earth pressure
          • Consolidation theory
          • Bearing capacity
          • Stability of slopes

            Authorrsquos note

            In order not to short-circuit the learningprocess it is vital that the reader shouldattempt the problems before referring to thesolutions in this manual

            Chapter 1

            Basic characteristics of soils

            11

            Soil E consists of 98 coarse material (31 gravel size 67 sand size) and 2 finesIt is classified as SW well-graded gravelly SAND or in greater detail well-gradedslightly silty very gravelly SAND

            Soil F consists of 63 coarse material (2 gravel size 61 sand size) and 37non-plastic fines (ie between 35 and 65 fines) therefore the soil is classified as MSsandy SILT

            Soil G consists of 73 fine material (ie between 65 and 100 fines) and 27 sandsize The liquid limit is 32 and the plasticity index is 8 (ie 32 24) plotting marginallybelow the A-line in the ML zone on the plasticity chart Thus the classification is MLSILT (M-SOIL) of low plasticity (The plasticity chart is given in Figure 17)

            Figure Q11

            Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

            12

            From Equation 117

            1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

            191frac14 155

            e frac14 055

            Using Equation 113

            Sr frac14 wGs

            efrac14 0095 270

            055frac14 0466 eth466THORN

            Using Equation 119

            sat frac14 Gs thorn e1thorn e w frac14 325

            155 100 frac14 210Mg=m3

            From Equation 114

            w frac14 e

            Gsfrac14 055

            270frac14 0204 eth204THORN

            13

            Equations similar to 117ndash120 apply in the case of unit weights thus

            d frac14 Gs

            1thorn e w frac14272

            170 98 frac14 157 kN=m3

            sat frac14 Gs thorn e1thorn e w frac14 342

            170 98 frac14 197 kN=m3

            Using Equation 121

            0 frac14 Gs 1

            1thorn e w frac14 172

            170 98 frac14 99 kN=m3

            Using Equation 118a with Srfrac14 075

            frac14 Gs thorn Sre

            1thorn e w frac14 3245

            170 98 frac14 187 kN=m3

            2 Basic characteristics of soils

            Using Equation 113

            w frac14 Sre

            Gsfrac14 075 070

            272frac14 0193 eth193THORN

            The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

            14

            Volume of specimenfrac14

            438276 frac14 86 200mm3

            Bulk density ethTHORN frac14 Mass

            Volumefrac14 1680

            86 200 103frac14 195Mg=m3

            Water content ethwTHORN frac14 1680 1305

            1305frac14 0287 eth287THORN

            From Equation 117

            1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

            195frac14 180

            e frac14 080

            Using Equation 113

            Sr frac14 wGs

            efrac14 0287 273

            080frac14 098 eth98THORN

            15

            Using Equation 124

            d frac14

            1thorn w frac14215

            112frac14 192Mg=m3

            From Equation 117

            1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

            215frac14 138

            e frac14 038

            Using Equation 113

            Sr frac14 wGs

            efrac14 012 265

            038frac14 0837 eth837THORN

            Basic characteristics of soils 3

            Using Equation 115

            Afrac14 e wGs

            1thorn e frac14038 0318

            138frac14 0045 eth45THORN

            The zero air voids dry density is given by Equation 125

            d frac14 Gs

            1thorn wGsw frac14 265

            1thorn eth0135 265THORN 100 frac14 195Mg=m3

            ie a dry density of 200Mgm3 would not be possible

            16

            Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

            (Mgm3) d10(Mgm3)

            2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

            In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

            wopt frac14 15 and dmaxfrac14 183Mg=m3

            Figure Q16

            4 Basic characteristics of soils

            Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

            d5 d10

            respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

            17

            From Equation 120

            e frac14 Gswd 1

            The maximum and minimum values of void ratio are given by

            emax frac14 Gsw

            dmin

            1

            emin frac14 Gswdmax

            1

            From Equation 123

            ID frac14 Gsweth1=dmin 1=dTHORN

            Gsweth1=dmin 1=dmax

            THORN

            frac14 frac121 ethdmin=dTHORN1=dmin

            frac121 ethdmin=dmax

            THORN1=dmin

            frac14 d dmin

            dmax dmin

            dmax

            d

            frac14 172 154

            181 154

            181

            172

            frac14 070 eth70THORN

            Basic characteristics of soils 5

            Chapter 2

            Seepage

            21

            The coefficient of permeability is determined from the equation

            k frac14 23al

            At1log

            h0

            h1

            where

            a frac14

            4 00052 m2 l frac14 02m

            A frac14

            4 012 m2 t1 frac14 3 602 s

            logh0

            h1frac14 log

            100

            035frac14 0456

            k frac14 23 00052 02 0456

            012 3 602frac14 49 108 m=s

            22

            The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

            q frac14 kh Nf

            Ndfrac14 106 400 37

            11frac14 13 106 m3=s per m

            Figure Q22

            23

            The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

            q frac14 kh Nf

            Ndfrac14 5 105 300 35

            9frac14 58 105 m3=s per m

            The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

            Point h (m) h z (m) u frac14 w(h z)(kNm2)

            1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

            eg for Point 1

            h1 frac14 7

            9 300 frac14 233m

            h1 z1 frac14 233 eth250THORN frac14 483m

            Figure Q23

            Seepage 7

            u1 frac14 98 483 frac14 47 kN=m2

            The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

            24

            The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

            q frac14 kh Nf

            Ndfrac14 40 107 550 10

            11frac14 20 106 m3=s per m

            25

            The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

            q frac14 kh Nf

            Ndfrac14 20 106 500 42

            9frac14 47 106 m3=s per m

            Figure Q24

            8 Seepage

            26

            The scale transformation factor in the x direction is given by Equation 221 ie

            xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

            ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

            Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

            k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

            pfrac14

            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

            p 107 frac14 30 107 m=s

            Thus the quantity of seepage is given by

            q frac14 k0h Nf

            Ndfrac14 30 107 1400 325

            12frac14 11 106 m3=s per m

            Figure Q25

            Seepage 9

            27

            The scale transformation factor in the x direction is

            xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

            ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

            Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

            k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

            pfrac14

            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

            p 106 frac14 45 106 m=s

            The focus of the basic parabola is at point A The parabola passes through point Gsuch that

            GC frac14 03HC frac14 03 30 frac14 90m

            Thus the coordinates of G are

            x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

            480 frac14 x0 2002

            4x0

            Figure Q26

            10 Seepage

            Hence

            x0 frac14 20m

            Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

            x frac14 20 z2

            80

            x 20 0 50 100 200 300z 0 400 748 980 1327 1600

            The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

            No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

            q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

            28

            The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

            Figure Q27

            Seepage 11

            q frac14 kh Nf

            Ndfrac14 45 105 28 33

            7

            frac14 59 105 m3=s per m

            29

            The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

            kx frac14 H1k1 thornH2k2

            H1 thornH2frac14 106

            10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

            kz frac14 H1 thornH2

            H1

            k1thornH2

            k2

            frac14 10

            5

            eth2 106THORN thorn5

            eth16 106THORNfrac14 36 106 m=s

            Then the scale transformation factor is given by

            xt frac14 xffiffiffiffiffikz

            pffiffiffiffiffikx

            p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

            Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

            Figure Q28

            12 Seepage

            k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

            qfrac14

            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

            p 106 frac14 57 106 m=s

            Then the quantity of seepage is given by

            q frac14 k0h Nf

            Ndfrac14 57 106 350 56

            11

            frac14 10 105 m3=s per m

            Figure Q29

            Seepage 13

            Chapter 3

            Effective stress

            31

            Buoyant unit weight

            0 frac14 sat w frac14 20 98 frac14 102 kN=m3

            Effective vertical stress

            0v frac14 5 102 frac14 51 kN=m2 or

            Total vertical stress

            v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

            Pore water pressure

            u frac14 7 98 frac14 686 kN=m2

            Effective vertical stress

            0v frac14 v u frac14 1196 686 frac14 51 kN=m2

            32

            Buoyant unit weight

            0 frac14 sat w frac14 20 98 frac14 102 kN=m3

            Effective vertical stress

            0v frac14 5 102 frac14 51 kN=m2 or

            Total vertical stress

            v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

            Pore water pressure

            u frac14 205 98 frac14 2009 kN=m2

            Effective vertical stress

            0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

            33

            At top of the clay

            v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

            u frac14 2 98 frac14 196 kN=m2

            0v frac14 v u frac14 710 196 frac14 514 kN=m2

            Alternatively

            0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

            0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

            At bottom of the clay

            v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

            u frac14 12 98 frac14 1176 kN=m2

            0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

            NB The alternative method of calculation is not applicable because of the artesiancondition

            Figure Q3132

            Effective stress 15

            34

            0 frac14 20 98 frac14 102 kN=m3

            At 8m depth

            0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

            35

            0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

            0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

            Figure Q33

            Figure Q34

            16 Effective stress

            (a) Immediately after WT rise

            At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

            0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

            At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

            0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

            (b) Several years after WT rise

            At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

            At 8m depth

            0v frac14 940 kN=m2 (as above)

            At 12m depth

            0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

            Figure Q35

            Effective stress 17

            36

            Total weight

            ab frac14 210 kN

            Effective weight

            ac frac14 112 kN

            Resultant boundary water force

            be frac14 119 kN

            Seepage force

            ce frac14 34 kN

            Resultant body force

            ae frac14 99 kN eth73 to horizontalTHORN

            (Refer to Figure Q36)

            Figure Q36

            18 Effective stress

            37

            Situation (1)(a)

            frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

            u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

            0 frac14 u frac14 694 392 frac14 302 kN=m2

            (b)

            i frac14 2

            4frac14 05

            j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

            Situation (2)(a)

            frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

            u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

            0 frac14 u frac14 498 392 frac14 106 kN=m2

            (b)

            i frac14 2

            4frac14 05

            j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

            38

            The flow net is drawn in Figure Q24

            Loss in total head between adjacent equipotentials

            h frac14 550

            Ndfrac14 550

            11frac14 050m

            Exit hydraulic gradient

            ie frac14 h

            sfrac14 050

            070frac14 071

            Effective stress 19

            The critical hydraulic gradient is given by Equation 39

            ic frac14 0

            wfrac14 102

            98frac14 104

            Therefore factor of safety against lsquoboilingrsquo (Equation 311)

            F frac14 iciefrac14 104

            071frac14 15

            Total head at C

            hC frac14 nd

            Ndh frac14 24

            11 550 frac14 120m

            Elevation head at C

            zC frac14 250m

            Pore water pressure at C

            uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

            Therefore effective vertical stress at C

            0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

            For point D

            hD frac14 73

            11 550 frac14 365m

            zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

            0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

            39

            The flow net is drawn in Figure Q25

            For a soil prism 150 300m adjacent to the piling

            hm frac14 26

            9 500 frac14 145m

            20 Effective stress

            Factor of safety against lsquoheavingrsquo (Equation 310)

            F frac14 ic

            imfrac14 0d

            whmfrac14 97 300

            98 145frac14 20

            With a filter

            F frac14 0d thorn wwhm

            3 frac14 eth97 300THORN thorn w98 145

            w frac14 135 kN=m2

            Depth of filterfrac14 13521frac14 065m (if above water level)

            Effective stress 21

            Chapter 4

            Shear strength

            41

            frac14 295 kN=m2

            u frac14 120 kN=m2

            0 frac14 u frac14 295 120 frac14 175 kN=m2

            f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

            42

            03 (kNm2) 1 3 (kNm2) 01 (kNm2)

            100 452 552200 908 1108400 1810 2210800 3624 4424

            The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

            Figure Q42

            43

            3 (kNm2) 1 3 (kNm2) 1 (kNm2)

            200 222 422400 218 618600 220 820

            The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

            44

            The modified shear strength parameters are

            0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

            a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

            The coordinates of the stress point representing failure conditions in the test are

            1

            2eth1 2THORN frac14 1

            2 170 frac14 85 kN=m2

            1

            2eth1 thorn 3THORN frac14 1

            2eth270thorn 100THORN frac14 185 kN=m2

            The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

            uf frac14 36 kN=m2

            Figure Q43

            Figure Q44

            Shear strength 23

            45

            3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

            150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

            The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

            The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

            46

            03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

            200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

            The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

            Figure Q45

            24 Shear strength

            47

            The torque required to produce shear failure is given by

            T frac14 dh cud

            2thorn 2

            Z d=2

            0

            2r drcur

            frac14 cud2h

            2thorn 4cu

            Z d=2

            0

            r2dr

            frac14 cud2h

            2thorn d

            3

            6

            Then

            35 frac14 cu52 10

            2thorn 53

            6

            103

            cu frac14 76 kN=m3

            400

            0 400 800 1200 1600

            τ (k

            Nm

            2 )

            σprime (kNm2)

            34deg

            315deg29deg

            (a)

            (b)

            0 400

            400

            800 1200 1600

            Failure envelope

            300 500

            σprime (kNm2)

            τ (k

            Nm

            2 )

            20 (kNm2)

            31deg

            Figure Q46

            Shear strength 25

            48

            The relevant stress values are calculated as follows

            3 frac14 600 kN=m2

            1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

            2(1 3) 0 40 79 107 139 159

            1

            2(01 thorn 03) 400 411 402 389 351 326

            1

            2(1 thorn 3) 600 640 679 707 739 759

            The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

            Af frac14 433 200

            319frac14 073

            49

            B frac14 u33

            frac14 144

            350 200frac14 096

            a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

            0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

            10 283 65 023

            Figure Q48

            26 Shear strength

            The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

            Figure Q49

            Shear strength 27

            Chapter 5

            Stresses and displacements

            51

            Vertical stress is given by

            z frac14 Qz2Ip frac14 5000

            52Ip

            Values of Ip are obtained from Table 51

            r (m) rz Ip z (kNm2)

            0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

            10 20 0009 2

            The variation of z with radial distance (r) is plotted in Figure Q51

            Figure Q51

            52

            Below the centre load (Figure Q52)

            r

            zfrac14 0 for the 7500-kN load

            Ip frac14 0478

            r

            zfrac14 5

            4frac14 125 for the 10 000- and 9000-kN loads

            Ip frac14 0045

            Then

            z frac14X Q

            z2Ip

            frac14 7500 0478

            42thorn 10 000 0045

            42thorn 9000 0045

            42

            frac14 224thorn 28thorn 25 frac14 277 kN=m2

            53

            The vertical stress under a corner of a rectangular area is given by

            z frac14 qIr

            where values of Ir are obtained from Figure 510 In this case

            z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

            z

            Figure Q52

            Stresses and displacements 29

            z (m) m n Ir z (kNm2)

            0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

            10 010 0005 5

            z is plotted against z in Figure Q53

            54

            (a)

            m frac14 125

            12frac14 104

            n frac14 18

            12frac14 150

            From Figure 510 Irfrac14 0196

            z frac14 2 175 0196 frac14 68 kN=m2

            Figure Q53

            30 Stresses and displacements

            (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

            z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

            55

            Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

            Px frac14 2Q

            1

            m2 thorn 1frac14 2 150

            125frac14 76 kN=m

            Equation 517 is used to obtain the pressure distribution

            px frac14 4Q

            h

            m2n

            ethm2 thorn n2THORN2 frac14150

            m2n

            ethm2 thorn n2THORN2 ethkN=m2THORN

            Figure Q54

            Stresses and displacements 31

            n m2n

            (m2 thorn n2)2

            px(kNm2)

            0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

            The pressure distribution is plotted in Figure Q55

            56

            H

            Bfrac14 10

            2frac14 5

            L

            Bfrac14 4

            2frac14 2

            D

            Bfrac14 1

            2frac14 05

            Hence from Figure 515

            131 frac14 082

            130 frac14 094

            Figure Q55

            32 Stresses and displacements

            The immediate settlement is given by Equation 528

            si frac14 130131qB

            Eu

            frac14 094 082 200 2

            45frac14 7mm

            Stresses and displacements 33

            Chapter 6

            Lateral earth pressure

            61

            For 0 frac14 37 the active pressure coefficient is given by

            Ka frac14 1 sin 37

            1thorn sin 37frac14 025

            The total active thrust (Equation 66a with c0 frac14 0) is

            Pa frac14 1

            2KaH

            2 frac14 1

            2 025 17 62 frac14 765 kN=m

            If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

            K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

            and the thrust on the wall is

            P0 frac14 1

            2K0H

            2 frac14 1

            2 040 17 62 frac14 122 kN=m

            62

            The active pressure coefficients for the three soil types are as follows

            Ka1 frac141 sin 35

            1thorn sin 35frac14 0271

            Ka2 frac141 sin 27

            1thorn sin 27frac14 0375

            ffiffiffiffiffiffiffiKa2

            p frac14 0613

            Ka3 frac141 sin 42

            1thorn sin 42frac14 0198

            Distribution of active pressure (plotted in Figure Q62)

            Depth (m) Soil Active pressure (kNm2)

            3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

            12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

            At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

            Total thrust frac14 571 kNm

            Point of application is (4893571) m from the top of the wall ie 857m

            Force (kN) Arm (m) Moment (kN m)

            (1)1

            2 0271 16 32 frac14 195 20 390

            (2) 0271 16 3 2 frac14 260 40 1040

            (3)1

            2 0271 92 22 frac14 50 433 217

            (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

            (5)1

            2 0375 102 32 frac14 172 70 1204

            (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

            (7)1

            2 0198 112 42 frac14 177 1067 1889

            (8)1

            2 98 92 frac14 3969 90 35721

            5713 48934

            Figure Q62

            Lateral earth pressure 35

            63

            (a) For u frac14 0 Ka frac14 Kp frac14 1

            Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

            frac14 245

            At the lower end of the piling

            pa frac14 Kaqthorn Kasatz Kaccu

            frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

            frac14 115 kN=m2

            pp frac14 Kpsatzthorn Kpccu

            frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

            frac14 202 kN=m2

            (b) For 0 frac14 26 and frac14 1

            20

            Ka frac14 035

            Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

            pfrac14 145 ethEquation 619THORN

            Kp frac14 37

            Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

            pfrac14 47 ethEquation 624THORN

            At the lower end of the piling

            pa frac14 Kaqthorn Ka0z Kacc

            0

            frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

            frac14 187 kN=m2

            pp frac14 Kp0zthorn Kpcc

            0

            frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

            frac14 198 kN=m2

            36 Lateral earth pressure

            64

            (a) For 0 frac14 38 Ka frac14 024

            0 frac14 20 98 frac14 102 kN=m3

            The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

            Force (kN) Arm (m) Moment (kN m)

            (1) 024 10 66 frac14 159 33 525

            (2)1

            2 024 17 392 frac14 310 400 1240

            (3) 024 17 39 27 frac14 430 135 580

            (4)1

            2 024 102 272 frac14 89 090 80

            (5)1

            2 98 272 frac14 357 090 321

            Hfrac14 1345 MH frac14 2746

            (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

            (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

            XM frac14MV MH frac14 7790 kNm

            Lever arm of base resultant

            M

            Vfrac14 779

            488frac14 160

            Eccentricity of base resultant

            e frac14 200 160 frac14 040m

            39 m

            27 m

            40 m

            04 m

            04 m

            26 m

            (7)

            (9)

            (1)(2)

            (3)

            (4)

            (5)

            (8)(6)

            (10)

            WT

            10 kNm2

            Hydrostatic

            Figure Q64

            Lateral earth pressure 37

            Base pressures (Equation 627)

            p frac14 VB

            1 6e

            B

            frac14 488

            4eth1 060THORN

            frac14 195 kN=m2 and 49 kN=m2

            Factor of safety against sliding (Equation 628)

            F frac14 V tan

            Hfrac14 488 tan 25

            1345frac14 17

            (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

            Hfrac14 1633 kN

            V frac14 4879 kN

            MH frac14 3453 kNm

            MV frac14 10536 kNm

            The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

            65

            For 0 frac14 36 Ka frac14 026 and Kp frac14 385

            Kp

            Ffrac14 385

            2

            0 frac14 20 98 frac14 102 kN=m3

            The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

            Force (kN) Arm (m) Moment (kN m)

            (1)1

            2 026 17 452 frac14 448 dthorn 15 448dthorn 672

            (2) 026 17 45 d frac14 199d d2 995d2

            (3)1

            2 026 102 d2 frac14 133d2 d3 044d3

            (4)1

            2 385

            2 17 152 frac14 368 dthorn 05 368d 184

            (5)385

            2 17 15 d frac14 491d d2 2455d2

            (6)1

            2 385

            2 102 d2 frac14 982d2 d3 327d3

            38 Lateral earth pressure

            XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

            d3 thorn 516d2 283d 1724 frac14 0

            d frac14 179m

            Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

            Over additional 20 embedded depth

            pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

            Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

            66

            The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

            Ka frac14 sin 69=sin 105

            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

            pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

            26664

            37775

            2

            frac14 050

            The total active thrust (acting at 25 above the normal) is given by Equation 616

            Pa frac14 1

            2 050 19 7502 frac14 267 kN=m

            Figure Q65

            Lateral earth pressure 39

            Horizontal component

            Ph frac14 267 cos 40 frac14 205 kN=m

            Vertical component

            Pv frac14 267 sin 40 frac14 172 kN=m

            Consider moments about the toe of the wall (Figure Q66) (per m)

            Force (kN) Arm (m) Moment (kN m)

            (1)1

            2 175 650 235 frac14 1337 258 345

            (2) 050 650 235 frac14 764 175 134

            (3)1

            2 070 650 235 frac14 535 127 68

            (4) 100 400 235 frac14 940 200 188

            (5) 1

            2 080 050 235 frac14 47 027 1

            Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

            Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

            Lever arm of base resultant

            M

            Vfrac14 795

            525frac14 151m

            Eccentricity of base resultant

            e frac14 200 151 frac14 049m

            Figure Q66

            40 Lateral earth pressure

            Base pressures (Equation 627)

            p frac14 525

            41 6 049

            4

            frac14 228 kN=m2 and 35 kN=m2

            The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

            The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

            The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

            67

            For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

            Force (kN) Arm (m) Moment (kNm)

            (1)1

            2 027 17 52 frac14 574 183 1050

            (2) 027 17 5 3 frac14 689 500 3445

            (3)1

            2 027 102 32 frac14 124 550 682

            (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

            (5)1

            2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

            (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

            (7) 1

            2 267

            2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

            (8) 2 10ffiffiffiffiffiffiffiffiffi267p

            2 d frac14 163d d2thorn 650 82d2 1060d

            Tie rod force per m frac14 T 0 0

            XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

            d3 thorn 77d2 269d 1438 frac14 0

            d frac14 467m

            Depth of penetration frac14 12d frac14 560m

            Lateral earth pressure 41

            Algebraic sum of forces for d frac14 467m isX

            F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

            T frac14 905 kN=m

            Force in each tie rod frac14 25T frac14 226 kN

            68

            (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

            0 frac14 21 98 frac14 112 kN=m3

            The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

            uC frac14 150

            165 15 98 frac14 134 kN=m2

            The average seepage pressure is

            j frac14 15

            165 98 frac14 09 kN=m3

            Hence

            0 thorn j frac14 112thorn 09 frac14 121 kN=m3

            0 j frac14 112 09 frac14 103 kN=m3

            Figure Q67

            42 Lateral earth pressure

            Consider moments about the anchor point A (per m)

            Force (kN) Arm (m) Moment (kN m)

            (1) 10 026 150 frac14 390 60 2340

            (2)1

            2 026 18 452 frac14 474 15 711

            (3) 026 18 45 105 frac14 2211 825 18240

            (4)1

            2 026 121 1052 frac14 1734 100 17340

            (5)1

            2 134 15 frac14 101 40 404

            (6) 134 30 frac14 402 60 2412

            (7)1

            2 134 60 frac14 402 95 3819

            571 4527(8) Ppm

            115 115PPm

            XM frac14 0

            Ppm frac144527

            115frac14 394 kN=m

            Available passive resistance

            Pp frac14 1

            2 385 103 62 frac14 714 kN=m

            Factor of safety

            Fp frac14 Pp

            Ppm

            frac14 714

            394frac14 18

            Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

            Figure Q68

            Lateral earth pressure 43

            (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

            Consider moments (per m) about the tie point A

            Force (kN) Arm (m)

            (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

            (2)1

            2 033 18 452 frac14 601 15

            (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

            (4)1

            2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

            (5)1

            2 134 15 frac14 101 40

            (6) 134 30 frac14 402 60

            (7)1

            2 134 d frac14 67d d3thorn 75

            (8) 1

            2 30 103 d2 frac141545d2 2d3thorn 75

            Moment (kN m)

            (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

            XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

            d3 thorn 827d2 466d 1518 frac14 0

            By trial

            d frac14 544m

            The minimum depth of embedment required is 544m

            69

            For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

            0 frac14 20 98 frac14 102 kN=m3

            The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

            44 Lateral earth pressure

            uC frac14 147

            173 26 98 frac14 216 kN=m2

            and the average seepage pressure around the wall is

            j frac14 26

            173 98 frac14 15 kN=m3

            Consider moments about the prop (A) (per m)

            Force (kN) Arm (m) Moment (kN m)

            (1)1

            2 03 17 272 frac14 186 020 37

            (2) 03 17 27 53 frac14 730 335 2445

            (3)1

            2 03 (102thorn 15) 532 frac14 493 423 2085

            (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

            (5)1

            2 216 26 frac14 281 243 684

            (6) 216 27 frac14 583 465 2712

            (7)1

            2 216 60 frac14 648 800 5184

            3055(8)

            1

            2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

            Factor of safety

            Fr frac14 6885

            3055frac14 225

            Figure Q69

            Lateral earth pressure 45

            610

            For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

            p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

            Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

            Using the recommendations of Twine and Roscoe

            p frac14 02H frac14 02 19 9 frac14 342 kN=m2

            Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

            611

            frac14 18 kN=m3 0 frac14 34

            H frac14 350m nH frac14 335m mH frac14 185m

            Consider a trial value of F frac14 20 Refer to Figure 635

            0m frac14 tan1tan 34

            20

            frac14 186

            Then

            frac14 45 thorn 0m2frac14 543

            W frac14 1

            2 18 3502 cot 543 frac14 792 kN=m

            Figure Q610

            46 Lateral earth pressure

            P frac14 1

            2 s 3352 frac14 561s kN=m

            U frac14 1

            2 98 1852 cosec 543 frac14 206 kN=m

            Equations 630 and 631 then become

            561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

            792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

            ie

            561s 0616N 405 frac14 0

            792 0857N thorn 563 frac14 0

            N frac14 848

            0857frac14 989 kN=m

            Then

            561s 609 405 frac14 0

            s frac14 649

            561frac14 116 kN=m3

            The calculations for trial values of F of 20 15 and 10 are summarized below

            F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

            20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

            s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

            Figure Q611

            Lateral earth pressure 47

            612

            For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

            45 thorn 0

            2frac14 63

            For the retained material between the surface and a depth of 36m

            Pa frac14 1

            2 030 18 362 frac14 350 kN=m

            Weight of reinforced fill between the surface and a depth of 36m is

            Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

            eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

            Eccentricity of Rv

            e frac14 263 250 frac14 013m

            The average vertical stress at a depth of 36m is

            z frac14 Rv

            L 2efrac14 324

            474frac14 68 kN=m2

            (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

            Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

            Tensile stress in the element frac14 138 103

            65 3frac14 71N=mm2

            Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

            Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

            Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

            The weight of ABC is

            W frac14 1

            2 18 52 265 frac14 124 kN=m

            From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

            48 Lateral earth pressure

            (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

            Tp frac14 032 68 120 065 frac14 170 kN

            Tr frac14 213 420

            418frac14 214 kN

            Again the tensile failure and slipping limit states are satisfied for this element

            Figure Q612

            Lateral earth pressure 49

            Chapter 7

            Consolidation theory

            71

            Total change in thickness

            H frac14 782 602 frac14 180mm

            Average thickness frac14 1530thorn 180

            2frac14 1620mm

            Length of drainage path d frac14 1620

            2frac14 810mm

            Root time plot (Figure Q71a)

            ffiffiffiffiffiffit90p frac14 33

            t90 frac14 109min

            cv frac14 0848d2

            t90frac14 0848 8102

            109 1440 365

            106frac14 27m2=year

            r0 frac14 782 764

            782 602frac14 018

            180frac14 0100

            rp frac14 10eth764 645THORN9eth782 602THORN frac14

            10 119

            9 180frac14 0735

            rs frac14 1 eth0100thorn 0735THORN frac14 0165

            Log time plot (Figure Q71b)

            t50 frac14 26min

            cv frac14 0196d2

            t50frac14 0196 8102

            26 1440 365

            106frac14 26m2=year

            r0 frac14 782 763

            782 602frac14 019

            180frac14 0106

            rp frac14 763 623

            782 602frac14 140

            180frac14 0778

            rs frac14 1 eth0106thorn 0778THORN frac14 0116

            Figure Q71(a)

            Figure Q71(b)

            Final void ratio

            e1 frac14 w1Gs frac14 0232 272 frac14 0631

            e

            Hfrac14 1thorn e0

            H0frac14 1thorn e1 thorne

            H0

            ie

            e

            180frac14 1631thorne

            1710

            e frac14 2936

            1530frac14 0192

            Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

            mv frac14 1

            1thorn e0 e0 e101 00

            frac14 1

            1823 0192

            0107frac14 098m2=MN

            k frac14 cvmvw frac14 265 098 98

            60 1440 365 103frac14 81 1010 m=s

            72

            Using Equation 77 (one-dimensional method)

            sc frac14 e0 e11thorn e0 H

            Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

            Figure Q72

            52 Consolidation theory

            Settlement

            Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

            1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

            318

            Notes 5 92y 460thorn 84

            Heave

            Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

            1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

            38

            73

            U frac14 f ethTvTHORN frac14 f cvt

            d2

            Hence if cv is constant

            t1

            t2frac14 d

            21

            d22

            where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

            d1 frac14 95mm and d2 frac14 2500mm

            for U frac14 050 t2 frac14 t1 d22

            d21

            frac14 20

            60 24 365 25002

            952frac14 263 years

            for U lt 060 Tv frac14

            4U2 (Equation 724(a))

            t030 frac14 t050 0302

            0502

            frac14 263 036 frac14 095 years

            Consolidation theory 53

            74

            The layer is open

            d frac14 8

            2frac14 4m

            Tv frac14 cvtd2frac14 24 3

            42frac14 0450

            ui frac14 frac14 84 kN=m2

            The excess pore water pressure is given by Equation 721

            ue frac14Xmfrac141mfrac140

            2ui

            Msin

            Mz

            d

            expethM2TvTHORN

            In this case z frac14 d

            sinMz

            d

            frac14 sinM

            where

            M frac14

            23

            25

            2

            M sin M M2Tv exp (M2Tv)

            2thorn1 1110 0329

            3

            21 9993 457 105

            ue frac14 2 84 2

            1 0329 ethother terms negligibleTHORN

            frac14 352 kN=m2

            75

            The layer is open

            d frac14 6

            2frac14 3m

            Tv frac14 cvtd2frac14 10 3

            32frac14 0333

            The layer thickness will be divided into six equal parts ie m frac14 6

            54 Consolidation theory

            For an open layer

            Tv frac14 4n

            m2

            n frac14 0333 62

            4frac14 300

            The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

            ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

            i j

            0 1 2 3 4 5 6 7 8 9 10 11 12

            0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

            The initial and 3-year isochrones are plotted in Figure Q75

            Area under initial isochrone frac14 180 units

            Area under 3-year isochrone frac14 63 units

            The average degree of consolidation is given by Equation 725Thus

            U frac14 1 63

            180frac14 065

            Figure Q75

            Consolidation theory 55

            76

            At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

            0 frac14 2w frac14 2 98 frac14 196 kN=m2

            The final consolidation settlement (one-dimensional method) is

            sc frac14 mv0H frac14 083 196 8 frac14 130mm

            Corrected time t frac14 2 1

            2

            40

            52

            frac14 1615 years

            Tv frac14 cvtd2frac14 44 1615

            42frac14 0444

            From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

            77

            The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

            Figure Q77

            56 Consolidation theory

            Point m n Ir (kNm2) sc (mm)

            13020frac14 15 20

            20frac14 10 0194 (4) 113 124

            260

            20frac14 30

            20

            20frac14 10 0204 (2) 59 65

            360

            20frac14 30

            40

            20frac14 20 0238 (1) 35 38

            430

            20frac14 15

            40

            20frac14 20 0224 (2) 65 72

            Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

            78

            Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

            (a) Immediate settlement

            H

            Bfrac14 30

            35frac14 086

            D

            Bfrac14 2

            35frac14 006

            Figure Q78

            Consolidation theory 57

            From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

            si frac14 130131qB

            Eufrac14 10 032 105 35

            40frac14 30mm

            (b) Consolidation settlement

            Layer z (m) Dz Ic (kNm2) syod (mm)

            1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

            3150

            Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

            Now

            H

            Bfrac14 30

            35frac14 086 and A frac14 065

            from Figure 712 13 frac14 079

            sc frac14 13sod frac14 079 315 frac14 250mm

            Total settlement

            s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

            79

            Without sand drains

            Uv frac14 025

            Tv frac14 0049 ethfrom Figure 718THORN

            t frac14 Tvd2

            cvfrac14 0049 82

            cvWith sand drains

            R frac14 0564S frac14 0564 3 frac14 169m

            n frac14 Rrfrac14 169

            015frac14 113

            Tr frac14 cht

            4R2frac14 ch

            4 1692 0049 82

            cvethand ch frac14 cvTHORN

            frac14 0275

            Ur frac14 073 (from Figure 730)

            58 Consolidation theory

            Using Equation 740

            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

            U frac14 080

            710

            Without sand drains

            Uv frac14 090

            Tv frac14 0848

            t frac14 Tvd2

            cvfrac14 0848 102

            96frac14 88 years

            With sand drains

            R frac14 0564S frac14 0564 4 frac14 226m

            n frac14 Rrfrac14 226

            015frac14 15

            Tr

            Tvfrac14 chcv

            d2

            4R2ethsame tTHORN

            Tr

            Tvfrac14 140

            96 102

            4 2262frac14 714 eth1THORN

            Using Equation 740

            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

            An iterative solution is required using (1) and (2) an initial value ofUv being estimated

            Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

            040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

            Thus

            Uv frac14 0295 and Ur frac14 086

            t frac14 88 00683

            0848frac14 07 years

            Consolidation theory 59

            Chapter 8

            Bearing capacity

            81

            (a) The ultimate bearing capacity is given by Equation 83

            qf frac14 cNc thorn DNq thorn 1

            2BN

            For u frac14 0

            Nc frac14 514 Nq frac14 1 N frac14 0

            qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

            The net ultimate bearing capacity is

            qnf frac14 qf D frac14 540 kN=m2

            The net foundation pressure is

            qn frac14 q D frac14 425

            2 eth21 1THORN frac14 192 kN=m2

            The factor of safety (Equation 86) is

            F frac14 qnfqnfrac14 540

            192frac14 28

            (b) For 0 frac14 28

            Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

            qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

            2 112 2 13

            frac14 260thorn 168thorn 146 frac14 574 kN=m2

            qnf frac14 574 112 frac14 563 kN=m2

            F frac14 563

            192frac14 29

            (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

            82

            For 0 frac14 38

            Nq frac14 49 N frac14 67

            qnf frac14 DethNq 1THORN thorn 1

            2BN ethfrom Equation 83THORN

            frac14 eth18 075 48THORN thorn 1

            2 18 15 67

            frac14 648thorn 905 frac14 1553 kN=m2

            qn frac14 500

            15 eth18 075THORN frac14 320 kN=m2

            F frac14 qnfqnfrac14 1553

            320frac14 48

            0d frac14 tan1tan 38

            125

            frac14 32 therefore Nq frac14 23 and N frac14 25

            Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

            2 18 15 25

            frac14 15eth310thorn 337THORNfrac14 970 kN=m

            Design load (action) Vd frac14 500 kN=m

            The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

            83

            D

            Bfrac14 350

            225frac14 155

            From Figure 85 for a square foundation

            Nc frac14 81

            Bearing capacity 61

            For a rectangular foundation (L frac14 450m B frac14 225m)

            Nc frac14 084thorn 016B

            L

            81 frac14 745

            Using Equation 810

            qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

            For F frac14 3

            qn frac14 1006

            3frac14 335 kN=m2

            q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

            Design load frac14 405 450 225 frac14 4100 kN

            Design undrained strength cud frac14 135

            14frac14 96 kN=m2

            Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

            frac14 7241 kN

            Design load Vd frac14 4100 kN

            Rd gt Vd therefore the bearing resistance limit state is satisfied

            84

            For 0 frac14 40

            Nq frac14 64 N frac14 95

            qnf frac14 DethNq 1THORN thorn 04BN

            (a) Water table 5m below ground level

            qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

            qn frac14 400 17 frac14 383 kN=m2

            F frac14 2686

            383frac14 70

            (b) Water table 1m below ground level (ie at foundation level)

            0 frac14 20 98 frac14 102 kN=m3

            62 Bearing capacity

            qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

            F frac14 2040

            383frac14 53

            (c) Water table at ground level with upward hydraulic gradient 02

            eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

            qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

            F frac14 1296

            392frac14 33

            85

            The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

            Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

            Design value of 0 frac14 tan1tan 39

            125

            frac14 33

            For 0 frac14 33 Nq frac14 26 and N frac14 29

            Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

            Rd gt Vd therefore the bearing resistance limit state is satisfied

            86

            (a) Undrained shear for u frac14 0

            Nc frac14 514 Nq frac14 1 N frac14 0

            qnf frac14 12cuNc

            frac14 12 100 514 frac14 617 kN=m2

            qn frac14 qnfFfrac14 617

            3frac14 206 kN=m2

            q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

            Bearing capacity 63

            Drained shear for 0 frac14 32

            Nq frac14 23 N frac14 25

            0 frac14 21 98 frac14 112 kN=m3

            qnf frac14 0DethNq 1THORN thorn 040BN

            frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

            frac14 694 kN=m2

            q frac14 694

            3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

            Design load frac14 42 227 frac14 3632 kN

            (b) Design undrained strength cud frac14 100

            14frac14 71 kNm2

            Design bearing resistance Rd frac14 12cudNe area

            frac14 12 71 514 42

            frac14 7007 kN

            For drained shear 0d frac14 tan1tan 32

            125

            frac14 26

            Nq frac14 12 N frac14 10

            Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

            (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

            Layer z (m) m n Ir 0 (kNm2) sod (mm)

            1 2 100 0175 0700qn 0182qn

            2 6 033 0044 0176qn 0046qn

            3 10 020 0017 0068qn 0018qn

            0246qn

            Diameter of equivalent circle B frac14 45m

            H

            Bfrac14 12

            45frac14 27 and A frac14 042

            13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

            64 Bearing capacity

            For sc frac14 30mm

            qn frac14 30

            0147frac14 204 kN=m2

            q frac14 204thorn 21 frac14 225 kN=m2

            Design load frac14 42 225 frac14 3600 kN

            The design load is 3600 kN settlement being the limiting criterion

            87

            D

            Bfrac14 8

            4frac14 20

            From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

            F frac14 cuNc

            Dfrac14 40 71

            20 8frac14 18

            88

            Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

            Design value of 0 frac14 tan1tan 38

            125

            frac14 32

            Figure Q86

            Bearing capacity 65

            For 0 frac14 32 Nq frac14 23 and N frac14 25

            Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

            The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

            Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

            For B frac14 250m qn frac14 3750

            2502 17 frac14 583 kN=m2

            From Figure 510 m frac14 n frac14 126

            6frac14 021

            Ir frac14 0019

            Stress increment frac14 4 0019 583 frac14 44 kN=m2

            Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

            The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

            The settlement is less than 20mm therefore the serviceability limit state is satisfied

            89

            Depth (m) N 0v (kNm2) CN N1

            070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

            Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

            (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

            Cw frac14 05thorn 0530

            47

            frac14 082

            66 Bearing capacity

            Thus

            qa frac14 150 082 frac14 120 kN=m2

            (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

            Thus

            qa frac14 90 15 frac14 135 kN=m2

            (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

            Ic frac14 171

            1014frac14 0068

            From Equation 819(a) with s frac14 25mm

            q frac14 25

            3507 0068frac14 150 kN=m2

            810

            Peak value of strain influence factor occurs at a depth of 27m and is given by

            Izp frac14 05thorn 01130

            16 27

            05

            frac14 067

            Refer to Figure Q810

            E frac14 25qc

            Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

            Ez (mm3MN)

            1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

            0203

            C1 frac14 1 0500qnfrac14 1 05 12 16

            130frac14 093

            C2 frac14 1 ethsayTHORN

            s frac14 C1C2qnX Iz

            Ez frac14 093 1 130 0203 frac14 25mm

            Bearing capacity 67

            811

            At pile base level

            cu frac14 220 kN=m2

            qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

            Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

            00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

            qs frac14 00 frac14 07 1226 frac14 86 kN=m2

            Then

            Qf frac14 Abqb thorn Asqs

            frac14

            4 32 1980

            thorn eth 105 139 86THORN

            frac14 13 996thorn 3941 frac14 17 937 kN

            0 01 02 03 04 05 06 07

            0 2 4 6 8 10 12 14

            1

            2

            3

            4

            5

            6

            7

            8

            (1)

            (2)

            (3)

            (4)

            (5)

            qc

            qc

            Iz

            Iz

            (MNm2)

            z (m)

            Figure Q810

            68 Bearing capacity

            Allowable load

            ethaTHORN Qf

            2frac14 17 937

            2frac14 8968 kN

            ethbTHORN Abqb

            3thorn Asqs frac14 13 996

            3thorn 3941 frac14 8606 kN

            ie allowable load frac14 8600 kN

            Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

            According to the limit state method

            Characteristic undrained strength at base level cuk frac14 220

            150kN=m2

            Characteristic base resistance qbk frac14 9cuk frac14 9 220

            150frac14 1320 kN=m2

            Characteristic shaft resistance qsk frac14 00150

            frac14 86

            150frac14 57 kN=m2

            Characteristic base and shaft resistances

            Rbk frac14

            4 32 1320 frac14 9330 kN

            Rsk frac14 105 139 86

            150frac14 2629 kN

            For a bored pile the partial factors are b frac14 160 and s frac14 130

            Design bearing resistance Rcd frac14 9330

            160thorn 2629

            130

            frac14 5831thorn 2022

            frac14 7850 kN

            Adding ethDAb W) the design bearing resistance becomes 9650 kN

            812

            ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

            qs frac14 cu frac14 040 105 frac14 42 kN=m2

            For a single pile

            Qf frac14 Abqb thorn Asqs

            frac14

            4 062 1305

            thorn eth 06 15 42THORN

            frac14 369thorn 1187 frac14 1556 kN

            Bearing capacity 69

            Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

            eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

            (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

            (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

            qbkfrac14 9cuk frac14 9 220

            150frac14 1320 kN=m2

            qskfrac14cuk frac14 040 105

            150frac14 28 kN=m2

            Rbkfrac14

            4 0602 1320 frac14 373 kN

            Rskfrac14 060 15 28 frac14 791 kN

            Rcdfrac14 373

            160thorn 791

            130frac14 233thorn 608 frac14 841 kN

            Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

            Rcd gt Fcd therefore the bearing resistance limit state is satisfied

            (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

            q frac14 21 000

            1762frac14 68 kN=m2

            Immediate settlement

            H

            Bfrac14 15

            176frac14 085

            D

            Bfrac14 13

            176frac14 074

            L

            Bfrac14 1

            Hence from Figure 515

            130 frac14 078 and 131 frac14 041

            70 Bearing capacity

            Thus using Equation 528

            si frac14 078 041 68 176

            65frac14 6mm

            Consolidation settlement

            Layer z (m) Area (m2) (kNm2) mvH (mm)

            1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

            434 (sod)

            Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

            sc frac14 056 434 frac14 24mm

            The total settlement is (6thorn 24) frac14 30mm

            813

            At base level N frac14 26 Then using Equation 830

            qb frac14 40NDb

            Bfrac14 40 26 2

            025frac14 8320 kN=m2

            ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

            Figure Q812

            Bearing capacity 71

            Over the length embedded in sand

            N frac14 21 ie18thorn 24

            2

            Using Equation 831

            qs frac14 2N frac14 2 21 frac14 42 kN=m2

            For a single pile

            Qf frac14 Abqb thorn Asqs

            frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

            For the pile group assuming a group efficiency of 12

            XQf frac14 12 9 604 frac14 6523 kN

            Then the load factor is

            F frac14 6523

            2000thorn 1000frac14 21

            (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

            Characteristic base resistance per unit area qbk frac14 8320

            150frac14 5547 kNm2

            Characteristic shaft resistance per unit area qsk frac14 42

            150frac14 28 kNm2

            Characteristic base and shaft resistances for a single pile

            Rbk frac14 0252 5547 frac14 347 kN

            Rsk frac14 4 025 2 28 frac14 56 kN

            For a driven pile the partial factors are b frac14 s frac14 130

            Design bearing resistance Rcd frac14 347

            130thorn 56

            130frac14 310 kN

            For the pile group Rcd frac14 12 9 310 frac14 3348 kN

            Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

            (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

            From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

            72 Bearing capacity

            N frac14 24thorn 26thorn 34

            3frac14 28

            Ic frac14 171

            2814frac14 0016 ethEquation 818THORN

            s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

            The settlement is less than 20mm therefore the serviceability limit state is satisfied

            814

            Using Equation 841

            Tf frac14 DLcu thorn

            4ethD2 d2THORNcuNc

            frac14 eth 02 5 06 110THORN thorn

            4eth022 012THORN110 9

            frac14 207thorn 23 frac14 230 kN

            Figure Q813

            Bearing capacity 73

            Chapter 9

            Stability of slopes

            91

            Referring to Figure Q91

            W frac14 417 19 frac14 792 kN=m

            Q frac14 20 28 frac14 56 kN=m

            Arc lengthAB frac14

            180 73 90 frac14 115m

            Arc length BC frac14

            180 28 90 frac14 44m

            The factor of safety is given by

            F frac14 rethcuLaTHORNWd1 thornQd2 frac14

            90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

            Depth of tension crack z0 frac14 2cu

            frac14 2 20

            19frac14 21m

            Arc length BD frac14

            180 13

            1

            2 90 frac14 21m

            F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

            The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

            Design resisting moment frac14 rXethcudLaTHORN frac14 90

            14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

            Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

            The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

            92

            u frac14 0

            Depth factor D frac14 11

            9frac14 122

            Using Equation 92 with F frac14 10

            Ns frac14 cu

            FHfrac14 30

            10 19 9frac14 0175

            Hence from Figure 93

            frac14 50

            For F frac14 12

            Ns frac14 30

            12 19 9frac14 0146

            frac14 27

            93

            Refer to Figure Q93

            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

            1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

            74 m

            214 1deg

            213 1deg

            39 m

            WB

            D

            C

            28 m

            21 m

            A

            Q

            Soil (1)Soil (2)

            73deg

            Figure Q91

            Stability of slopes 75

            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

            9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

            599 256 328 1372

            Figure Q93

            76 Stability of slopes

            XW cos frac14 b

            Xh cos frac14 21 2 599 frac14 2516 kN=mX

            W sin frac14 bX

            h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

            Arc length La frac14

            180 57

            1

            2 326 frac14 327m

            The factor of safety is given by

            F frac14 c0La thorn tan0ethW cos ulTHORN

            W sin

            frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

            frac14 091

            According to the limit state method

            0d frac14 tan1tan 32

            125

            frac14 265

            c0 frac14 8

            160frac14 5 kN=m2

            Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

            Design disturbing moment frac14 1075 kN=m

            The design resisting moment is less than the design disturbing moment therefore a slipwill occur

            94

            F frac14 1

            W sin

            Xfc0bthorn ethW ubTHORN tan0g sec

            1thorn ethtan tan0=FTHORN

            c0 frac14 8 kN=m2

            0 frac14 32

            c0b frac14 8 2 frac14 16 kN=m

            W frac14 bh frac14 21 2 h frac14 42h kN=m

            Try F frac14 100

            tan0

            Ffrac14 0625

            Stability of slopes 77

            Values of u are as obtained in Figure Q93

            SliceNo

            h(m)

            W frac14 bh(kNm)

            W sin(kNm)

            ub(kNm)

            c0bthorn (W ub) tan0(kNm)

            sec

            1thorn (tan tan0)FProduct(kNm)

            1 05 21 6 2 8 24 1078 262 13 55 31

            23 33 30 1042 31

            3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

            224 92 72 0931 67

            6 50 210 11 40 100 85 0907 777 55 231 14

            12 58 112 90 0889 80

            8 60 252 1812

            80 114 102 0874 899 63 265 22 99 116 109 0861 94

            10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

            2154 88 116 0853 99

            14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

            1074 1091

            F frac14 1091

            1074frac14 102 (assumed value 100)

            Thus

            F frac14 101

            95

            F frac14 1

            W sin

            XfWeth1 ruTHORN tan0g sec

            1thorn ethtan tan0THORN=F

            0 frac14 33

            ru frac14 020

            W frac14 bh frac14 20 5 h frac14 100h kN=m

            eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

            Try F frac14 110

            tan 0

            Ffrac14 tan 33

            110frac14 0590

            78 Stability of slopes

            Referring to Figure Q95

            SliceNo

            h(m)

            W frac14 bh(kNm)

            W sin(kNm)

            W(1 ru) tan0(kNm)

            sec

            1thorn ( tan tan0)FProduct(kNm)

            1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

            2120 234 0892 209

            4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

            1185 1271

            Figure Q95

            Stability of slopes 79

            F frac14 1271

            1185frac14 107

            The trial value was 110 therefore take F to be 108

            96

            (a) Water table at surface the factor of safety is given by Equation 912

            F frac14 0

            sat

            tan0

            tan

            ptie 15 frac14 92

            19

            tan 36

            tan

            tan frac14 0234

            frac14 13

            Water table well below surface the factor of safety is given by Equation 911

            F frac14 tan0

            tan

            frac14 tan 36

            tan 13

            frac14 31

            (b) 0d frac14 tan1tan 36

            125

            frac14 30

            Depth of potential failure surface frac14 z

            Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

            frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

            frac14 504z kN

            Design disturbing moment per unit area Sd frac14 sat sin cos

            frac14 19 z sin 13 cos 13

            frac14 416z kN

            Rd gtSd therefore the limit state for overall stability is satisfied

            80 Stability of slopes

            • Book Cover
            • Title
            • Contents
            • Basic characteristics of soils
            • Seepage
            • Effective stress
            • Shear strength
            • Stresses and displacements
            • Lateral earth pressure
            • Consolidation theory
            • Bearing capacity
            • Stability of slopes

              Chapter 1

              Basic characteristics of soils

              11

              Soil E consists of 98 coarse material (31 gravel size 67 sand size) and 2 finesIt is classified as SW well-graded gravelly SAND or in greater detail well-gradedslightly silty very gravelly SAND

              Soil F consists of 63 coarse material (2 gravel size 61 sand size) and 37non-plastic fines (ie between 35 and 65 fines) therefore the soil is classified as MSsandy SILT

              Soil G consists of 73 fine material (ie between 65 and 100 fines) and 27 sandsize The liquid limit is 32 and the plasticity index is 8 (ie 32 24) plotting marginallybelow the A-line in the ML zone on the plasticity chart Thus the classification is MLSILT (M-SOIL) of low plasticity (The plasticity chart is given in Figure 17)

              Figure Q11

              Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

              12

              From Equation 117

              1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

              191frac14 155

              e frac14 055

              Using Equation 113

              Sr frac14 wGs

              efrac14 0095 270

              055frac14 0466 eth466THORN

              Using Equation 119

              sat frac14 Gs thorn e1thorn e w frac14 325

              155 100 frac14 210Mg=m3

              From Equation 114

              w frac14 e

              Gsfrac14 055

              270frac14 0204 eth204THORN

              13

              Equations similar to 117ndash120 apply in the case of unit weights thus

              d frac14 Gs

              1thorn e w frac14272

              170 98 frac14 157 kN=m3

              sat frac14 Gs thorn e1thorn e w frac14 342

              170 98 frac14 197 kN=m3

              Using Equation 121

              0 frac14 Gs 1

              1thorn e w frac14 172

              170 98 frac14 99 kN=m3

              Using Equation 118a with Srfrac14 075

              frac14 Gs thorn Sre

              1thorn e w frac14 3245

              170 98 frac14 187 kN=m3

              2 Basic characteristics of soils

              Using Equation 113

              w frac14 Sre

              Gsfrac14 075 070

              272frac14 0193 eth193THORN

              The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

              14

              Volume of specimenfrac14

              438276 frac14 86 200mm3

              Bulk density ethTHORN frac14 Mass

              Volumefrac14 1680

              86 200 103frac14 195Mg=m3

              Water content ethwTHORN frac14 1680 1305

              1305frac14 0287 eth287THORN

              From Equation 117

              1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

              195frac14 180

              e frac14 080

              Using Equation 113

              Sr frac14 wGs

              efrac14 0287 273

              080frac14 098 eth98THORN

              15

              Using Equation 124

              d frac14

              1thorn w frac14215

              112frac14 192Mg=m3

              From Equation 117

              1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

              215frac14 138

              e frac14 038

              Using Equation 113

              Sr frac14 wGs

              efrac14 012 265

              038frac14 0837 eth837THORN

              Basic characteristics of soils 3

              Using Equation 115

              Afrac14 e wGs

              1thorn e frac14038 0318

              138frac14 0045 eth45THORN

              The zero air voids dry density is given by Equation 125

              d frac14 Gs

              1thorn wGsw frac14 265

              1thorn eth0135 265THORN 100 frac14 195Mg=m3

              ie a dry density of 200Mgm3 would not be possible

              16

              Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

              (Mgm3) d10(Mgm3)

              2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

              In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

              wopt frac14 15 and dmaxfrac14 183Mg=m3

              Figure Q16

              4 Basic characteristics of soils

              Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

              d5 d10

              respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

              17

              From Equation 120

              e frac14 Gswd 1

              The maximum and minimum values of void ratio are given by

              emax frac14 Gsw

              dmin

              1

              emin frac14 Gswdmax

              1

              From Equation 123

              ID frac14 Gsweth1=dmin 1=dTHORN

              Gsweth1=dmin 1=dmax

              THORN

              frac14 frac121 ethdmin=dTHORN1=dmin

              frac121 ethdmin=dmax

              THORN1=dmin

              frac14 d dmin

              dmax dmin

              dmax

              d

              frac14 172 154

              181 154

              181

              172

              frac14 070 eth70THORN

              Basic characteristics of soils 5

              Chapter 2

              Seepage

              21

              The coefficient of permeability is determined from the equation

              k frac14 23al

              At1log

              h0

              h1

              where

              a frac14

              4 00052 m2 l frac14 02m

              A frac14

              4 012 m2 t1 frac14 3 602 s

              logh0

              h1frac14 log

              100

              035frac14 0456

              k frac14 23 00052 02 0456

              012 3 602frac14 49 108 m=s

              22

              The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

              q frac14 kh Nf

              Ndfrac14 106 400 37

              11frac14 13 106 m3=s per m

              Figure Q22

              23

              The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

              q frac14 kh Nf

              Ndfrac14 5 105 300 35

              9frac14 58 105 m3=s per m

              The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

              Point h (m) h z (m) u frac14 w(h z)(kNm2)

              1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

              eg for Point 1

              h1 frac14 7

              9 300 frac14 233m

              h1 z1 frac14 233 eth250THORN frac14 483m

              Figure Q23

              Seepage 7

              u1 frac14 98 483 frac14 47 kN=m2

              The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

              24

              The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

              q frac14 kh Nf

              Ndfrac14 40 107 550 10

              11frac14 20 106 m3=s per m

              25

              The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

              q frac14 kh Nf

              Ndfrac14 20 106 500 42

              9frac14 47 106 m3=s per m

              Figure Q24

              8 Seepage

              26

              The scale transformation factor in the x direction is given by Equation 221 ie

              xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

              ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

              Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

              k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

              pfrac14

              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

              p 107 frac14 30 107 m=s

              Thus the quantity of seepage is given by

              q frac14 k0h Nf

              Ndfrac14 30 107 1400 325

              12frac14 11 106 m3=s per m

              Figure Q25

              Seepage 9

              27

              The scale transformation factor in the x direction is

              xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

              ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

              Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

              k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

              pfrac14

              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

              p 106 frac14 45 106 m=s

              The focus of the basic parabola is at point A The parabola passes through point Gsuch that

              GC frac14 03HC frac14 03 30 frac14 90m

              Thus the coordinates of G are

              x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

              480 frac14 x0 2002

              4x0

              Figure Q26

              10 Seepage

              Hence

              x0 frac14 20m

              Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

              x frac14 20 z2

              80

              x 20 0 50 100 200 300z 0 400 748 980 1327 1600

              The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

              No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

              q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

              28

              The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

              Figure Q27

              Seepage 11

              q frac14 kh Nf

              Ndfrac14 45 105 28 33

              7

              frac14 59 105 m3=s per m

              29

              The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

              kx frac14 H1k1 thornH2k2

              H1 thornH2frac14 106

              10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

              kz frac14 H1 thornH2

              H1

              k1thornH2

              k2

              frac14 10

              5

              eth2 106THORN thorn5

              eth16 106THORNfrac14 36 106 m=s

              Then the scale transformation factor is given by

              xt frac14 xffiffiffiffiffikz

              pffiffiffiffiffikx

              p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

              Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

              Figure Q28

              12 Seepage

              k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

              qfrac14

              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

              p 106 frac14 57 106 m=s

              Then the quantity of seepage is given by

              q frac14 k0h Nf

              Ndfrac14 57 106 350 56

              11

              frac14 10 105 m3=s per m

              Figure Q29

              Seepage 13

              Chapter 3

              Effective stress

              31

              Buoyant unit weight

              0 frac14 sat w frac14 20 98 frac14 102 kN=m3

              Effective vertical stress

              0v frac14 5 102 frac14 51 kN=m2 or

              Total vertical stress

              v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

              Pore water pressure

              u frac14 7 98 frac14 686 kN=m2

              Effective vertical stress

              0v frac14 v u frac14 1196 686 frac14 51 kN=m2

              32

              Buoyant unit weight

              0 frac14 sat w frac14 20 98 frac14 102 kN=m3

              Effective vertical stress

              0v frac14 5 102 frac14 51 kN=m2 or

              Total vertical stress

              v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

              Pore water pressure

              u frac14 205 98 frac14 2009 kN=m2

              Effective vertical stress

              0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

              33

              At top of the clay

              v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

              u frac14 2 98 frac14 196 kN=m2

              0v frac14 v u frac14 710 196 frac14 514 kN=m2

              Alternatively

              0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

              0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

              At bottom of the clay

              v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

              u frac14 12 98 frac14 1176 kN=m2

              0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

              NB The alternative method of calculation is not applicable because of the artesiancondition

              Figure Q3132

              Effective stress 15

              34

              0 frac14 20 98 frac14 102 kN=m3

              At 8m depth

              0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

              35

              0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

              0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

              Figure Q33

              Figure Q34

              16 Effective stress

              (a) Immediately after WT rise

              At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

              0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

              At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

              0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

              (b) Several years after WT rise

              At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

              At 8m depth

              0v frac14 940 kN=m2 (as above)

              At 12m depth

              0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

              Figure Q35

              Effective stress 17

              36

              Total weight

              ab frac14 210 kN

              Effective weight

              ac frac14 112 kN

              Resultant boundary water force

              be frac14 119 kN

              Seepage force

              ce frac14 34 kN

              Resultant body force

              ae frac14 99 kN eth73 to horizontalTHORN

              (Refer to Figure Q36)

              Figure Q36

              18 Effective stress

              37

              Situation (1)(a)

              frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

              u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

              0 frac14 u frac14 694 392 frac14 302 kN=m2

              (b)

              i frac14 2

              4frac14 05

              j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

              Situation (2)(a)

              frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

              u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

              0 frac14 u frac14 498 392 frac14 106 kN=m2

              (b)

              i frac14 2

              4frac14 05

              j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

              38

              The flow net is drawn in Figure Q24

              Loss in total head between adjacent equipotentials

              h frac14 550

              Ndfrac14 550

              11frac14 050m

              Exit hydraulic gradient

              ie frac14 h

              sfrac14 050

              070frac14 071

              Effective stress 19

              The critical hydraulic gradient is given by Equation 39

              ic frac14 0

              wfrac14 102

              98frac14 104

              Therefore factor of safety against lsquoboilingrsquo (Equation 311)

              F frac14 iciefrac14 104

              071frac14 15

              Total head at C

              hC frac14 nd

              Ndh frac14 24

              11 550 frac14 120m

              Elevation head at C

              zC frac14 250m

              Pore water pressure at C

              uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

              Therefore effective vertical stress at C

              0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

              For point D

              hD frac14 73

              11 550 frac14 365m

              zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

              0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

              39

              The flow net is drawn in Figure Q25

              For a soil prism 150 300m adjacent to the piling

              hm frac14 26

              9 500 frac14 145m

              20 Effective stress

              Factor of safety against lsquoheavingrsquo (Equation 310)

              F frac14 ic

              imfrac14 0d

              whmfrac14 97 300

              98 145frac14 20

              With a filter

              F frac14 0d thorn wwhm

              3 frac14 eth97 300THORN thorn w98 145

              w frac14 135 kN=m2

              Depth of filterfrac14 13521frac14 065m (if above water level)

              Effective stress 21

              Chapter 4

              Shear strength

              41

              frac14 295 kN=m2

              u frac14 120 kN=m2

              0 frac14 u frac14 295 120 frac14 175 kN=m2

              f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

              42

              03 (kNm2) 1 3 (kNm2) 01 (kNm2)

              100 452 552200 908 1108400 1810 2210800 3624 4424

              The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

              Figure Q42

              43

              3 (kNm2) 1 3 (kNm2) 1 (kNm2)

              200 222 422400 218 618600 220 820

              The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

              44

              The modified shear strength parameters are

              0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

              a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

              The coordinates of the stress point representing failure conditions in the test are

              1

              2eth1 2THORN frac14 1

              2 170 frac14 85 kN=m2

              1

              2eth1 thorn 3THORN frac14 1

              2eth270thorn 100THORN frac14 185 kN=m2

              The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

              uf frac14 36 kN=m2

              Figure Q43

              Figure Q44

              Shear strength 23

              45

              3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

              150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

              The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

              The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

              46

              03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

              200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

              The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

              Figure Q45

              24 Shear strength

              47

              The torque required to produce shear failure is given by

              T frac14 dh cud

              2thorn 2

              Z d=2

              0

              2r drcur

              frac14 cud2h

              2thorn 4cu

              Z d=2

              0

              r2dr

              frac14 cud2h

              2thorn d

              3

              6

              Then

              35 frac14 cu52 10

              2thorn 53

              6

              103

              cu frac14 76 kN=m3

              400

              0 400 800 1200 1600

              τ (k

              Nm

              2 )

              σprime (kNm2)

              34deg

              315deg29deg

              (a)

              (b)

              0 400

              400

              800 1200 1600

              Failure envelope

              300 500

              σprime (kNm2)

              τ (k

              Nm

              2 )

              20 (kNm2)

              31deg

              Figure Q46

              Shear strength 25

              48

              The relevant stress values are calculated as follows

              3 frac14 600 kN=m2

              1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

              2(1 3) 0 40 79 107 139 159

              1

              2(01 thorn 03) 400 411 402 389 351 326

              1

              2(1 thorn 3) 600 640 679 707 739 759

              The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

              Af frac14 433 200

              319frac14 073

              49

              B frac14 u33

              frac14 144

              350 200frac14 096

              a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

              0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

              10 283 65 023

              Figure Q48

              26 Shear strength

              The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

              Figure Q49

              Shear strength 27

              Chapter 5

              Stresses and displacements

              51

              Vertical stress is given by

              z frac14 Qz2Ip frac14 5000

              52Ip

              Values of Ip are obtained from Table 51

              r (m) rz Ip z (kNm2)

              0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

              10 20 0009 2

              The variation of z with radial distance (r) is plotted in Figure Q51

              Figure Q51

              52

              Below the centre load (Figure Q52)

              r

              zfrac14 0 for the 7500-kN load

              Ip frac14 0478

              r

              zfrac14 5

              4frac14 125 for the 10 000- and 9000-kN loads

              Ip frac14 0045

              Then

              z frac14X Q

              z2Ip

              frac14 7500 0478

              42thorn 10 000 0045

              42thorn 9000 0045

              42

              frac14 224thorn 28thorn 25 frac14 277 kN=m2

              53

              The vertical stress under a corner of a rectangular area is given by

              z frac14 qIr

              where values of Ir are obtained from Figure 510 In this case

              z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

              z

              Figure Q52

              Stresses and displacements 29

              z (m) m n Ir z (kNm2)

              0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

              10 010 0005 5

              z is plotted against z in Figure Q53

              54

              (a)

              m frac14 125

              12frac14 104

              n frac14 18

              12frac14 150

              From Figure 510 Irfrac14 0196

              z frac14 2 175 0196 frac14 68 kN=m2

              Figure Q53

              30 Stresses and displacements

              (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

              z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

              55

              Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

              Px frac14 2Q

              1

              m2 thorn 1frac14 2 150

              125frac14 76 kN=m

              Equation 517 is used to obtain the pressure distribution

              px frac14 4Q

              h

              m2n

              ethm2 thorn n2THORN2 frac14150

              m2n

              ethm2 thorn n2THORN2 ethkN=m2THORN

              Figure Q54

              Stresses and displacements 31

              n m2n

              (m2 thorn n2)2

              px(kNm2)

              0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

              The pressure distribution is plotted in Figure Q55

              56

              H

              Bfrac14 10

              2frac14 5

              L

              Bfrac14 4

              2frac14 2

              D

              Bfrac14 1

              2frac14 05

              Hence from Figure 515

              131 frac14 082

              130 frac14 094

              Figure Q55

              32 Stresses and displacements

              The immediate settlement is given by Equation 528

              si frac14 130131qB

              Eu

              frac14 094 082 200 2

              45frac14 7mm

              Stresses and displacements 33

              Chapter 6

              Lateral earth pressure

              61

              For 0 frac14 37 the active pressure coefficient is given by

              Ka frac14 1 sin 37

              1thorn sin 37frac14 025

              The total active thrust (Equation 66a with c0 frac14 0) is

              Pa frac14 1

              2KaH

              2 frac14 1

              2 025 17 62 frac14 765 kN=m

              If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

              K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

              and the thrust on the wall is

              P0 frac14 1

              2K0H

              2 frac14 1

              2 040 17 62 frac14 122 kN=m

              62

              The active pressure coefficients for the three soil types are as follows

              Ka1 frac141 sin 35

              1thorn sin 35frac14 0271

              Ka2 frac141 sin 27

              1thorn sin 27frac14 0375

              ffiffiffiffiffiffiffiKa2

              p frac14 0613

              Ka3 frac141 sin 42

              1thorn sin 42frac14 0198

              Distribution of active pressure (plotted in Figure Q62)

              Depth (m) Soil Active pressure (kNm2)

              3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

              12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

              At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

              Total thrust frac14 571 kNm

              Point of application is (4893571) m from the top of the wall ie 857m

              Force (kN) Arm (m) Moment (kN m)

              (1)1

              2 0271 16 32 frac14 195 20 390

              (2) 0271 16 3 2 frac14 260 40 1040

              (3)1

              2 0271 92 22 frac14 50 433 217

              (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

              (5)1

              2 0375 102 32 frac14 172 70 1204

              (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

              (7)1

              2 0198 112 42 frac14 177 1067 1889

              (8)1

              2 98 92 frac14 3969 90 35721

              5713 48934

              Figure Q62

              Lateral earth pressure 35

              63

              (a) For u frac14 0 Ka frac14 Kp frac14 1

              Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

              frac14 245

              At the lower end of the piling

              pa frac14 Kaqthorn Kasatz Kaccu

              frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

              frac14 115 kN=m2

              pp frac14 Kpsatzthorn Kpccu

              frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

              frac14 202 kN=m2

              (b) For 0 frac14 26 and frac14 1

              20

              Ka frac14 035

              Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

              pfrac14 145 ethEquation 619THORN

              Kp frac14 37

              Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

              pfrac14 47 ethEquation 624THORN

              At the lower end of the piling

              pa frac14 Kaqthorn Ka0z Kacc

              0

              frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

              frac14 187 kN=m2

              pp frac14 Kp0zthorn Kpcc

              0

              frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

              frac14 198 kN=m2

              36 Lateral earth pressure

              64

              (a) For 0 frac14 38 Ka frac14 024

              0 frac14 20 98 frac14 102 kN=m3

              The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

              Force (kN) Arm (m) Moment (kN m)

              (1) 024 10 66 frac14 159 33 525

              (2)1

              2 024 17 392 frac14 310 400 1240

              (3) 024 17 39 27 frac14 430 135 580

              (4)1

              2 024 102 272 frac14 89 090 80

              (5)1

              2 98 272 frac14 357 090 321

              Hfrac14 1345 MH frac14 2746

              (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

              (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

              XM frac14MV MH frac14 7790 kNm

              Lever arm of base resultant

              M

              Vfrac14 779

              488frac14 160

              Eccentricity of base resultant

              e frac14 200 160 frac14 040m

              39 m

              27 m

              40 m

              04 m

              04 m

              26 m

              (7)

              (9)

              (1)(2)

              (3)

              (4)

              (5)

              (8)(6)

              (10)

              WT

              10 kNm2

              Hydrostatic

              Figure Q64

              Lateral earth pressure 37

              Base pressures (Equation 627)

              p frac14 VB

              1 6e

              B

              frac14 488

              4eth1 060THORN

              frac14 195 kN=m2 and 49 kN=m2

              Factor of safety against sliding (Equation 628)

              F frac14 V tan

              Hfrac14 488 tan 25

              1345frac14 17

              (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

              Hfrac14 1633 kN

              V frac14 4879 kN

              MH frac14 3453 kNm

              MV frac14 10536 kNm

              The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

              65

              For 0 frac14 36 Ka frac14 026 and Kp frac14 385

              Kp

              Ffrac14 385

              2

              0 frac14 20 98 frac14 102 kN=m3

              The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

              Force (kN) Arm (m) Moment (kN m)

              (1)1

              2 026 17 452 frac14 448 dthorn 15 448dthorn 672

              (2) 026 17 45 d frac14 199d d2 995d2

              (3)1

              2 026 102 d2 frac14 133d2 d3 044d3

              (4)1

              2 385

              2 17 152 frac14 368 dthorn 05 368d 184

              (5)385

              2 17 15 d frac14 491d d2 2455d2

              (6)1

              2 385

              2 102 d2 frac14 982d2 d3 327d3

              38 Lateral earth pressure

              XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

              d3 thorn 516d2 283d 1724 frac14 0

              d frac14 179m

              Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

              Over additional 20 embedded depth

              pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

              Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

              66

              The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

              Ka frac14 sin 69=sin 105

              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

              pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

              26664

              37775

              2

              frac14 050

              The total active thrust (acting at 25 above the normal) is given by Equation 616

              Pa frac14 1

              2 050 19 7502 frac14 267 kN=m

              Figure Q65

              Lateral earth pressure 39

              Horizontal component

              Ph frac14 267 cos 40 frac14 205 kN=m

              Vertical component

              Pv frac14 267 sin 40 frac14 172 kN=m

              Consider moments about the toe of the wall (Figure Q66) (per m)

              Force (kN) Arm (m) Moment (kN m)

              (1)1

              2 175 650 235 frac14 1337 258 345

              (2) 050 650 235 frac14 764 175 134

              (3)1

              2 070 650 235 frac14 535 127 68

              (4) 100 400 235 frac14 940 200 188

              (5) 1

              2 080 050 235 frac14 47 027 1

              Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

              Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

              Lever arm of base resultant

              M

              Vfrac14 795

              525frac14 151m

              Eccentricity of base resultant

              e frac14 200 151 frac14 049m

              Figure Q66

              40 Lateral earth pressure

              Base pressures (Equation 627)

              p frac14 525

              41 6 049

              4

              frac14 228 kN=m2 and 35 kN=m2

              The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

              The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

              The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

              67

              For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

              Force (kN) Arm (m) Moment (kNm)

              (1)1

              2 027 17 52 frac14 574 183 1050

              (2) 027 17 5 3 frac14 689 500 3445

              (3)1

              2 027 102 32 frac14 124 550 682

              (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

              (5)1

              2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

              (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

              (7) 1

              2 267

              2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

              (8) 2 10ffiffiffiffiffiffiffiffiffi267p

              2 d frac14 163d d2thorn 650 82d2 1060d

              Tie rod force per m frac14 T 0 0

              XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

              d3 thorn 77d2 269d 1438 frac14 0

              d frac14 467m

              Depth of penetration frac14 12d frac14 560m

              Lateral earth pressure 41

              Algebraic sum of forces for d frac14 467m isX

              F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

              T frac14 905 kN=m

              Force in each tie rod frac14 25T frac14 226 kN

              68

              (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

              0 frac14 21 98 frac14 112 kN=m3

              The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

              uC frac14 150

              165 15 98 frac14 134 kN=m2

              The average seepage pressure is

              j frac14 15

              165 98 frac14 09 kN=m3

              Hence

              0 thorn j frac14 112thorn 09 frac14 121 kN=m3

              0 j frac14 112 09 frac14 103 kN=m3

              Figure Q67

              42 Lateral earth pressure

              Consider moments about the anchor point A (per m)

              Force (kN) Arm (m) Moment (kN m)

              (1) 10 026 150 frac14 390 60 2340

              (2)1

              2 026 18 452 frac14 474 15 711

              (3) 026 18 45 105 frac14 2211 825 18240

              (4)1

              2 026 121 1052 frac14 1734 100 17340

              (5)1

              2 134 15 frac14 101 40 404

              (6) 134 30 frac14 402 60 2412

              (7)1

              2 134 60 frac14 402 95 3819

              571 4527(8) Ppm

              115 115PPm

              XM frac14 0

              Ppm frac144527

              115frac14 394 kN=m

              Available passive resistance

              Pp frac14 1

              2 385 103 62 frac14 714 kN=m

              Factor of safety

              Fp frac14 Pp

              Ppm

              frac14 714

              394frac14 18

              Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

              Figure Q68

              Lateral earth pressure 43

              (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

              Consider moments (per m) about the tie point A

              Force (kN) Arm (m)

              (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

              (2)1

              2 033 18 452 frac14 601 15

              (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

              (4)1

              2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

              (5)1

              2 134 15 frac14 101 40

              (6) 134 30 frac14 402 60

              (7)1

              2 134 d frac14 67d d3thorn 75

              (8) 1

              2 30 103 d2 frac141545d2 2d3thorn 75

              Moment (kN m)

              (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

              XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

              d3 thorn 827d2 466d 1518 frac14 0

              By trial

              d frac14 544m

              The minimum depth of embedment required is 544m

              69

              For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

              0 frac14 20 98 frac14 102 kN=m3

              The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

              44 Lateral earth pressure

              uC frac14 147

              173 26 98 frac14 216 kN=m2

              and the average seepage pressure around the wall is

              j frac14 26

              173 98 frac14 15 kN=m3

              Consider moments about the prop (A) (per m)

              Force (kN) Arm (m) Moment (kN m)

              (1)1

              2 03 17 272 frac14 186 020 37

              (2) 03 17 27 53 frac14 730 335 2445

              (3)1

              2 03 (102thorn 15) 532 frac14 493 423 2085

              (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

              (5)1

              2 216 26 frac14 281 243 684

              (6) 216 27 frac14 583 465 2712

              (7)1

              2 216 60 frac14 648 800 5184

              3055(8)

              1

              2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

              Factor of safety

              Fr frac14 6885

              3055frac14 225

              Figure Q69

              Lateral earth pressure 45

              610

              For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

              p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

              Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

              Using the recommendations of Twine and Roscoe

              p frac14 02H frac14 02 19 9 frac14 342 kN=m2

              Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

              611

              frac14 18 kN=m3 0 frac14 34

              H frac14 350m nH frac14 335m mH frac14 185m

              Consider a trial value of F frac14 20 Refer to Figure 635

              0m frac14 tan1tan 34

              20

              frac14 186

              Then

              frac14 45 thorn 0m2frac14 543

              W frac14 1

              2 18 3502 cot 543 frac14 792 kN=m

              Figure Q610

              46 Lateral earth pressure

              P frac14 1

              2 s 3352 frac14 561s kN=m

              U frac14 1

              2 98 1852 cosec 543 frac14 206 kN=m

              Equations 630 and 631 then become

              561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

              792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

              ie

              561s 0616N 405 frac14 0

              792 0857N thorn 563 frac14 0

              N frac14 848

              0857frac14 989 kN=m

              Then

              561s 609 405 frac14 0

              s frac14 649

              561frac14 116 kN=m3

              The calculations for trial values of F of 20 15 and 10 are summarized below

              F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

              20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

              s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

              Figure Q611

              Lateral earth pressure 47

              612

              For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

              45 thorn 0

              2frac14 63

              For the retained material between the surface and a depth of 36m

              Pa frac14 1

              2 030 18 362 frac14 350 kN=m

              Weight of reinforced fill between the surface and a depth of 36m is

              Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

              eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

              Eccentricity of Rv

              e frac14 263 250 frac14 013m

              The average vertical stress at a depth of 36m is

              z frac14 Rv

              L 2efrac14 324

              474frac14 68 kN=m2

              (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

              Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

              Tensile stress in the element frac14 138 103

              65 3frac14 71N=mm2

              Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

              Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

              Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

              The weight of ABC is

              W frac14 1

              2 18 52 265 frac14 124 kN=m

              From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

              48 Lateral earth pressure

              (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

              Tp frac14 032 68 120 065 frac14 170 kN

              Tr frac14 213 420

              418frac14 214 kN

              Again the tensile failure and slipping limit states are satisfied for this element

              Figure Q612

              Lateral earth pressure 49

              Chapter 7

              Consolidation theory

              71

              Total change in thickness

              H frac14 782 602 frac14 180mm

              Average thickness frac14 1530thorn 180

              2frac14 1620mm

              Length of drainage path d frac14 1620

              2frac14 810mm

              Root time plot (Figure Q71a)

              ffiffiffiffiffiffit90p frac14 33

              t90 frac14 109min

              cv frac14 0848d2

              t90frac14 0848 8102

              109 1440 365

              106frac14 27m2=year

              r0 frac14 782 764

              782 602frac14 018

              180frac14 0100

              rp frac14 10eth764 645THORN9eth782 602THORN frac14

              10 119

              9 180frac14 0735

              rs frac14 1 eth0100thorn 0735THORN frac14 0165

              Log time plot (Figure Q71b)

              t50 frac14 26min

              cv frac14 0196d2

              t50frac14 0196 8102

              26 1440 365

              106frac14 26m2=year

              r0 frac14 782 763

              782 602frac14 019

              180frac14 0106

              rp frac14 763 623

              782 602frac14 140

              180frac14 0778

              rs frac14 1 eth0106thorn 0778THORN frac14 0116

              Figure Q71(a)

              Figure Q71(b)

              Final void ratio

              e1 frac14 w1Gs frac14 0232 272 frac14 0631

              e

              Hfrac14 1thorn e0

              H0frac14 1thorn e1 thorne

              H0

              ie

              e

              180frac14 1631thorne

              1710

              e frac14 2936

              1530frac14 0192

              Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

              mv frac14 1

              1thorn e0 e0 e101 00

              frac14 1

              1823 0192

              0107frac14 098m2=MN

              k frac14 cvmvw frac14 265 098 98

              60 1440 365 103frac14 81 1010 m=s

              72

              Using Equation 77 (one-dimensional method)

              sc frac14 e0 e11thorn e0 H

              Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

              Figure Q72

              52 Consolidation theory

              Settlement

              Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

              1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

              318

              Notes 5 92y 460thorn 84

              Heave

              Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

              1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

              38

              73

              U frac14 f ethTvTHORN frac14 f cvt

              d2

              Hence if cv is constant

              t1

              t2frac14 d

              21

              d22

              where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

              d1 frac14 95mm and d2 frac14 2500mm

              for U frac14 050 t2 frac14 t1 d22

              d21

              frac14 20

              60 24 365 25002

              952frac14 263 years

              for U lt 060 Tv frac14

              4U2 (Equation 724(a))

              t030 frac14 t050 0302

              0502

              frac14 263 036 frac14 095 years

              Consolidation theory 53

              74

              The layer is open

              d frac14 8

              2frac14 4m

              Tv frac14 cvtd2frac14 24 3

              42frac14 0450

              ui frac14 frac14 84 kN=m2

              The excess pore water pressure is given by Equation 721

              ue frac14Xmfrac141mfrac140

              2ui

              Msin

              Mz

              d

              expethM2TvTHORN

              In this case z frac14 d

              sinMz

              d

              frac14 sinM

              where

              M frac14

              23

              25

              2

              M sin M M2Tv exp (M2Tv)

              2thorn1 1110 0329

              3

              21 9993 457 105

              ue frac14 2 84 2

              1 0329 ethother terms negligibleTHORN

              frac14 352 kN=m2

              75

              The layer is open

              d frac14 6

              2frac14 3m

              Tv frac14 cvtd2frac14 10 3

              32frac14 0333

              The layer thickness will be divided into six equal parts ie m frac14 6

              54 Consolidation theory

              For an open layer

              Tv frac14 4n

              m2

              n frac14 0333 62

              4frac14 300

              The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

              ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

              i j

              0 1 2 3 4 5 6 7 8 9 10 11 12

              0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

              The initial and 3-year isochrones are plotted in Figure Q75

              Area under initial isochrone frac14 180 units

              Area under 3-year isochrone frac14 63 units

              The average degree of consolidation is given by Equation 725Thus

              U frac14 1 63

              180frac14 065

              Figure Q75

              Consolidation theory 55

              76

              At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

              0 frac14 2w frac14 2 98 frac14 196 kN=m2

              The final consolidation settlement (one-dimensional method) is

              sc frac14 mv0H frac14 083 196 8 frac14 130mm

              Corrected time t frac14 2 1

              2

              40

              52

              frac14 1615 years

              Tv frac14 cvtd2frac14 44 1615

              42frac14 0444

              From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

              77

              The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

              Figure Q77

              56 Consolidation theory

              Point m n Ir (kNm2) sc (mm)

              13020frac14 15 20

              20frac14 10 0194 (4) 113 124

              260

              20frac14 30

              20

              20frac14 10 0204 (2) 59 65

              360

              20frac14 30

              40

              20frac14 20 0238 (1) 35 38

              430

              20frac14 15

              40

              20frac14 20 0224 (2) 65 72

              Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

              78

              Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

              (a) Immediate settlement

              H

              Bfrac14 30

              35frac14 086

              D

              Bfrac14 2

              35frac14 006

              Figure Q78

              Consolidation theory 57

              From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

              si frac14 130131qB

              Eufrac14 10 032 105 35

              40frac14 30mm

              (b) Consolidation settlement

              Layer z (m) Dz Ic (kNm2) syod (mm)

              1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

              3150

              Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

              Now

              H

              Bfrac14 30

              35frac14 086 and A frac14 065

              from Figure 712 13 frac14 079

              sc frac14 13sod frac14 079 315 frac14 250mm

              Total settlement

              s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

              79

              Without sand drains

              Uv frac14 025

              Tv frac14 0049 ethfrom Figure 718THORN

              t frac14 Tvd2

              cvfrac14 0049 82

              cvWith sand drains

              R frac14 0564S frac14 0564 3 frac14 169m

              n frac14 Rrfrac14 169

              015frac14 113

              Tr frac14 cht

              4R2frac14 ch

              4 1692 0049 82

              cvethand ch frac14 cvTHORN

              frac14 0275

              Ur frac14 073 (from Figure 730)

              58 Consolidation theory

              Using Equation 740

              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

              U frac14 080

              710

              Without sand drains

              Uv frac14 090

              Tv frac14 0848

              t frac14 Tvd2

              cvfrac14 0848 102

              96frac14 88 years

              With sand drains

              R frac14 0564S frac14 0564 4 frac14 226m

              n frac14 Rrfrac14 226

              015frac14 15

              Tr

              Tvfrac14 chcv

              d2

              4R2ethsame tTHORN

              Tr

              Tvfrac14 140

              96 102

              4 2262frac14 714 eth1THORN

              Using Equation 740

              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

              An iterative solution is required using (1) and (2) an initial value ofUv being estimated

              Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

              040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

              Thus

              Uv frac14 0295 and Ur frac14 086

              t frac14 88 00683

              0848frac14 07 years

              Consolidation theory 59

              Chapter 8

              Bearing capacity

              81

              (a) The ultimate bearing capacity is given by Equation 83

              qf frac14 cNc thorn DNq thorn 1

              2BN

              For u frac14 0

              Nc frac14 514 Nq frac14 1 N frac14 0

              qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

              The net ultimate bearing capacity is

              qnf frac14 qf D frac14 540 kN=m2

              The net foundation pressure is

              qn frac14 q D frac14 425

              2 eth21 1THORN frac14 192 kN=m2

              The factor of safety (Equation 86) is

              F frac14 qnfqnfrac14 540

              192frac14 28

              (b) For 0 frac14 28

              Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

              qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

              2 112 2 13

              frac14 260thorn 168thorn 146 frac14 574 kN=m2

              qnf frac14 574 112 frac14 563 kN=m2

              F frac14 563

              192frac14 29

              (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

              82

              For 0 frac14 38

              Nq frac14 49 N frac14 67

              qnf frac14 DethNq 1THORN thorn 1

              2BN ethfrom Equation 83THORN

              frac14 eth18 075 48THORN thorn 1

              2 18 15 67

              frac14 648thorn 905 frac14 1553 kN=m2

              qn frac14 500

              15 eth18 075THORN frac14 320 kN=m2

              F frac14 qnfqnfrac14 1553

              320frac14 48

              0d frac14 tan1tan 38

              125

              frac14 32 therefore Nq frac14 23 and N frac14 25

              Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

              2 18 15 25

              frac14 15eth310thorn 337THORNfrac14 970 kN=m

              Design load (action) Vd frac14 500 kN=m

              The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

              83

              D

              Bfrac14 350

              225frac14 155

              From Figure 85 for a square foundation

              Nc frac14 81

              Bearing capacity 61

              For a rectangular foundation (L frac14 450m B frac14 225m)

              Nc frac14 084thorn 016B

              L

              81 frac14 745

              Using Equation 810

              qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

              For F frac14 3

              qn frac14 1006

              3frac14 335 kN=m2

              q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

              Design load frac14 405 450 225 frac14 4100 kN

              Design undrained strength cud frac14 135

              14frac14 96 kN=m2

              Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

              frac14 7241 kN

              Design load Vd frac14 4100 kN

              Rd gt Vd therefore the bearing resistance limit state is satisfied

              84

              For 0 frac14 40

              Nq frac14 64 N frac14 95

              qnf frac14 DethNq 1THORN thorn 04BN

              (a) Water table 5m below ground level

              qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

              qn frac14 400 17 frac14 383 kN=m2

              F frac14 2686

              383frac14 70

              (b) Water table 1m below ground level (ie at foundation level)

              0 frac14 20 98 frac14 102 kN=m3

              62 Bearing capacity

              qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

              F frac14 2040

              383frac14 53

              (c) Water table at ground level with upward hydraulic gradient 02

              eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

              qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

              F frac14 1296

              392frac14 33

              85

              The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

              Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

              Design value of 0 frac14 tan1tan 39

              125

              frac14 33

              For 0 frac14 33 Nq frac14 26 and N frac14 29

              Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

              Rd gt Vd therefore the bearing resistance limit state is satisfied

              86

              (a) Undrained shear for u frac14 0

              Nc frac14 514 Nq frac14 1 N frac14 0

              qnf frac14 12cuNc

              frac14 12 100 514 frac14 617 kN=m2

              qn frac14 qnfFfrac14 617

              3frac14 206 kN=m2

              q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

              Bearing capacity 63

              Drained shear for 0 frac14 32

              Nq frac14 23 N frac14 25

              0 frac14 21 98 frac14 112 kN=m3

              qnf frac14 0DethNq 1THORN thorn 040BN

              frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

              frac14 694 kN=m2

              q frac14 694

              3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

              Design load frac14 42 227 frac14 3632 kN

              (b) Design undrained strength cud frac14 100

              14frac14 71 kNm2

              Design bearing resistance Rd frac14 12cudNe area

              frac14 12 71 514 42

              frac14 7007 kN

              For drained shear 0d frac14 tan1tan 32

              125

              frac14 26

              Nq frac14 12 N frac14 10

              Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

              (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

              Layer z (m) m n Ir 0 (kNm2) sod (mm)

              1 2 100 0175 0700qn 0182qn

              2 6 033 0044 0176qn 0046qn

              3 10 020 0017 0068qn 0018qn

              0246qn

              Diameter of equivalent circle B frac14 45m

              H

              Bfrac14 12

              45frac14 27 and A frac14 042

              13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

              64 Bearing capacity

              For sc frac14 30mm

              qn frac14 30

              0147frac14 204 kN=m2

              q frac14 204thorn 21 frac14 225 kN=m2

              Design load frac14 42 225 frac14 3600 kN

              The design load is 3600 kN settlement being the limiting criterion

              87

              D

              Bfrac14 8

              4frac14 20

              From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

              F frac14 cuNc

              Dfrac14 40 71

              20 8frac14 18

              88

              Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

              Design value of 0 frac14 tan1tan 38

              125

              frac14 32

              Figure Q86

              Bearing capacity 65

              For 0 frac14 32 Nq frac14 23 and N frac14 25

              Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

              The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

              Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

              For B frac14 250m qn frac14 3750

              2502 17 frac14 583 kN=m2

              From Figure 510 m frac14 n frac14 126

              6frac14 021

              Ir frac14 0019

              Stress increment frac14 4 0019 583 frac14 44 kN=m2

              Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

              The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

              The settlement is less than 20mm therefore the serviceability limit state is satisfied

              89

              Depth (m) N 0v (kNm2) CN N1

              070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

              Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

              (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

              Cw frac14 05thorn 0530

              47

              frac14 082

              66 Bearing capacity

              Thus

              qa frac14 150 082 frac14 120 kN=m2

              (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

              Thus

              qa frac14 90 15 frac14 135 kN=m2

              (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

              Ic frac14 171

              1014frac14 0068

              From Equation 819(a) with s frac14 25mm

              q frac14 25

              3507 0068frac14 150 kN=m2

              810

              Peak value of strain influence factor occurs at a depth of 27m and is given by

              Izp frac14 05thorn 01130

              16 27

              05

              frac14 067

              Refer to Figure Q810

              E frac14 25qc

              Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

              Ez (mm3MN)

              1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

              0203

              C1 frac14 1 0500qnfrac14 1 05 12 16

              130frac14 093

              C2 frac14 1 ethsayTHORN

              s frac14 C1C2qnX Iz

              Ez frac14 093 1 130 0203 frac14 25mm

              Bearing capacity 67

              811

              At pile base level

              cu frac14 220 kN=m2

              qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

              Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

              00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

              qs frac14 00 frac14 07 1226 frac14 86 kN=m2

              Then

              Qf frac14 Abqb thorn Asqs

              frac14

              4 32 1980

              thorn eth 105 139 86THORN

              frac14 13 996thorn 3941 frac14 17 937 kN

              0 01 02 03 04 05 06 07

              0 2 4 6 8 10 12 14

              1

              2

              3

              4

              5

              6

              7

              8

              (1)

              (2)

              (3)

              (4)

              (5)

              qc

              qc

              Iz

              Iz

              (MNm2)

              z (m)

              Figure Q810

              68 Bearing capacity

              Allowable load

              ethaTHORN Qf

              2frac14 17 937

              2frac14 8968 kN

              ethbTHORN Abqb

              3thorn Asqs frac14 13 996

              3thorn 3941 frac14 8606 kN

              ie allowable load frac14 8600 kN

              Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

              According to the limit state method

              Characteristic undrained strength at base level cuk frac14 220

              150kN=m2

              Characteristic base resistance qbk frac14 9cuk frac14 9 220

              150frac14 1320 kN=m2

              Characteristic shaft resistance qsk frac14 00150

              frac14 86

              150frac14 57 kN=m2

              Characteristic base and shaft resistances

              Rbk frac14

              4 32 1320 frac14 9330 kN

              Rsk frac14 105 139 86

              150frac14 2629 kN

              For a bored pile the partial factors are b frac14 160 and s frac14 130

              Design bearing resistance Rcd frac14 9330

              160thorn 2629

              130

              frac14 5831thorn 2022

              frac14 7850 kN

              Adding ethDAb W) the design bearing resistance becomes 9650 kN

              812

              ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

              qs frac14 cu frac14 040 105 frac14 42 kN=m2

              For a single pile

              Qf frac14 Abqb thorn Asqs

              frac14

              4 062 1305

              thorn eth 06 15 42THORN

              frac14 369thorn 1187 frac14 1556 kN

              Bearing capacity 69

              Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

              eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

              (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

              (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

              qbkfrac14 9cuk frac14 9 220

              150frac14 1320 kN=m2

              qskfrac14cuk frac14 040 105

              150frac14 28 kN=m2

              Rbkfrac14

              4 0602 1320 frac14 373 kN

              Rskfrac14 060 15 28 frac14 791 kN

              Rcdfrac14 373

              160thorn 791

              130frac14 233thorn 608 frac14 841 kN

              Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

              Rcd gt Fcd therefore the bearing resistance limit state is satisfied

              (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

              q frac14 21 000

              1762frac14 68 kN=m2

              Immediate settlement

              H

              Bfrac14 15

              176frac14 085

              D

              Bfrac14 13

              176frac14 074

              L

              Bfrac14 1

              Hence from Figure 515

              130 frac14 078 and 131 frac14 041

              70 Bearing capacity

              Thus using Equation 528

              si frac14 078 041 68 176

              65frac14 6mm

              Consolidation settlement

              Layer z (m) Area (m2) (kNm2) mvH (mm)

              1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

              434 (sod)

              Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

              sc frac14 056 434 frac14 24mm

              The total settlement is (6thorn 24) frac14 30mm

              813

              At base level N frac14 26 Then using Equation 830

              qb frac14 40NDb

              Bfrac14 40 26 2

              025frac14 8320 kN=m2

              ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

              Figure Q812

              Bearing capacity 71

              Over the length embedded in sand

              N frac14 21 ie18thorn 24

              2

              Using Equation 831

              qs frac14 2N frac14 2 21 frac14 42 kN=m2

              For a single pile

              Qf frac14 Abqb thorn Asqs

              frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

              For the pile group assuming a group efficiency of 12

              XQf frac14 12 9 604 frac14 6523 kN

              Then the load factor is

              F frac14 6523

              2000thorn 1000frac14 21

              (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

              Characteristic base resistance per unit area qbk frac14 8320

              150frac14 5547 kNm2

              Characteristic shaft resistance per unit area qsk frac14 42

              150frac14 28 kNm2

              Characteristic base and shaft resistances for a single pile

              Rbk frac14 0252 5547 frac14 347 kN

              Rsk frac14 4 025 2 28 frac14 56 kN

              For a driven pile the partial factors are b frac14 s frac14 130

              Design bearing resistance Rcd frac14 347

              130thorn 56

              130frac14 310 kN

              For the pile group Rcd frac14 12 9 310 frac14 3348 kN

              Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

              (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

              From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

              72 Bearing capacity

              N frac14 24thorn 26thorn 34

              3frac14 28

              Ic frac14 171

              2814frac14 0016 ethEquation 818THORN

              s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

              The settlement is less than 20mm therefore the serviceability limit state is satisfied

              814

              Using Equation 841

              Tf frac14 DLcu thorn

              4ethD2 d2THORNcuNc

              frac14 eth 02 5 06 110THORN thorn

              4eth022 012THORN110 9

              frac14 207thorn 23 frac14 230 kN

              Figure Q813

              Bearing capacity 73

              Chapter 9

              Stability of slopes

              91

              Referring to Figure Q91

              W frac14 417 19 frac14 792 kN=m

              Q frac14 20 28 frac14 56 kN=m

              Arc lengthAB frac14

              180 73 90 frac14 115m

              Arc length BC frac14

              180 28 90 frac14 44m

              The factor of safety is given by

              F frac14 rethcuLaTHORNWd1 thornQd2 frac14

              90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

              Depth of tension crack z0 frac14 2cu

              frac14 2 20

              19frac14 21m

              Arc length BD frac14

              180 13

              1

              2 90 frac14 21m

              F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

              The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

              Design resisting moment frac14 rXethcudLaTHORN frac14 90

              14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

              Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

              The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

              92

              u frac14 0

              Depth factor D frac14 11

              9frac14 122

              Using Equation 92 with F frac14 10

              Ns frac14 cu

              FHfrac14 30

              10 19 9frac14 0175

              Hence from Figure 93

              frac14 50

              For F frac14 12

              Ns frac14 30

              12 19 9frac14 0146

              frac14 27

              93

              Refer to Figure Q93

              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

              1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

              74 m

              214 1deg

              213 1deg

              39 m

              WB

              D

              C

              28 m

              21 m

              A

              Q

              Soil (1)Soil (2)

              73deg

              Figure Q91

              Stability of slopes 75

              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

              9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

              599 256 328 1372

              Figure Q93

              76 Stability of slopes

              XW cos frac14 b

              Xh cos frac14 21 2 599 frac14 2516 kN=mX

              W sin frac14 bX

              h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

              Arc length La frac14

              180 57

              1

              2 326 frac14 327m

              The factor of safety is given by

              F frac14 c0La thorn tan0ethW cos ulTHORN

              W sin

              frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

              frac14 091

              According to the limit state method

              0d frac14 tan1tan 32

              125

              frac14 265

              c0 frac14 8

              160frac14 5 kN=m2

              Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

              Design disturbing moment frac14 1075 kN=m

              The design resisting moment is less than the design disturbing moment therefore a slipwill occur

              94

              F frac14 1

              W sin

              Xfc0bthorn ethW ubTHORN tan0g sec

              1thorn ethtan tan0=FTHORN

              c0 frac14 8 kN=m2

              0 frac14 32

              c0b frac14 8 2 frac14 16 kN=m

              W frac14 bh frac14 21 2 h frac14 42h kN=m

              Try F frac14 100

              tan0

              Ffrac14 0625

              Stability of slopes 77

              Values of u are as obtained in Figure Q93

              SliceNo

              h(m)

              W frac14 bh(kNm)

              W sin(kNm)

              ub(kNm)

              c0bthorn (W ub) tan0(kNm)

              sec

              1thorn (tan tan0)FProduct(kNm)

              1 05 21 6 2 8 24 1078 262 13 55 31

              23 33 30 1042 31

              3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

              224 92 72 0931 67

              6 50 210 11 40 100 85 0907 777 55 231 14

              12 58 112 90 0889 80

              8 60 252 1812

              80 114 102 0874 899 63 265 22 99 116 109 0861 94

              10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

              2154 88 116 0853 99

              14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

              1074 1091

              F frac14 1091

              1074frac14 102 (assumed value 100)

              Thus

              F frac14 101

              95

              F frac14 1

              W sin

              XfWeth1 ruTHORN tan0g sec

              1thorn ethtan tan0THORN=F

              0 frac14 33

              ru frac14 020

              W frac14 bh frac14 20 5 h frac14 100h kN=m

              eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

              Try F frac14 110

              tan 0

              Ffrac14 tan 33

              110frac14 0590

              78 Stability of slopes

              Referring to Figure Q95

              SliceNo

              h(m)

              W frac14 bh(kNm)

              W sin(kNm)

              W(1 ru) tan0(kNm)

              sec

              1thorn ( tan tan0)FProduct(kNm)

              1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

              2120 234 0892 209

              4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

              1185 1271

              Figure Q95

              Stability of slopes 79

              F frac14 1271

              1185frac14 107

              The trial value was 110 therefore take F to be 108

              96

              (a) Water table at surface the factor of safety is given by Equation 912

              F frac14 0

              sat

              tan0

              tan

              ptie 15 frac14 92

              19

              tan 36

              tan

              tan frac14 0234

              frac14 13

              Water table well below surface the factor of safety is given by Equation 911

              F frac14 tan0

              tan

              frac14 tan 36

              tan 13

              frac14 31

              (b) 0d frac14 tan1tan 36

              125

              frac14 30

              Depth of potential failure surface frac14 z

              Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

              frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

              frac14 504z kN

              Design disturbing moment per unit area Sd frac14 sat sin cos

              frac14 19 z sin 13 cos 13

              frac14 416z kN

              Rd gtSd therefore the limit state for overall stability is satisfied

              80 Stability of slopes

              • Book Cover
              • Title
              • Contents
              • Basic characteristics of soils
              • Seepage
              • Effective stress
              • Shear strength
              • Stresses and displacements
              • Lateral earth pressure
              • Consolidation theory
              • Bearing capacity
              • Stability of slopes

                Soil H consists of 99 fine material (58 clay size 47 silt size) The liquid limit is78 and the plasticity index is 47 (ie 78 31) plotting above the A-line in the CV zoneon the plasticity chart Thus the classification is CV CLAY of very high plasticity

                12

                From Equation 117

                1thorn e frac14 Gseth1thorn wTHORN wfrac14 270 1095 100

                191frac14 155

                e frac14 055

                Using Equation 113

                Sr frac14 wGs

                efrac14 0095 270

                055frac14 0466 eth466THORN

                Using Equation 119

                sat frac14 Gs thorn e1thorn e w frac14 325

                155 100 frac14 210Mg=m3

                From Equation 114

                w frac14 e

                Gsfrac14 055

                270frac14 0204 eth204THORN

                13

                Equations similar to 117ndash120 apply in the case of unit weights thus

                d frac14 Gs

                1thorn e w frac14272

                170 98 frac14 157 kN=m3

                sat frac14 Gs thorn e1thorn e w frac14 342

                170 98 frac14 197 kN=m3

                Using Equation 121

                0 frac14 Gs 1

                1thorn e w frac14 172

                170 98 frac14 99 kN=m3

                Using Equation 118a with Srfrac14 075

                frac14 Gs thorn Sre

                1thorn e w frac14 3245

                170 98 frac14 187 kN=m3

                2 Basic characteristics of soils

                Using Equation 113

                w frac14 Sre

                Gsfrac14 075 070

                272frac14 0193 eth193THORN

                The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

                14

                Volume of specimenfrac14

                438276 frac14 86 200mm3

                Bulk density ethTHORN frac14 Mass

                Volumefrac14 1680

                86 200 103frac14 195Mg=m3

                Water content ethwTHORN frac14 1680 1305

                1305frac14 0287 eth287THORN

                From Equation 117

                1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

                195frac14 180

                e frac14 080

                Using Equation 113

                Sr frac14 wGs

                efrac14 0287 273

                080frac14 098 eth98THORN

                15

                Using Equation 124

                d frac14

                1thorn w frac14215

                112frac14 192Mg=m3

                From Equation 117

                1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

                215frac14 138

                e frac14 038

                Using Equation 113

                Sr frac14 wGs

                efrac14 012 265

                038frac14 0837 eth837THORN

                Basic characteristics of soils 3

                Using Equation 115

                Afrac14 e wGs

                1thorn e frac14038 0318

                138frac14 0045 eth45THORN

                The zero air voids dry density is given by Equation 125

                d frac14 Gs

                1thorn wGsw frac14 265

                1thorn eth0135 265THORN 100 frac14 195Mg=m3

                ie a dry density of 200Mgm3 would not be possible

                16

                Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

                (Mgm3) d10(Mgm3)

                2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

                In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

                wopt frac14 15 and dmaxfrac14 183Mg=m3

                Figure Q16

                4 Basic characteristics of soils

                Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

                d5 d10

                respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

                17

                From Equation 120

                e frac14 Gswd 1

                The maximum and minimum values of void ratio are given by

                emax frac14 Gsw

                dmin

                1

                emin frac14 Gswdmax

                1

                From Equation 123

                ID frac14 Gsweth1=dmin 1=dTHORN

                Gsweth1=dmin 1=dmax

                THORN

                frac14 frac121 ethdmin=dTHORN1=dmin

                frac121 ethdmin=dmax

                THORN1=dmin

                frac14 d dmin

                dmax dmin

                dmax

                d

                frac14 172 154

                181 154

                181

                172

                frac14 070 eth70THORN

                Basic characteristics of soils 5

                Chapter 2

                Seepage

                21

                The coefficient of permeability is determined from the equation

                k frac14 23al

                At1log

                h0

                h1

                where

                a frac14

                4 00052 m2 l frac14 02m

                A frac14

                4 012 m2 t1 frac14 3 602 s

                logh0

                h1frac14 log

                100

                035frac14 0456

                k frac14 23 00052 02 0456

                012 3 602frac14 49 108 m=s

                22

                The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

                q frac14 kh Nf

                Ndfrac14 106 400 37

                11frac14 13 106 m3=s per m

                Figure Q22

                23

                The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

                q frac14 kh Nf

                Ndfrac14 5 105 300 35

                9frac14 58 105 m3=s per m

                The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

                Point h (m) h z (m) u frac14 w(h z)(kNm2)

                1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

                eg for Point 1

                h1 frac14 7

                9 300 frac14 233m

                h1 z1 frac14 233 eth250THORN frac14 483m

                Figure Q23

                Seepage 7

                u1 frac14 98 483 frac14 47 kN=m2

                The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

                24

                The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

                q frac14 kh Nf

                Ndfrac14 40 107 550 10

                11frac14 20 106 m3=s per m

                25

                The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

                q frac14 kh Nf

                Ndfrac14 20 106 500 42

                9frac14 47 106 m3=s per m

                Figure Q24

                8 Seepage

                26

                The scale transformation factor in the x direction is given by Equation 221 ie

                xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

                Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

                k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                pfrac14

                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

                p 107 frac14 30 107 m=s

                Thus the quantity of seepage is given by

                q frac14 k0h Nf

                Ndfrac14 30 107 1400 325

                12frac14 11 106 m3=s per m

                Figure Q25

                Seepage 9

                27

                The scale transformation factor in the x direction is

                xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

                Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

                k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                pfrac14

                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

                p 106 frac14 45 106 m=s

                The focus of the basic parabola is at point A The parabola passes through point Gsuch that

                GC frac14 03HC frac14 03 30 frac14 90m

                Thus the coordinates of G are

                x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

                480 frac14 x0 2002

                4x0

                Figure Q26

                10 Seepage

                Hence

                x0 frac14 20m

                Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

                x frac14 20 z2

                80

                x 20 0 50 100 200 300z 0 400 748 980 1327 1600

                The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

                No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

                q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

                28

                The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

                Figure Q27

                Seepage 11

                q frac14 kh Nf

                Ndfrac14 45 105 28 33

                7

                frac14 59 105 m3=s per m

                29

                The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

                kx frac14 H1k1 thornH2k2

                H1 thornH2frac14 106

                10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

                kz frac14 H1 thornH2

                H1

                k1thornH2

                k2

                frac14 10

                5

                eth2 106THORN thorn5

                eth16 106THORNfrac14 36 106 m=s

                Then the scale transformation factor is given by

                xt frac14 xffiffiffiffiffikz

                pffiffiffiffiffikx

                p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

                Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

                Figure Q28

                12 Seepage

                k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                qfrac14

                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

                p 106 frac14 57 106 m=s

                Then the quantity of seepage is given by

                q frac14 k0h Nf

                Ndfrac14 57 106 350 56

                11

                frac14 10 105 m3=s per m

                Figure Q29

                Seepage 13

                Chapter 3

                Effective stress

                31

                Buoyant unit weight

                0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                Effective vertical stress

                0v frac14 5 102 frac14 51 kN=m2 or

                Total vertical stress

                v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

                Pore water pressure

                u frac14 7 98 frac14 686 kN=m2

                Effective vertical stress

                0v frac14 v u frac14 1196 686 frac14 51 kN=m2

                32

                Buoyant unit weight

                0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                Effective vertical stress

                0v frac14 5 102 frac14 51 kN=m2 or

                Total vertical stress

                v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

                Pore water pressure

                u frac14 205 98 frac14 2009 kN=m2

                Effective vertical stress

                0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

                33

                At top of the clay

                v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

                u frac14 2 98 frac14 196 kN=m2

                0v frac14 v u frac14 710 196 frac14 514 kN=m2

                Alternatively

                0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

                At bottom of the clay

                v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

                u frac14 12 98 frac14 1176 kN=m2

                0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

                NB The alternative method of calculation is not applicable because of the artesiancondition

                Figure Q3132

                Effective stress 15

                34

                0 frac14 20 98 frac14 102 kN=m3

                At 8m depth

                0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                35

                0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                Figure Q33

                Figure Q34

                16 Effective stress

                (a) Immediately after WT rise

                At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                (b) Several years after WT rise

                At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                At 8m depth

                0v frac14 940 kN=m2 (as above)

                At 12m depth

                0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                Figure Q35

                Effective stress 17

                36

                Total weight

                ab frac14 210 kN

                Effective weight

                ac frac14 112 kN

                Resultant boundary water force

                be frac14 119 kN

                Seepage force

                ce frac14 34 kN

                Resultant body force

                ae frac14 99 kN eth73 to horizontalTHORN

                (Refer to Figure Q36)

                Figure Q36

                18 Effective stress

                37

                Situation (1)(a)

                frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                0 frac14 u frac14 694 392 frac14 302 kN=m2

                (b)

                i frac14 2

                4frac14 05

                j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                Situation (2)(a)

                frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                0 frac14 u frac14 498 392 frac14 106 kN=m2

                (b)

                i frac14 2

                4frac14 05

                j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                38

                The flow net is drawn in Figure Q24

                Loss in total head between adjacent equipotentials

                h frac14 550

                Ndfrac14 550

                11frac14 050m

                Exit hydraulic gradient

                ie frac14 h

                sfrac14 050

                070frac14 071

                Effective stress 19

                The critical hydraulic gradient is given by Equation 39

                ic frac14 0

                wfrac14 102

                98frac14 104

                Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                F frac14 iciefrac14 104

                071frac14 15

                Total head at C

                hC frac14 nd

                Ndh frac14 24

                11 550 frac14 120m

                Elevation head at C

                zC frac14 250m

                Pore water pressure at C

                uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                Therefore effective vertical stress at C

                0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                For point D

                hD frac14 73

                11 550 frac14 365m

                zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                39

                The flow net is drawn in Figure Q25

                For a soil prism 150 300m adjacent to the piling

                hm frac14 26

                9 500 frac14 145m

                20 Effective stress

                Factor of safety against lsquoheavingrsquo (Equation 310)

                F frac14 ic

                imfrac14 0d

                whmfrac14 97 300

                98 145frac14 20

                With a filter

                F frac14 0d thorn wwhm

                3 frac14 eth97 300THORN thorn w98 145

                w frac14 135 kN=m2

                Depth of filterfrac14 13521frac14 065m (if above water level)

                Effective stress 21

                Chapter 4

                Shear strength

                41

                frac14 295 kN=m2

                u frac14 120 kN=m2

                0 frac14 u frac14 295 120 frac14 175 kN=m2

                f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                42

                03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                100 452 552200 908 1108400 1810 2210800 3624 4424

                The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                Figure Q42

                43

                3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                200 222 422400 218 618600 220 820

                The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                44

                The modified shear strength parameters are

                0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                The coordinates of the stress point representing failure conditions in the test are

                1

                2eth1 2THORN frac14 1

                2 170 frac14 85 kN=m2

                1

                2eth1 thorn 3THORN frac14 1

                2eth270thorn 100THORN frac14 185 kN=m2

                The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                uf frac14 36 kN=m2

                Figure Q43

                Figure Q44

                Shear strength 23

                45

                3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                46

                03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                Figure Q45

                24 Shear strength

                47

                The torque required to produce shear failure is given by

                T frac14 dh cud

                2thorn 2

                Z d=2

                0

                2r drcur

                frac14 cud2h

                2thorn 4cu

                Z d=2

                0

                r2dr

                frac14 cud2h

                2thorn d

                3

                6

                Then

                35 frac14 cu52 10

                2thorn 53

                6

                103

                cu frac14 76 kN=m3

                400

                0 400 800 1200 1600

                τ (k

                Nm

                2 )

                σprime (kNm2)

                34deg

                315deg29deg

                (a)

                (b)

                0 400

                400

                800 1200 1600

                Failure envelope

                300 500

                σprime (kNm2)

                τ (k

                Nm

                2 )

                20 (kNm2)

                31deg

                Figure Q46

                Shear strength 25

                48

                The relevant stress values are calculated as follows

                3 frac14 600 kN=m2

                1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                2(1 3) 0 40 79 107 139 159

                1

                2(01 thorn 03) 400 411 402 389 351 326

                1

                2(1 thorn 3) 600 640 679 707 739 759

                The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                Af frac14 433 200

                319frac14 073

                49

                B frac14 u33

                frac14 144

                350 200frac14 096

                a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                10 283 65 023

                Figure Q48

                26 Shear strength

                The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                Figure Q49

                Shear strength 27

                Chapter 5

                Stresses and displacements

                51

                Vertical stress is given by

                z frac14 Qz2Ip frac14 5000

                52Ip

                Values of Ip are obtained from Table 51

                r (m) rz Ip z (kNm2)

                0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                10 20 0009 2

                The variation of z with radial distance (r) is plotted in Figure Q51

                Figure Q51

                52

                Below the centre load (Figure Q52)

                r

                zfrac14 0 for the 7500-kN load

                Ip frac14 0478

                r

                zfrac14 5

                4frac14 125 for the 10 000- and 9000-kN loads

                Ip frac14 0045

                Then

                z frac14X Q

                z2Ip

                frac14 7500 0478

                42thorn 10 000 0045

                42thorn 9000 0045

                42

                frac14 224thorn 28thorn 25 frac14 277 kN=m2

                53

                The vertical stress under a corner of a rectangular area is given by

                z frac14 qIr

                where values of Ir are obtained from Figure 510 In this case

                z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                z

                Figure Q52

                Stresses and displacements 29

                z (m) m n Ir z (kNm2)

                0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                10 010 0005 5

                z is plotted against z in Figure Q53

                54

                (a)

                m frac14 125

                12frac14 104

                n frac14 18

                12frac14 150

                From Figure 510 Irfrac14 0196

                z frac14 2 175 0196 frac14 68 kN=m2

                Figure Q53

                30 Stresses and displacements

                (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                55

                Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                Px frac14 2Q

                1

                m2 thorn 1frac14 2 150

                125frac14 76 kN=m

                Equation 517 is used to obtain the pressure distribution

                px frac14 4Q

                h

                m2n

                ethm2 thorn n2THORN2 frac14150

                m2n

                ethm2 thorn n2THORN2 ethkN=m2THORN

                Figure Q54

                Stresses and displacements 31

                n m2n

                (m2 thorn n2)2

                px(kNm2)

                0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                The pressure distribution is plotted in Figure Q55

                56

                H

                Bfrac14 10

                2frac14 5

                L

                Bfrac14 4

                2frac14 2

                D

                Bfrac14 1

                2frac14 05

                Hence from Figure 515

                131 frac14 082

                130 frac14 094

                Figure Q55

                32 Stresses and displacements

                The immediate settlement is given by Equation 528

                si frac14 130131qB

                Eu

                frac14 094 082 200 2

                45frac14 7mm

                Stresses and displacements 33

                Chapter 6

                Lateral earth pressure

                61

                For 0 frac14 37 the active pressure coefficient is given by

                Ka frac14 1 sin 37

                1thorn sin 37frac14 025

                The total active thrust (Equation 66a with c0 frac14 0) is

                Pa frac14 1

                2KaH

                2 frac14 1

                2 025 17 62 frac14 765 kN=m

                If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                and the thrust on the wall is

                P0 frac14 1

                2K0H

                2 frac14 1

                2 040 17 62 frac14 122 kN=m

                62

                The active pressure coefficients for the three soil types are as follows

                Ka1 frac141 sin 35

                1thorn sin 35frac14 0271

                Ka2 frac141 sin 27

                1thorn sin 27frac14 0375

                ffiffiffiffiffiffiffiKa2

                p frac14 0613

                Ka3 frac141 sin 42

                1thorn sin 42frac14 0198

                Distribution of active pressure (plotted in Figure Q62)

                Depth (m) Soil Active pressure (kNm2)

                3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                Total thrust frac14 571 kNm

                Point of application is (4893571) m from the top of the wall ie 857m

                Force (kN) Arm (m) Moment (kN m)

                (1)1

                2 0271 16 32 frac14 195 20 390

                (2) 0271 16 3 2 frac14 260 40 1040

                (3)1

                2 0271 92 22 frac14 50 433 217

                (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                (5)1

                2 0375 102 32 frac14 172 70 1204

                (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                (7)1

                2 0198 112 42 frac14 177 1067 1889

                (8)1

                2 98 92 frac14 3969 90 35721

                5713 48934

                Figure Q62

                Lateral earth pressure 35

                63

                (a) For u frac14 0 Ka frac14 Kp frac14 1

                Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                frac14 245

                At the lower end of the piling

                pa frac14 Kaqthorn Kasatz Kaccu

                frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                frac14 115 kN=m2

                pp frac14 Kpsatzthorn Kpccu

                frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                frac14 202 kN=m2

                (b) For 0 frac14 26 and frac14 1

                20

                Ka frac14 035

                Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                pfrac14 145 ethEquation 619THORN

                Kp frac14 37

                Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                pfrac14 47 ethEquation 624THORN

                At the lower end of the piling

                pa frac14 Kaqthorn Ka0z Kacc

                0

                frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                frac14 187 kN=m2

                pp frac14 Kp0zthorn Kpcc

                0

                frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                frac14 198 kN=m2

                36 Lateral earth pressure

                64

                (a) For 0 frac14 38 Ka frac14 024

                0 frac14 20 98 frac14 102 kN=m3

                The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                Force (kN) Arm (m) Moment (kN m)

                (1) 024 10 66 frac14 159 33 525

                (2)1

                2 024 17 392 frac14 310 400 1240

                (3) 024 17 39 27 frac14 430 135 580

                (4)1

                2 024 102 272 frac14 89 090 80

                (5)1

                2 98 272 frac14 357 090 321

                Hfrac14 1345 MH frac14 2746

                (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                XM frac14MV MH frac14 7790 kNm

                Lever arm of base resultant

                M

                Vfrac14 779

                488frac14 160

                Eccentricity of base resultant

                e frac14 200 160 frac14 040m

                39 m

                27 m

                40 m

                04 m

                04 m

                26 m

                (7)

                (9)

                (1)(2)

                (3)

                (4)

                (5)

                (8)(6)

                (10)

                WT

                10 kNm2

                Hydrostatic

                Figure Q64

                Lateral earth pressure 37

                Base pressures (Equation 627)

                p frac14 VB

                1 6e

                B

                frac14 488

                4eth1 060THORN

                frac14 195 kN=m2 and 49 kN=m2

                Factor of safety against sliding (Equation 628)

                F frac14 V tan

                Hfrac14 488 tan 25

                1345frac14 17

                (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                Hfrac14 1633 kN

                V frac14 4879 kN

                MH frac14 3453 kNm

                MV frac14 10536 kNm

                The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                65

                For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                Kp

                Ffrac14 385

                2

                0 frac14 20 98 frac14 102 kN=m3

                The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                Force (kN) Arm (m) Moment (kN m)

                (1)1

                2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                (2) 026 17 45 d frac14 199d d2 995d2

                (3)1

                2 026 102 d2 frac14 133d2 d3 044d3

                (4)1

                2 385

                2 17 152 frac14 368 dthorn 05 368d 184

                (5)385

                2 17 15 d frac14 491d d2 2455d2

                (6)1

                2 385

                2 102 d2 frac14 982d2 d3 327d3

                38 Lateral earth pressure

                XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                d3 thorn 516d2 283d 1724 frac14 0

                d frac14 179m

                Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                Over additional 20 embedded depth

                pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                66

                The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                Ka frac14 sin 69=sin 105

                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                26664

                37775

                2

                frac14 050

                The total active thrust (acting at 25 above the normal) is given by Equation 616

                Pa frac14 1

                2 050 19 7502 frac14 267 kN=m

                Figure Q65

                Lateral earth pressure 39

                Horizontal component

                Ph frac14 267 cos 40 frac14 205 kN=m

                Vertical component

                Pv frac14 267 sin 40 frac14 172 kN=m

                Consider moments about the toe of the wall (Figure Q66) (per m)

                Force (kN) Arm (m) Moment (kN m)

                (1)1

                2 175 650 235 frac14 1337 258 345

                (2) 050 650 235 frac14 764 175 134

                (3)1

                2 070 650 235 frac14 535 127 68

                (4) 100 400 235 frac14 940 200 188

                (5) 1

                2 080 050 235 frac14 47 027 1

                Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                Lever arm of base resultant

                M

                Vfrac14 795

                525frac14 151m

                Eccentricity of base resultant

                e frac14 200 151 frac14 049m

                Figure Q66

                40 Lateral earth pressure

                Base pressures (Equation 627)

                p frac14 525

                41 6 049

                4

                frac14 228 kN=m2 and 35 kN=m2

                The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                67

                For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                Force (kN) Arm (m) Moment (kNm)

                (1)1

                2 027 17 52 frac14 574 183 1050

                (2) 027 17 5 3 frac14 689 500 3445

                (3)1

                2 027 102 32 frac14 124 550 682

                (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                (5)1

                2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                (7) 1

                2 267

                2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                2 d frac14 163d d2thorn 650 82d2 1060d

                Tie rod force per m frac14 T 0 0

                XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                d3 thorn 77d2 269d 1438 frac14 0

                d frac14 467m

                Depth of penetration frac14 12d frac14 560m

                Lateral earth pressure 41

                Algebraic sum of forces for d frac14 467m isX

                F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                T frac14 905 kN=m

                Force in each tie rod frac14 25T frac14 226 kN

                68

                (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                0 frac14 21 98 frac14 112 kN=m3

                The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                uC frac14 150

                165 15 98 frac14 134 kN=m2

                The average seepage pressure is

                j frac14 15

                165 98 frac14 09 kN=m3

                Hence

                0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                0 j frac14 112 09 frac14 103 kN=m3

                Figure Q67

                42 Lateral earth pressure

                Consider moments about the anchor point A (per m)

                Force (kN) Arm (m) Moment (kN m)

                (1) 10 026 150 frac14 390 60 2340

                (2)1

                2 026 18 452 frac14 474 15 711

                (3) 026 18 45 105 frac14 2211 825 18240

                (4)1

                2 026 121 1052 frac14 1734 100 17340

                (5)1

                2 134 15 frac14 101 40 404

                (6) 134 30 frac14 402 60 2412

                (7)1

                2 134 60 frac14 402 95 3819

                571 4527(8) Ppm

                115 115PPm

                XM frac14 0

                Ppm frac144527

                115frac14 394 kN=m

                Available passive resistance

                Pp frac14 1

                2 385 103 62 frac14 714 kN=m

                Factor of safety

                Fp frac14 Pp

                Ppm

                frac14 714

                394frac14 18

                Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                Figure Q68

                Lateral earth pressure 43

                (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                Consider moments (per m) about the tie point A

                Force (kN) Arm (m)

                (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                (2)1

                2 033 18 452 frac14 601 15

                (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                (4)1

                2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                (5)1

                2 134 15 frac14 101 40

                (6) 134 30 frac14 402 60

                (7)1

                2 134 d frac14 67d d3thorn 75

                (8) 1

                2 30 103 d2 frac141545d2 2d3thorn 75

                Moment (kN m)

                (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                d3 thorn 827d2 466d 1518 frac14 0

                By trial

                d frac14 544m

                The minimum depth of embedment required is 544m

                69

                For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                0 frac14 20 98 frac14 102 kN=m3

                The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                44 Lateral earth pressure

                uC frac14 147

                173 26 98 frac14 216 kN=m2

                and the average seepage pressure around the wall is

                j frac14 26

                173 98 frac14 15 kN=m3

                Consider moments about the prop (A) (per m)

                Force (kN) Arm (m) Moment (kN m)

                (1)1

                2 03 17 272 frac14 186 020 37

                (2) 03 17 27 53 frac14 730 335 2445

                (3)1

                2 03 (102thorn 15) 532 frac14 493 423 2085

                (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                (5)1

                2 216 26 frac14 281 243 684

                (6) 216 27 frac14 583 465 2712

                (7)1

                2 216 60 frac14 648 800 5184

                3055(8)

                1

                2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                Factor of safety

                Fr frac14 6885

                3055frac14 225

                Figure Q69

                Lateral earth pressure 45

                610

                For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                Using the recommendations of Twine and Roscoe

                p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                611

                frac14 18 kN=m3 0 frac14 34

                H frac14 350m nH frac14 335m mH frac14 185m

                Consider a trial value of F frac14 20 Refer to Figure 635

                0m frac14 tan1tan 34

                20

                frac14 186

                Then

                frac14 45 thorn 0m2frac14 543

                W frac14 1

                2 18 3502 cot 543 frac14 792 kN=m

                Figure Q610

                46 Lateral earth pressure

                P frac14 1

                2 s 3352 frac14 561s kN=m

                U frac14 1

                2 98 1852 cosec 543 frac14 206 kN=m

                Equations 630 and 631 then become

                561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                ie

                561s 0616N 405 frac14 0

                792 0857N thorn 563 frac14 0

                N frac14 848

                0857frac14 989 kN=m

                Then

                561s 609 405 frac14 0

                s frac14 649

                561frac14 116 kN=m3

                The calculations for trial values of F of 20 15 and 10 are summarized below

                F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                Figure Q611

                Lateral earth pressure 47

                612

                For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                45 thorn 0

                2frac14 63

                For the retained material between the surface and a depth of 36m

                Pa frac14 1

                2 030 18 362 frac14 350 kN=m

                Weight of reinforced fill between the surface and a depth of 36m is

                Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                Eccentricity of Rv

                e frac14 263 250 frac14 013m

                The average vertical stress at a depth of 36m is

                z frac14 Rv

                L 2efrac14 324

                474frac14 68 kN=m2

                (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                Tensile stress in the element frac14 138 103

                65 3frac14 71N=mm2

                Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                The weight of ABC is

                W frac14 1

                2 18 52 265 frac14 124 kN=m

                From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                48 Lateral earth pressure

                (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                Tp frac14 032 68 120 065 frac14 170 kN

                Tr frac14 213 420

                418frac14 214 kN

                Again the tensile failure and slipping limit states are satisfied for this element

                Figure Q612

                Lateral earth pressure 49

                Chapter 7

                Consolidation theory

                71

                Total change in thickness

                H frac14 782 602 frac14 180mm

                Average thickness frac14 1530thorn 180

                2frac14 1620mm

                Length of drainage path d frac14 1620

                2frac14 810mm

                Root time plot (Figure Q71a)

                ffiffiffiffiffiffit90p frac14 33

                t90 frac14 109min

                cv frac14 0848d2

                t90frac14 0848 8102

                109 1440 365

                106frac14 27m2=year

                r0 frac14 782 764

                782 602frac14 018

                180frac14 0100

                rp frac14 10eth764 645THORN9eth782 602THORN frac14

                10 119

                9 180frac14 0735

                rs frac14 1 eth0100thorn 0735THORN frac14 0165

                Log time plot (Figure Q71b)

                t50 frac14 26min

                cv frac14 0196d2

                t50frac14 0196 8102

                26 1440 365

                106frac14 26m2=year

                r0 frac14 782 763

                782 602frac14 019

                180frac14 0106

                rp frac14 763 623

                782 602frac14 140

                180frac14 0778

                rs frac14 1 eth0106thorn 0778THORN frac14 0116

                Figure Q71(a)

                Figure Q71(b)

                Final void ratio

                e1 frac14 w1Gs frac14 0232 272 frac14 0631

                e

                Hfrac14 1thorn e0

                H0frac14 1thorn e1 thorne

                H0

                ie

                e

                180frac14 1631thorne

                1710

                e frac14 2936

                1530frac14 0192

                Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                mv frac14 1

                1thorn e0 e0 e101 00

                frac14 1

                1823 0192

                0107frac14 098m2=MN

                k frac14 cvmvw frac14 265 098 98

                60 1440 365 103frac14 81 1010 m=s

                72

                Using Equation 77 (one-dimensional method)

                sc frac14 e0 e11thorn e0 H

                Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                Figure Q72

                52 Consolidation theory

                Settlement

                Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                318

                Notes 5 92y 460thorn 84

                Heave

                Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                38

                73

                U frac14 f ethTvTHORN frac14 f cvt

                d2

                Hence if cv is constant

                t1

                t2frac14 d

                21

                d22

                where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                d1 frac14 95mm and d2 frac14 2500mm

                for U frac14 050 t2 frac14 t1 d22

                d21

                frac14 20

                60 24 365 25002

                952frac14 263 years

                for U lt 060 Tv frac14

                4U2 (Equation 724(a))

                t030 frac14 t050 0302

                0502

                frac14 263 036 frac14 095 years

                Consolidation theory 53

                74

                The layer is open

                d frac14 8

                2frac14 4m

                Tv frac14 cvtd2frac14 24 3

                42frac14 0450

                ui frac14 frac14 84 kN=m2

                The excess pore water pressure is given by Equation 721

                ue frac14Xmfrac141mfrac140

                2ui

                Msin

                Mz

                d

                expethM2TvTHORN

                In this case z frac14 d

                sinMz

                d

                frac14 sinM

                where

                M frac14

                23

                25

                2

                M sin M M2Tv exp (M2Tv)

                2thorn1 1110 0329

                3

                21 9993 457 105

                ue frac14 2 84 2

                1 0329 ethother terms negligibleTHORN

                frac14 352 kN=m2

                75

                The layer is open

                d frac14 6

                2frac14 3m

                Tv frac14 cvtd2frac14 10 3

                32frac14 0333

                The layer thickness will be divided into six equal parts ie m frac14 6

                54 Consolidation theory

                For an open layer

                Tv frac14 4n

                m2

                n frac14 0333 62

                4frac14 300

                The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                i j

                0 1 2 3 4 5 6 7 8 9 10 11 12

                0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                The initial and 3-year isochrones are plotted in Figure Q75

                Area under initial isochrone frac14 180 units

                Area under 3-year isochrone frac14 63 units

                The average degree of consolidation is given by Equation 725Thus

                U frac14 1 63

                180frac14 065

                Figure Q75

                Consolidation theory 55

                76

                At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                0 frac14 2w frac14 2 98 frac14 196 kN=m2

                The final consolidation settlement (one-dimensional method) is

                sc frac14 mv0H frac14 083 196 8 frac14 130mm

                Corrected time t frac14 2 1

                2

                40

                52

                frac14 1615 years

                Tv frac14 cvtd2frac14 44 1615

                42frac14 0444

                From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                77

                The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                Figure Q77

                56 Consolidation theory

                Point m n Ir (kNm2) sc (mm)

                13020frac14 15 20

                20frac14 10 0194 (4) 113 124

                260

                20frac14 30

                20

                20frac14 10 0204 (2) 59 65

                360

                20frac14 30

                40

                20frac14 20 0238 (1) 35 38

                430

                20frac14 15

                40

                20frac14 20 0224 (2) 65 72

                Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                78

                Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                (a) Immediate settlement

                H

                Bfrac14 30

                35frac14 086

                D

                Bfrac14 2

                35frac14 006

                Figure Q78

                Consolidation theory 57

                From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                si frac14 130131qB

                Eufrac14 10 032 105 35

                40frac14 30mm

                (b) Consolidation settlement

                Layer z (m) Dz Ic (kNm2) syod (mm)

                1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                3150

                Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                Now

                H

                Bfrac14 30

                35frac14 086 and A frac14 065

                from Figure 712 13 frac14 079

                sc frac14 13sod frac14 079 315 frac14 250mm

                Total settlement

                s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                79

                Without sand drains

                Uv frac14 025

                Tv frac14 0049 ethfrom Figure 718THORN

                t frac14 Tvd2

                cvfrac14 0049 82

                cvWith sand drains

                R frac14 0564S frac14 0564 3 frac14 169m

                n frac14 Rrfrac14 169

                015frac14 113

                Tr frac14 cht

                4R2frac14 ch

                4 1692 0049 82

                cvethand ch frac14 cvTHORN

                frac14 0275

                Ur frac14 073 (from Figure 730)

                58 Consolidation theory

                Using Equation 740

                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                U frac14 080

                710

                Without sand drains

                Uv frac14 090

                Tv frac14 0848

                t frac14 Tvd2

                cvfrac14 0848 102

                96frac14 88 years

                With sand drains

                R frac14 0564S frac14 0564 4 frac14 226m

                n frac14 Rrfrac14 226

                015frac14 15

                Tr

                Tvfrac14 chcv

                d2

                4R2ethsame tTHORN

                Tr

                Tvfrac14 140

                96 102

                4 2262frac14 714 eth1THORN

                Using Equation 740

                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                Thus

                Uv frac14 0295 and Ur frac14 086

                t frac14 88 00683

                0848frac14 07 years

                Consolidation theory 59

                Chapter 8

                Bearing capacity

                81

                (a) The ultimate bearing capacity is given by Equation 83

                qf frac14 cNc thorn DNq thorn 1

                2BN

                For u frac14 0

                Nc frac14 514 Nq frac14 1 N frac14 0

                qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                The net ultimate bearing capacity is

                qnf frac14 qf D frac14 540 kN=m2

                The net foundation pressure is

                qn frac14 q D frac14 425

                2 eth21 1THORN frac14 192 kN=m2

                The factor of safety (Equation 86) is

                F frac14 qnfqnfrac14 540

                192frac14 28

                (b) For 0 frac14 28

                Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                2 112 2 13

                frac14 260thorn 168thorn 146 frac14 574 kN=m2

                qnf frac14 574 112 frac14 563 kN=m2

                F frac14 563

                192frac14 29

                (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                82

                For 0 frac14 38

                Nq frac14 49 N frac14 67

                qnf frac14 DethNq 1THORN thorn 1

                2BN ethfrom Equation 83THORN

                frac14 eth18 075 48THORN thorn 1

                2 18 15 67

                frac14 648thorn 905 frac14 1553 kN=m2

                qn frac14 500

                15 eth18 075THORN frac14 320 kN=m2

                F frac14 qnfqnfrac14 1553

                320frac14 48

                0d frac14 tan1tan 38

                125

                frac14 32 therefore Nq frac14 23 and N frac14 25

                Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                2 18 15 25

                frac14 15eth310thorn 337THORNfrac14 970 kN=m

                Design load (action) Vd frac14 500 kN=m

                The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                83

                D

                Bfrac14 350

                225frac14 155

                From Figure 85 for a square foundation

                Nc frac14 81

                Bearing capacity 61

                For a rectangular foundation (L frac14 450m B frac14 225m)

                Nc frac14 084thorn 016B

                L

                81 frac14 745

                Using Equation 810

                qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                For F frac14 3

                qn frac14 1006

                3frac14 335 kN=m2

                q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                Design load frac14 405 450 225 frac14 4100 kN

                Design undrained strength cud frac14 135

                14frac14 96 kN=m2

                Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                frac14 7241 kN

                Design load Vd frac14 4100 kN

                Rd gt Vd therefore the bearing resistance limit state is satisfied

                84

                For 0 frac14 40

                Nq frac14 64 N frac14 95

                qnf frac14 DethNq 1THORN thorn 04BN

                (a) Water table 5m below ground level

                qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                qn frac14 400 17 frac14 383 kN=m2

                F frac14 2686

                383frac14 70

                (b) Water table 1m below ground level (ie at foundation level)

                0 frac14 20 98 frac14 102 kN=m3

                62 Bearing capacity

                qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                F frac14 2040

                383frac14 53

                (c) Water table at ground level with upward hydraulic gradient 02

                eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                F frac14 1296

                392frac14 33

                85

                The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                Design value of 0 frac14 tan1tan 39

                125

                frac14 33

                For 0 frac14 33 Nq frac14 26 and N frac14 29

                Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                Rd gt Vd therefore the bearing resistance limit state is satisfied

                86

                (a) Undrained shear for u frac14 0

                Nc frac14 514 Nq frac14 1 N frac14 0

                qnf frac14 12cuNc

                frac14 12 100 514 frac14 617 kN=m2

                qn frac14 qnfFfrac14 617

                3frac14 206 kN=m2

                q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                Bearing capacity 63

                Drained shear for 0 frac14 32

                Nq frac14 23 N frac14 25

                0 frac14 21 98 frac14 112 kN=m3

                qnf frac14 0DethNq 1THORN thorn 040BN

                frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                frac14 694 kN=m2

                q frac14 694

                3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                Design load frac14 42 227 frac14 3632 kN

                (b) Design undrained strength cud frac14 100

                14frac14 71 kNm2

                Design bearing resistance Rd frac14 12cudNe area

                frac14 12 71 514 42

                frac14 7007 kN

                For drained shear 0d frac14 tan1tan 32

                125

                frac14 26

                Nq frac14 12 N frac14 10

                Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                Layer z (m) m n Ir 0 (kNm2) sod (mm)

                1 2 100 0175 0700qn 0182qn

                2 6 033 0044 0176qn 0046qn

                3 10 020 0017 0068qn 0018qn

                0246qn

                Diameter of equivalent circle B frac14 45m

                H

                Bfrac14 12

                45frac14 27 and A frac14 042

                13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                64 Bearing capacity

                For sc frac14 30mm

                qn frac14 30

                0147frac14 204 kN=m2

                q frac14 204thorn 21 frac14 225 kN=m2

                Design load frac14 42 225 frac14 3600 kN

                The design load is 3600 kN settlement being the limiting criterion

                87

                D

                Bfrac14 8

                4frac14 20

                From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                F frac14 cuNc

                Dfrac14 40 71

                20 8frac14 18

                88

                Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                Design value of 0 frac14 tan1tan 38

                125

                frac14 32

                Figure Q86

                Bearing capacity 65

                For 0 frac14 32 Nq frac14 23 and N frac14 25

                Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                For B frac14 250m qn frac14 3750

                2502 17 frac14 583 kN=m2

                From Figure 510 m frac14 n frac14 126

                6frac14 021

                Ir frac14 0019

                Stress increment frac14 4 0019 583 frac14 44 kN=m2

                Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                89

                Depth (m) N 0v (kNm2) CN N1

                070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                Cw frac14 05thorn 0530

                47

                frac14 082

                66 Bearing capacity

                Thus

                qa frac14 150 082 frac14 120 kN=m2

                (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                Thus

                qa frac14 90 15 frac14 135 kN=m2

                (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                Ic frac14 171

                1014frac14 0068

                From Equation 819(a) with s frac14 25mm

                q frac14 25

                3507 0068frac14 150 kN=m2

                810

                Peak value of strain influence factor occurs at a depth of 27m and is given by

                Izp frac14 05thorn 01130

                16 27

                05

                frac14 067

                Refer to Figure Q810

                E frac14 25qc

                Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                Ez (mm3MN)

                1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                0203

                C1 frac14 1 0500qnfrac14 1 05 12 16

                130frac14 093

                C2 frac14 1 ethsayTHORN

                s frac14 C1C2qnX Iz

                Ez frac14 093 1 130 0203 frac14 25mm

                Bearing capacity 67

                811

                At pile base level

                cu frac14 220 kN=m2

                qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                Then

                Qf frac14 Abqb thorn Asqs

                frac14

                4 32 1980

                thorn eth 105 139 86THORN

                frac14 13 996thorn 3941 frac14 17 937 kN

                0 01 02 03 04 05 06 07

                0 2 4 6 8 10 12 14

                1

                2

                3

                4

                5

                6

                7

                8

                (1)

                (2)

                (3)

                (4)

                (5)

                qc

                qc

                Iz

                Iz

                (MNm2)

                z (m)

                Figure Q810

                68 Bearing capacity

                Allowable load

                ethaTHORN Qf

                2frac14 17 937

                2frac14 8968 kN

                ethbTHORN Abqb

                3thorn Asqs frac14 13 996

                3thorn 3941 frac14 8606 kN

                ie allowable load frac14 8600 kN

                Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                According to the limit state method

                Characteristic undrained strength at base level cuk frac14 220

                150kN=m2

                Characteristic base resistance qbk frac14 9cuk frac14 9 220

                150frac14 1320 kN=m2

                Characteristic shaft resistance qsk frac14 00150

                frac14 86

                150frac14 57 kN=m2

                Characteristic base and shaft resistances

                Rbk frac14

                4 32 1320 frac14 9330 kN

                Rsk frac14 105 139 86

                150frac14 2629 kN

                For a bored pile the partial factors are b frac14 160 and s frac14 130

                Design bearing resistance Rcd frac14 9330

                160thorn 2629

                130

                frac14 5831thorn 2022

                frac14 7850 kN

                Adding ethDAb W) the design bearing resistance becomes 9650 kN

                812

                ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                qs frac14 cu frac14 040 105 frac14 42 kN=m2

                For a single pile

                Qf frac14 Abqb thorn Asqs

                frac14

                4 062 1305

                thorn eth 06 15 42THORN

                frac14 369thorn 1187 frac14 1556 kN

                Bearing capacity 69

                Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                qbkfrac14 9cuk frac14 9 220

                150frac14 1320 kN=m2

                qskfrac14cuk frac14 040 105

                150frac14 28 kN=m2

                Rbkfrac14

                4 0602 1320 frac14 373 kN

                Rskfrac14 060 15 28 frac14 791 kN

                Rcdfrac14 373

                160thorn 791

                130frac14 233thorn 608 frac14 841 kN

                Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                q frac14 21 000

                1762frac14 68 kN=m2

                Immediate settlement

                H

                Bfrac14 15

                176frac14 085

                D

                Bfrac14 13

                176frac14 074

                L

                Bfrac14 1

                Hence from Figure 515

                130 frac14 078 and 131 frac14 041

                70 Bearing capacity

                Thus using Equation 528

                si frac14 078 041 68 176

                65frac14 6mm

                Consolidation settlement

                Layer z (m) Area (m2) (kNm2) mvH (mm)

                1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                434 (sod)

                Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                sc frac14 056 434 frac14 24mm

                The total settlement is (6thorn 24) frac14 30mm

                813

                At base level N frac14 26 Then using Equation 830

                qb frac14 40NDb

                Bfrac14 40 26 2

                025frac14 8320 kN=m2

                ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                Figure Q812

                Bearing capacity 71

                Over the length embedded in sand

                N frac14 21 ie18thorn 24

                2

                Using Equation 831

                qs frac14 2N frac14 2 21 frac14 42 kN=m2

                For a single pile

                Qf frac14 Abqb thorn Asqs

                frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                For the pile group assuming a group efficiency of 12

                XQf frac14 12 9 604 frac14 6523 kN

                Then the load factor is

                F frac14 6523

                2000thorn 1000frac14 21

                (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                Characteristic base resistance per unit area qbk frac14 8320

                150frac14 5547 kNm2

                Characteristic shaft resistance per unit area qsk frac14 42

                150frac14 28 kNm2

                Characteristic base and shaft resistances for a single pile

                Rbk frac14 0252 5547 frac14 347 kN

                Rsk frac14 4 025 2 28 frac14 56 kN

                For a driven pile the partial factors are b frac14 s frac14 130

                Design bearing resistance Rcd frac14 347

                130thorn 56

                130frac14 310 kN

                For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                72 Bearing capacity

                N frac14 24thorn 26thorn 34

                3frac14 28

                Ic frac14 171

                2814frac14 0016 ethEquation 818THORN

                s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                814

                Using Equation 841

                Tf frac14 DLcu thorn

                4ethD2 d2THORNcuNc

                frac14 eth 02 5 06 110THORN thorn

                4eth022 012THORN110 9

                frac14 207thorn 23 frac14 230 kN

                Figure Q813

                Bearing capacity 73

                Chapter 9

                Stability of slopes

                91

                Referring to Figure Q91

                W frac14 417 19 frac14 792 kN=m

                Q frac14 20 28 frac14 56 kN=m

                Arc lengthAB frac14

                180 73 90 frac14 115m

                Arc length BC frac14

                180 28 90 frac14 44m

                The factor of safety is given by

                F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                Depth of tension crack z0 frac14 2cu

                frac14 2 20

                19frac14 21m

                Arc length BD frac14

                180 13

                1

                2 90 frac14 21m

                F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                Design resisting moment frac14 rXethcudLaTHORN frac14 90

                14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                92

                u frac14 0

                Depth factor D frac14 11

                9frac14 122

                Using Equation 92 with F frac14 10

                Ns frac14 cu

                FHfrac14 30

                10 19 9frac14 0175

                Hence from Figure 93

                frac14 50

                For F frac14 12

                Ns frac14 30

                12 19 9frac14 0146

                frac14 27

                93

                Refer to Figure Q93

                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                74 m

                214 1deg

                213 1deg

                39 m

                WB

                D

                C

                28 m

                21 m

                A

                Q

                Soil (1)Soil (2)

                73deg

                Figure Q91

                Stability of slopes 75

                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                599 256 328 1372

                Figure Q93

                76 Stability of slopes

                XW cos frac14 b

                Xh cos frac14 21 2 599 frac14 2516 kN=mX

                W sin frac14 bX

                h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                Arc length La frac14

                180 57

                1

                2 326 frac14 327m

                The factor of safety is given by

                F frac14 c0La thorn tan0ethW cos ulTHORN

                W sin

                frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                frac14 091

                According to the limit state method

                0d frac14 tan1tan 32

                125

                frac14 265

                c0 frac14 8

                160frac14 5 kN=m2

                Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                Design disturbing moment frac14 1075 kN=m

                The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                94

                F frac14 1

                W sin

                Xfc0bthorn ethW ubTHORN tan0g sec

                1thorn ethtan tan0=FTHORN

                c0 frac14 8 kN=m2

                0 frac14 32

                c0b frac14 8 2 frac14 16 kN=m

                W frac14 bh frac14 21 2 h frac14 42h kN=m

                Try F frac14 100

                tan0

                Ffrac14 0625

                Stability of slopes 77

                Values of u are as obtained in Figure Q93

                SliceNo

                h(m)

                W frac14 bh(kNm)

                W sin(kNm)

                ub(kNm)

                c0bthorn (W ub) tan0(kNm)

                sec

                1thorn (tan tan0)FProduct(kNm)

                1 05 21 6 2 8 24 1078 262 13 55 31

                23 33 30 1042 31

                3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                224 92 72 0931 67

                6 50 210 11 40 100 85 0907 777 55 231 14

                12 58 112 90 0889 80

                8 60 252 1812

                80 114 102 0874 899 63 265 22 99 116 109 0861 94

                10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                2154 88 116 0853 99

                14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                1074 1091

                F frac14 1091

                1074frac14 102 (assumed value 100)

                Thus

                F frac14 101

                95

                F frac14 1

                W sin

                XfWeth1 ruTHORN tan0g sec

                1thorn ethtan tan0THORN=F

                0 frac14 33

                ru frac14 020

                W frac14 bh frac14 20 5 h frac14 100h kN=m

                eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                Try F frac14 110

                tan 0

                Ffrac14 tan 33

                110frac14 0590

                78 Stability of slopes

                Referring to Figure Q95

                SliceNo

                h(m)

                W frac14 bh(kNm)

                W sin(kNm)

                W(1 ru) tan0(kNm)

                sec

                1thorn ( tan tan0)FProduct(kNm)

                1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                2120 234 0892 209

                4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                1185 1271

                Figure Q95

                Stability of slopes 79

                F frac14 1271

                1185frac14 107

                The trial value was 110 therefore take F to be 108

                96

                (a) Water table at surface the factor of safety is given by Equation 912

                F frac14 0

                sat

                tan0

                tan

                ptie 15 frac14 92

                19

                tan 36

                tan

                tan frac14 0234

                frac14 13

                Water table well below surface the factor of safety is given by Equation 911

                F frac14 tan0

                tan

                frac14 tan 36

                tan 13

                frac14 31

                (b) 0d frac14 tan1tan 36

                125

                frac14 30

                Depth of potential failure surface frac14 z

                Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                frac14 504z kN

                Design disturbing moment per unit area Sd frac14 sat sin cos

                frac14 19 z sin 13 cos 13

                frac14 416z kN

                Rd gtSd therefore the limit state for overall stability is satisfied

                80 Stability of slopes

                • Book Cover
                • Title
                • Contents
                • Basic characteristics of soils
                • Seepage
                • Effective stress
                • Shear strength
                • Stresses and displacements
                • Lateral earth pressure
                • Consolidation theory
                • Bearing capacity
                • Stability of slopes

                  Using Equation 113

                  w frac14 Sre

                  Gsfrac14 075 070

                  272frac14 0193 eth193THORN

                  The reader should not attempt to memorize the above equations Figure 110(b)should be drawn and from a knowledge of the definitions relevant expressions canbe written by inspection

                  14

                  Volume of specimenfrac14

                  438276 frac14 86 200mm3

                  Bulk density ethTHORN frac14 Mass

                  Volumefrac14 1680

                  86 200 103frac14 195Mg=m3

                  Water content ethwTHORN frac14 1680 1305

                  1305frac14 0287 eth287THORN

                  From Equation 117

                  1thorn e frac14 Gseth1thorn wTHORN wfrac14 273 1287 100

                  195frac14 180

                  e frac14 080

                  Using Equation 113

                  Sr frac14 wGs

                  efrac14 0287 273

                  080frac14 098 eth98THORN

                  15

                  Using Equation 124

                  d frac14

                  1thorn w frac14215

                  112frac14 192Mg=m3

                  From Equation 117

                  1thorn e frac14 Gseth1thorn wTHORN wfrac14 265 112 100

                  215frac14 138

                  e frac14 038

                  Using Equation 113

                  Sr frac14 wGs

                  efrac14 012 265

                  038frac14 0837 eth837THORN

                  Basic characteristics of soils 3

                  Using Equation 115

                  Afrac14 e wGs

                  1thorn e frac14038 0318

                  138frac14 0045 eth45THORN

                  The zero air voids dry density is given by Equation 125

                  d frac14 Gs

                  1thorn wGsw frac14 265

                  1thorn eth0135 265THORN 100 frac14 195Mg=m3

                  ie a dry density of 200Mgm3 would not be possible

                  16

                  Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

                  (Mgm3) d10(Mgm3)

                  2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

                  In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

                  wopt frac14 15 and dmaxfrac14 183Mg=m3

                  Figure Q16

                  4 Basic characteristics of soils

                  Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

                  d5 d10

                  respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

                  17

                  From Equation 120

                  e frac14 Gswd 1

                  The maximum and minimum values of void ratio are given by

                  emax frac14 Gsw

                  dmin

                  1

                  emin frac14 Gswdmax

                  1

                  From Equation 123

                  ID frac14 Gsweth1=dmin 1=dTHORN

                  Gsweth1=dmin 1=dmax

                  THORN

                  frac14 frac121 ethdmin=dTHORN1=dmin

                  frac121 ethdmin=dmax

                  THORN1=dmin

                  frac14 d dmin

                  dmax dmin

                  dmax

                  d

                  frac14 172 154

                  181 154

                  181

                  172

                  frac14 070 eth70THORN

                  Basic characteristics of soils 5

                  Chapter 2

                  Seepage

                  21

                  The coefficient of permeability is determined from the equation

                  k frac14 23al

                  At1log

                  h0

                  h1

                  where

                  a frac14

                  4 00052 m2 l frac14 02m

                  A frac14

                  4 012 m2 t1 frac14 3 602 s

                  logh0

                  h1frac14 log

                  100

                  035frac14 0456

                  k frac14 23 00052 02 0456

                  012 3 602frac14 49 108 m=s

                  22

                  The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

                  q frac14 kh Nf

                  Ndfrac14 106 400 37

                  11frac14 13 106 m3=s per m

                  Figure Q22

                  23

                  The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

                  q frac14 kh Nf

                  Ndfrac14 5 105 300 35

                  9frac14 58 105 m3=s per m

                  The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

                  Point h (m) h z (m) u frac14 w(h z)(kNm2)

                  1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

                  eg for Point 1

                  h1 frac14 7

                  9 300 frac14 233m

                  h1 z1 frac14 233 eth250THORN frac14 483m

                  Figure Q23

                  Seepage 7

                  u1 frac14 98 483 frac14 47 kN=m2

                  The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

                  24

                  The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

                  q frac14 kh Nf

                  Ndfrac14 40 107 550 10

                  11frac14 20 106 m3=s per m

                  25

                  The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

                  q frac14 kh Nf

                  Ndfrac14 20 106 500 42

                  9frac14 47 106 m3=s per m

                  Figure Q24

                  8 Seepage

                  26

                  The scale transformation factor in the x direction is given by Equation 221 ie

                  xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                  ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

                  Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

                  k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                  pfrac14

                  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

                  p 107 frac14 30 107 m=s

                  Thus the quantity of seepage is given by

                  q frac14 k0h Nf

                  Ndfrac14 30 107 1400 325

                  12frac14 11 106 m3=s per m

                  Figure Q25

                  Seepage 9

                  27

                  The scale transformation factor in the x direction is

                  xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                  ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

                  Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

                  k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                  pfrac14

                  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

                  p 106 frac14 45 106 m=s

                  The focus of the basic parabola is at point A The parabola passes through point Gsuch that

                  GC frac14 03HC frac14 03 30 frac14 90m

                  Thus the coordinates of G are

                  x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

                  480 frac14 x0 2002

                  4x0

                  Figure Q26

                  10 Seepage

                  Hence

                  x0 frac14 20m

                  Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

                  x frac14 20 z2

                  80

                  x 20 0 50 100 200 300z 0 400 748 980 1327 1600

                  The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

                  No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

                  q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

                  28

                  The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

                  Figure Q27

                  Seepage 11

                  q frac14 kh Nf

                  Ndfrac14 45 105 28 33

                  7

                  frac14 59 105 m3=s per m

                  29

                  The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

                  kx frac14 H1k1 thornH2k2

                  H1 thornH2frac14 106

                  10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

                  kz frac14 H1 thornH2

                  H1

                  k1thornH2

                  k2

                  frac14 10

                  5

                  eth2 106THORN thorn5

                  eth16 106THORNfrac14 36 106 m=s

                  Then the scale transformation factor is given by

                  xt frac14 xffiffiffiffiffikz

                  pffiffiffiffiffikx

                  p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

                  Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

                  Figure Q28

                  12 Seepage

                  k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                  qfrac14

                  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

                  p 106 frac14 57 106 m=s

                  Then the quantity of seepage is given by

                  q frac14 k0h Nf

                  Ndfrac14 57 106 350 56

                  11

                  frac14 10 105 m3=s per m

                  Figure Q29

                  Seepage 13

                  Chapter 3

                  Effective stress

                  31

                  Buoyant unit weight

                  0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                  Effective vertical stress

                  0v frac14 5 102 frac14 51 kN=m2 or

                  Total vertical stress

                  v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

                  Pore water pressure

                  u frac14 7 98 frac14 686 kN=m2

                  Effective vertical stress

                  0v frac14 v u frac14 1196 686 frac14 51 kN=m2

                  32

                  Buoyant unit weight

                  0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                  Effective vertical stress

                  0v frac14 5 102 frac14 51 kN=m2 or

                  Total vertical stress

                  v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

                  Pore water pressure

                  u frac14 205 98 frac14 2009 kN=m2

                  Effective vertical stress

                  0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

                  33

                  At top of the clay

                  v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

                  u frac14 2 98 frac14 196 kN=m2

                  0v frac14 v u frac14 710 196 frac14 514 kN=m2

                  Alternatively

                  0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                  0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

                  At bottom of the clay

                  v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

                  u frac14 12 98 frac14 1176 kN=m2

                  0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

                  NB The alternative method of calculation is not applicable because of the artesiancondition

                  Figure Q3132

                  Effective stress 15

                  34

                  0 frac14 20 98 frac14 102 kN=m3

                  At 8m depth

                  0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                  35

                  0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                  0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                  Figure Q33

                  Figure Q34

                  16 Effective stress

                  (a) Immediately after WT rise

                  At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                  0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                  At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                  0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                  (b) Several years after WT rise

                  At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                  At 8m depth

                  0v frac14 940 kN=m2 (as above)

                  At 12m depth

                  0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                  Figure Q35

                  Effective stress 17

                  36

                  Total weight

                  ab frac14 210 kN

                  Effective weight

                  ac frac14 112 kN

                  Resultant boundary water force

                  be frac14 119 kN

                  Seepage force

                  ce frac14 34 kN

                  Resultant body force

                  ae frac14 99 kN eth73 to horizontalTHORN

                  (Refer to Figure Q36)

                  Figure Q36

                  18 Effective stress

                  37

                  Situation (1)(a)

                  frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                  u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                  0 frac14 u frac14 694 392 frac14 302 kN=m2

                  (b)

                  i frac14 2

                  4frac14 05

                  j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                  Situation (2)(a)

                  frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                  u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                  0 frac14 u frac14 498 392 frac14 106 kN=m2

                  (b)

                  i frac14 2

                  4frac14 05

                  j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                  38

                  The flow net is drawn in Figure Q24

                  Loss in total head between adjacent equipotentials

                  h frac14 550

                  Ndfrac14 550

                  11frac14 050m

                  Exit hydraulic gradient

                  ie frac14 h

                  sfrac14 050

                  070frac14 071

                  Effective stress 19

                  The critical hydraulic gradient is given by Equation 39

                  ic frac14 0

                  wfrac14 102

                  98frac14 104

                  Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                  F frac14 iciefrac14 104

                  071frac14 15

                  Total head at C

                  hC frac14 nd

                  Ndh frac14 24

                  11 550 frac14 120m

                  Elevation head at C

                  zC frac14 250m

                  Pore water pressure at C

                  uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                  Therefore effective vertical stress at C

                  0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                  For point D

                  hD frac14 73

                  11 550 frac14 365m

                  zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                  0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                  39

                  The flow net is drawn in Figure Q25

                  For a soil prism 150 300m adjacent to the piling

                  hm frac14 26

                  9 500 frac14 145m

                  20 Effective stress

                  Factor of safety against lsquoheavingrsquo (Equation 310)

                  F frac14 ic

                  imfrac14 0d

                  whmfrac14 97 300

                  98 145frac14 20

                  With a filter

                  F frac14 0d thorn wwhm

                  3 frac14 eth97 300THORN thorn w98 145

                  w frac14 135 kN=m2

                  Depth of filterfrac14 13521frac14 065m (if above water level)

                  Effective stress 21

                  Chapter 4

                  Shear strength

                  41

                  frac14 295 kN=m2

                  u frac14 120 kN=m2

                  0 frac14 u frac14 295 120 frac14 175 kN=m2

                  f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                  42

                  03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                  100 452 552200 908 1108400 1810 2210800 3624 4424

                  The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                  Figure Q42

                  43

                  3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                  200 222 422400 218 618600 220 820

                  The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                  44

                  The modified shear strength parameters are

                  0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                  a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                  The coordinates of the stress point representing failure conditions in the test are

                  1

                  2eth1 2THORN frac14 1

                  2 170 frac14 85 kN=m2

                  1

                  2eth1 thorn 3THORN frac14 1

                  2eth270thorn 100THORN frac14 185 kN=m2

                  The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                  uf frac14 36 kN=m2

                  Figure Q43

                  Figure Q44

                  Shear strength 23

                  45

                  3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                  150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                  The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                  The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                  46

                  03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                  200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                  The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                  Figure Q45

                  24 Shear strength

                  47

                  The torque required to produce shear failure is given by

                  T frac14 dh cud

                  2thorn 2

                  Z d=2

                  0

                  2r drcur

                  frac14 cud2h

                  2thorn 4cu

                  Z d=2

                  0

                  r2dr

                  frac14 cud2h

                  2thorn d

                  3

                  6

                  Then

                  35 frac14 cu52 10

                  2thorn 53

                  6

                  103

                  cu frac14 76 kN=m3

                  400

                  0 400 800 1200 1600

                  τ (k

                  Nm

                  2 )

                  σprime (kNm2)

                  34deg

                  315deg29deg

                  (a)

                  (b)

                  0 400

                  400

                  800 1200 1600

                  Failure envelope

                  300 500

                  σprime (kNm2)

                  τ (k

                  Nm

                  2 )

                  20 (kNm2)

                  31deg

                  Figure Q46

                  Shear strength 25

                  48

                  The relevant stress values are calculated as follows

                  3 frac14 600 kN=m2

                  1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                  2(1 3) 0 40 79 107 139 159

                  1

                  2(01 thorn 03) 400 411 402 389 351 326

                  1

                  2(1 thorn 3) 600 640 679 707 739 759

                  The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                  Af frac14 433 200

                  319frac14 073

                  49

                  B frac14 u33

                  frac14 144

                  350 200frac14 096

                  a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                  0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                  10 283 65 023

                  Figure Q48

                  26 Shear strength

                  The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                  Figure Q49

                  Shear strength 27

                  Chapter 5

                  Stresses and displacements

                  51

                  Vertical stress is given by

                  z frac14 Qz2Ip frac14 5000

                  52Ip

                  Values of Ip are obtained from Table 51

                  r (m) rz Ip z (kNm2)

                  0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                  10 20 0009 2

                  The variation of z with radial distance (r) is plotted in Figure Q51

                  Figure Q51

                  52

                  Below the centre load (Figure Q52)

                  r

                  zfrac14 0 for the 7500-kN load

                  Ip frac14 0478

                  r

                  zfrac14 5

                  4frac14 125 for the 10 000- and 9000-kN loads

                  Ip frac14 0045

                  Then

                  z frac14X Q

                  z2Ip

                  frac14 7500 0478

                  42thorn 10 000 0045

                  42thorn 9000 0045

                  42

                  frac14 224thorn 28thorn 25 frac14 277 kN=m2

                  53

                  The vertical stress under a corner of a rectangular area is given by

                  z frac14 qIr

                  where values of Ir are obtained from Figure 510 In this case

                  z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                  z

                  Figure Q52

                  Stresses and displacements 29

                  z (m) m n Ir z (kNm2)

                  0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                  10 010 0005 5

                  z is plotted against z in Figure Q53

                  54

                  (a)

                  m frac14 125

                  12frac14 104

                  n frac14 18

                  12frac14 150

                  From Figure 510 Irfrac14 0196

                  z frac14 2 175 0196 frac14 68 kN=m2

                  Figure Q53

                  30 Stresses and displacements

                  (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                  z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                  55

                  Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                  Px frac14 2Q

                  1

                  m2 thorn 1frac14 2 150

                  125frac14 76 kN=m

                  Equation 517 is used to obtain the pressure distribution

                  px frac14 4Q

                  h

                  m2n

                  ethm2 thorn n2THORN2 frac14150

                  m2n

                  ethm2 thorn n2THORN2 ethkN=m2THORN

                  Figure Q54

                  Stresses and displacements 31

                  n m2n

                  (m2 thorn n2)2

                  px(kNm2)

                  0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                  The pressure distribution is plotted in Figure Q55

                  56

                  H

                  Bfrac14 10

                  2frac14 5

                  L

                  Bfrac14 4

                  2frac14 2

                  D

                  Bfrac14 1

                  2frac14 05

                  Hence from Figure 515

                  131 frac14 082

                  130 frac14 094

                  Figure Q55

                  32 Stresses and displacements

                  The immediate settlement is given by Equation 528

                  si frac14 130131qB

                  Eu

                  frac14 094 082 200 2

                  45frac14 7mm

                  Stresses and displacements 33

                  Chapter 6

                  Lateral earth pressure

                  61

                  For 0 frac14 37 the active pressure coefficient is given by

                  Ka frac14 1 sin 37

                  1thorn sin 37frac14 025

                  The total active thrust (Equation 66a with c0 frac14 0) is

                  Pa frac14 1

                  2KaH

                  2 frac14 1

                  2 025 17 62 frac14 765 kN=m

                  If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                  K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                  and the thrust on the wall is

                  P0 frac14 1

                  2K0H

                  2 frac14 1

                  2 040 17 62 frac14 122 kN=m

                  62

                  The active pressure coefficients for the three soil types are as follows

                  Ka1 frac141 sin 35

                  1thorn sin 35frac14 0271

                  Ka2 frac141 sin 27

                  1thorn sin 27frac14 0375

                  ffiffiffiffiffiffiffiKa2

                  p frac14 0613

                  Ka3 frac141 sin 42

                  1thorn sin 42frac14 0198

                  Distribution of active pressure (plotted in Figure Q62)

                  Depth (m) Soil Active pressure (kNm2)

                  3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                  12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                  At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                  Total thrust frac14 571 kNm

                  Point of application is (4893571) m from the top of the wall ie 857m

                  Force (kN) Arm (m) Moment (kN m)

                  (1)1

                  2 0271 16 32 frac14 195 20 390

                  (2) 0271 16 3 2 frac14 260 40 1040

                  (3)1

                  2 0271 92 22 frac14 50 433 217

                  (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                  (5)1

                  2 0375 102 32 frac14 172 70 1204

                  (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                  (7)1

                  2 0198 112 42 frac14 177 1067 1889

                  (8)1

                  2 98 92 frac14 3969 90 35721

                  5713 48934

                  Figure Q62

                  Lateral earth pressure 35

                  63

                  (a) For u frac14 0 Ka frac14 Kp frac14 1

                  Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                  frac14 245

                  At the lower end of the piling

                  pa frac14 Kaqthorn Kasatz Kaccu

                  frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                  frac14 115 kN=m2

                  pp frac14 Kpsatzthorn Kpccu

                  frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                  frac14 202 kN=m2

                  (b) For 0 frac14 26 and frac14 1

                  20

                  Ka frac14 035

                  Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                  pfrac14 145 ethEquation 619THORN

                  Kp frac14 37

                  Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                  pfrac14 47 ethEquation 624THORN

                  At the lower end of the piling

                  pa frac14 Kaqthorn Ka0z Kacc

                  0

                  frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                  frac14 187 kN=m2

                  pp frac14 Kp0zthorn Kpcc

                  0

                  frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                  frac14 198 kN=m2

                  36 Lateral earth pressure

                  64

                  (a) For 0 frac14 38 Ka frac14 024

                  0 frac14 20 98 frac14 102 kN=m3

                  The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                  Force (kN) Arm (m) Moment (kN m)

                  (1) 024 10 66 frac14 159 33 525

                  (2)1

                  2 024 17 392 frac14 310 400 1240

                  (3) 024 17 39 27 frac14 430 135 580

                  (4)1

                  2 024 102 272 frac14 89 090 80

                  (5)1

                  2 98 272 frac14 357 090 321

                  Hfrac14 1345 MH frac14 2746

                  (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                  (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                  XM frac14MV MH frac14 7790 kNm

                  Lever arm of base resultant

                  M

                  Vfrac14 779

                  488frac14 160

                  Eccentricity of base resultant

                  e frac14 200 160 frac14 040m

                  39 m

                  27 m

                  40 m

                  04 m

                  04 m

                  26 m

                  (7)

                  (9)

                  (1)(2)

                  (3)

                  (4)

                  (5)

                  (8)(6)

                  (10)

                  WT

                  10 kNm2

                  Hydrostatic

                  Figure Q64

                  Lateral earth pressure 37

                  Base pressures (Equation 627)

                  p frac14 VB

                  1 6e

                  B

                  frac14 488

                  4eth1 060THORN

                  frac14 195 kN=m2 and 49 kN=m2

                  Factor of safety against sliding (Equation 628)

                  F frac14 V tan

                  Hfrac14 488 tan 25

                  1345frac14 17

                  (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                  Hfrac14 1633 kN

                  V frac14 4879 kN

                  MH frac14 3453 kNm

                  MV frac14 10536 kNm

                  The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                  65

                  For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                  Kp

                  Ffrac14 385

                  2

                  0 frac14 20 98 frac14 102 kN=m3

                  The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                  Force (kN) Arm (m) Moment (kN m)

                  (1)1

                  2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                  (2) 026 17 45 d frac14 199d d2 995d2

                  (3)1

                  2 026 102 d2 frac14 133d2 d3 044d3

                  (4)1

                  2 385

                  2 17 152 frac14 368 dthorn 05 368d 184

                  (5)385

                  2 17 15 d frac14 491d d2 2455d2

                  (6)1

                  2 385

                  2 102 d2 frac14 982d2 d3 327d3

                  38 Lateral earth pressure

                  XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                  d3 thorn 516d2 283d 1724 frac14 0

                  d frac14 179m

                  Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                  Over additional 20 embedded depth

                  pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                  Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                  66

                  The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                  Ka frac14 sin 69=sin 105

                  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                  pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                  26664

                  37775

                  2

                  frac14 050

                  The total active thrust (acting at 25 above the normal) is given by Equation 616

                  Pa frac14 1

                  2 050 19 7502 frac14 267 kN=m

                  Figure Q65

                  Lateral earth pressure 39

                  Horizontal component

                  Ph frac14 267 cos 40 frac14 205 kN=m

                  Vertical component

                  Pv frac14 267 sin 40 frac14 172 kN=m

                  Consider moments about the toe of the wall (Figure Q66) (per m)

                  Force (kN) Arm (m) Moment (kN m)

                  (1)1

                  2 175 650 235 frac14 1337 258 345

                  (2) 050 650 235 frac14 764 175 134

                  (3)1

                  2 070 650 235 frac14 535 127 68

                  (4) 100 400 235 frac14 940 200 188

                  (5) 1

                  2 080 050 235 frac14 47 027 1

                  Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                  Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                  Lever arm of base resultant

                  M

                  Vfrac14 795

                  525frac14 151m

                  Eccentricity of base resultant

                  e frac14 200 151 frac14 049m

                  Figure Q66

                  40 Lateral earth pressure

                  Base pressures (Equation 627)

                  p frac14 525

                  41 6 049

                  4

                  frac14 228 kN=m2 and 35 kN=m2

                  The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                  The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                  The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                  67

                  For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                  Force (kN) Arm (m) Moment (kNm)

                  (1)1

                  2 027 17 52 frac14 574 183 1050

                  (2) 027 17 5 3 frac14 689 500 3445

                  (3)1

                  2 027 102 32 frac14 124 550 682

                  (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                  (5)1

                  2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                  (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                  (7) 1

                  2 267

                  2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                  (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                  2 d frac14 163d d2thorn 650 82d2 1060d

                  Tie rod force per m frac14 T 0 0

                  XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                  d3 thorn 77d2 269d 1438 frac14 0

                  d frac14 467m

                  Depth of penetration frac14 12d frac14 560m

                  Lateral earth pressure 41

                  Algebraic sum of forces for d frac14 467m isX

                  F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                  T frac14 905 kN=m

                  Force in each tie rod frac14 25T frac14 226 kN

                  68

                  (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                  0 frac14 21 98 frac14 112 kN=m3

                  The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                  uC frac14 150

                  165 15 98 frac14 134 kN=m2

                  The average seepage pressure is

                  j frac14 15

                  165 98 frac14 09 kN=m3

                  Hence

                  0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                  0 j frac14 112 09 frac14 103 kN=m3

                  Figure Q67

                  42 Lateral earth pressure

                  Consider moments about the anchor point A (per m)

                  Force (kN) Arm (m) Moment (kN m)

                  (1) 10 026 150 frac14 390 60 2340

                  (2)1

                  2 026 18 452 frac14 474 15 711

                  (3) 026 18 45 105 frac14 2211 825 18240

                  (4)1

                  2 026 121 1052 frac14 1734 100 17340

                  (5)1

                  2 134 15 frac14 101 40 404

                  (6) 134 30 frac14 402 60 2412

                  (7)1

                  2 134 60 frac14 402 95 3819

                  571 4527(8) Ppm

                  115 115PPm

                  XM frac14 0

                  Ppm frac144527

                  115frac14 394 kN=m

                  Available passive resistance

                  Pp frac14 1

                  2 385 103 62 frac14 714 kN=m

                  Factor of safety

                  Fp frac14 Pp

                  Ppm

                  frac14 714

                  394frac14 18

                  Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                  Figure Q68

                  Lateral earth pressure 43

                  (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                  Consider moments (per m) about the tie point A

                  Force (kN) Arm (m)

                  (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                  (2)1

                  2 033 18 452 frac14 601 15

                  (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                  (4)1

                  2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                  (5)1

                  2 134 15 frac14 101 40

                  (6) 134 30 frac14 402 60

                  (7)1

                  2 134 d frac14 67d d3thorn 75

                  (8) 1

                  2 30 103 d2 frac141545d2 2d3thorn 75

                  Moment (kN m)

                  (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                  XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                  d3 thorn 827d2 466d 1518 frac14 0

                  By trial

                  d frac14 544m

                  The minimum depth of embedment required is 544m

                  69

                  For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                  0 frac14 20 98 frac14 102 kN=m3

                  The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                  44 Lateral earth pressure

                  uC frac14 147

                  173 26 98 frac14 216 kN=m2

                  and the average seepage pressure around the wall is

                  j frac14 26

                  173 98 frac14 15 kN=m3

                  Consider moments about the prop (A) (per m)

                  Force (kN) Arm (m) Moment (kN m)

                  (1)1

                  2 03 17 272 frac14 186 020 37

                  (2) 03 17 27 53 frac14 730 335 2445

                  (3)1

                  2 03 (102thorn 15) 532 frac14 493 423 2085

                  (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                  (5)1

                  2 216 26 frac14 281 243 684

                  (6) 216 27 frac14 583 465 2712

                  (7)1

                  2 216 60 frac14 648 800 5184

                  3055(8)

                  1

                  2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                  Factor of safety

                  Fr frac14 6885

                  3055frac14 225

                  Figure Q69

                  Lateral earth pressure 45

                  610

                  For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                  p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                  Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                  Using the recommendations of Twine and Roscoe

                  p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                  Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                  611

                  frac14 18 kN=m3 0 frac14 34

                  H frac14 350m nH frac14 335m mH frac14 185m

                  Consider a trial value of F frac14 20 Refer to Figure 635

                  0m frac14 tan1tan 34

                  20

                  frac14 186

                  Then

                  frac14 45 thorn 0m2frac14 543

                  W frac14 1

                  2 18 3502 cot 543 frac14 792 kN=m

                  Figure Q610

                  46 Lateral earth pressure

                  P frac14 1

                  2 s 3352 frac14 561s kN=m

                  U frac14 1

                  2 98 1852 cosec 543 frac14 206 kN=m

                  Equations 630 and 631 then become

                  561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                  792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                  ie

                  561s 0616N 405 frac14 0

                  792 0857N thorn 563 frac14 0

                  N frac14 848

                  0857frac14 989 kN=m

                  Then

                  561s 609 405 frac14 0

                  s frac14 649

                  561frac14 116 kN=m3

                  The calculations for trial values of F of 20 15 and 10 are summarized below

                  F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                  20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                  s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                  Figure Q611

                  Lateral earth pressure 47

                  612

                  For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                  45 thorn 0

                  2frac14 63

                  For the retained material between the surface and a depth of 36m

                  Pa frac14 1

                  2 030 18 362 frac14 350 kN=m

                  Weight of reinforced fill between the surface and a depth of 36m is

                  Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                  eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                  Eccentricity of Rv

                  e frac14 263 250 frac14 013m

                  The average vertical stress at a depth of 36m is

                  z frac14 Rv

                  L 2efrac14 324

                  474frac14 68 kN=m2

                  (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                  Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                  Tensile stress in the element frac14 138 103

                  65 3frac14 71N=mm2

                  Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                  Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                  Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                  The weight of ABC is

                  W frac14 1

                  2 18 52 265 frac14 124 kN=m

                  From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                  48 Lateral earth pressure

                  (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                  Tp frac14 032 68 120 065 frac14 170 kN

                  Tr frac14 213 420

                  418frac14 214 kN

                  Again the tensile failure and slipping limit states are satisfied for this element

                  Figure Q612

                  Lateral earth pressure 49

                  Chapter 7

                  Consolidation theory

                  71

                  Total change in thickness

                  H frac14 782 602 frac14 180mm

                  Average thickness frac14 1530thorn 180

                  2frac14 1620mm

                  Length of drainage path d frac14 1620

                  2frac14 810mm

                  Root time plot (Figure Q71a)

                  ffiffiffiffiffiffit90p frac14 33

                  t90 frac14 109min

                  cv frac14 0848d2

                  t90frac14 0848 8102

                  109 1440 365

                  106frac14 27m2=year

                  r0 frac14 782 764

                  782 602frac14 018

                  180frac14 0100

                  rp frac14 10eth764 645THORN9eth782 602THORN frac14

                  10 119

                  9 180frac14 0735

                  rs frac14 1 eth0100thorn 0735THORN frac14 0165

                  Log time plot (Figure Q71b)

                  t50 frac14 26min

                  cv frac14 0196d2

                  t50frac14 0196 8102

                  26 1440 365

                  106frac14 26m2=year

                  r0 frac14 782 763

                  782 602frac14 019

                  180frac14 0106

                  rp frac14 763 623

                  782 602frac14 140

                  180frac14 0778

                  rs frac14 1 eth0106thorn 0778THORN frac14 0116

                  Figure Q71(a)

                  Figure Q71(b)

                  Final void ratio

                  e1 frac14 w1Gs frac14 0232 272 frac14 0631

                  e

                  Hfrac14 1thorn e0

                  H0frac14 1thorn e1 thorne

                  H0

                  ie

                  e

                  180frac14 1631thorne

                  1710

                  e frac14 2936

                  1530frac14 0192

                  Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                  mv frac14 1

                  1thorn e0 e0 e101 00

                  frac14 1

                  1823 0192

                  0107frac14 098m2=MN

                  k frac14 cvmvw frac14 265 098 98

                  60 1440 365 103frac14 81 1010 m=s

                  72

                  Using Equation 77 (one-dimensional method)

                  sc frac14 e0 e11thorn e0 H

                  Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                  Figure Q72

                  52 Consolidation theory

                  Settlement

                  Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                  1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                  318

                  Notes 5 92y 460thorn 84

                  Heave

                  Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                  1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                  38

                  73

                  U frac14 f ethTvTHORN frac14 f cvt

                  d2

                  Hence if cv is constant

                  t1

                  t2frac14 d

                  21

                  d22

                  where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                  d1 frac14 95mm and d2 frac14 2500mm

                  for U frac14 050 t2 frac14 t1 d22

                  d21

                  frac14 20

                  60 24 365 25002

                  952frac14 263 years

                  for U lt 060 Tv frac14

                  4U2 (Equation 724(a))

                  t030 frac14 t050 0302

                  0502

                  frac14 263 036 frac14 095 years

                  Consolidation theory 53

                  74

                  The layer is open

                  d frac14 8

                  2frac14 4m

                  Tv frac14 cvtd2frac14 24 3

                  42frac14 0450

                  ui frac14 frac14 84 kN=m2

                  The excess pore water pressure is given by Equation 721

                  ue frac14Xmfrac141mfrac140

                  2ui

                  Msin

                  Mz

                  d

                  expethM2TvTHORN

                  In this case z frac14 d

                  sinMz

                  d

                  frac14 sinM

                  where

                  M frac14

                  23

                  25

                  2

                  M sin M M2Tv exp (M2Tv)

                  2thorn1 1110 0329

                  3

                  21 9993 457 105

                  ue frac14 2 84 2

                  1 0329 ethother terms negligibleTHORN

                  frac14 352 kN=m2

                  75

                  The layer is open

                  d frac14 6

                  2frac14 3m

                  Tv frac14 cvtd2frac14 10 3

                  32frac14 0333

                  The layer thickness will be divided into six equal parts ie m frac14 6

                  54 Consolidation theory

                  For an open layer

                  Tv frac14 4n

                  m2

                  n frac14 0333 62

                  4frac14 300

                  The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                  ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                  i j

                  0 1 2 3 4 5 6 7 8 9 10 11 12

                  0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                  The initial and 3-year isochrones are plotted in Figure Q75

                  Area under initial isochrone frac14 180 units

                  Area under 3-year isochrone frac14 63 units

                  The average degree of consolidation is given by Equation 725Thus

                  U frac14 1 63

                  180frac14 065

                  Figure Q75

                  Consolidation theory 55

                  76

                  At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                  0 frac14 2w frac14 2 98 frac14 196 kN=m2

                  The final consolidation settlement (one-dimensional method) is

                  sc frac14 mv0H frac14 083 196 8 frac14 130mm

                  Corrected time t frac14 2 1

                  2

                  40

                  52

                  frac14 1615 years

                  Tv frac14 cvtd2frac14 44 1615

                  42frac14 0444

                  From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                  77

                  The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                  Figure Q77

                  56 Consolidation theory

                  Point m n Ir (kNm2) sc (mm)

                  13020frac14 15 20

                  20frac14 10 0194 (4) 113 124

                  260

                  20frac14 30

                  20

                  20frac14 10 0204 (2) 59 65

                  360

                  20frac14 30

                  40

                  20frac14 20 0238 (1) 35 38

                  430

                  20frac14 15

                  40

                  20frac14 20 0224 (2) 65 72

                  Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                  78

                  Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                  (a) Immediate settlement

                  H

                  Bfrac14 30

                  35frac14 086

                  D

                  Bfrac14 2

                  35frac14 006

                  Figure Q78

                  Consolidation theory 57

                  From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                  si frac14 130131qB

                  Eufrac14 10 032 105 35

                  40frac14 30mm

                  (b) Consolidation settlement

                  Layer z (m) Dz Ic (kNm2) syod (mm)

                  1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                  3150

                  Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                  Now

                  H

                  Bfrac14 30

                  35frac14 086 and A frac14 065

                  from Figure 712 13 frac14 079

                  sc frac14 13sod frac14 079 315 frac14 250mm

                  Total settlement

                  s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                  79

                  Without sand drains

                  Uv frac14 025

                  Tv frac14 0049 ethfrom Figure 718THORN

                  t frac14 Tvd2

                  cvfrac14 0049 82

                  cvWith sand drains

                  R frac14 0564S frac14 0564 3 frac14 169m

                  n frac14 Rrfrac14 169

                  015frac14 113

                  Tr frac14 cht

                  4R2frac14 ch

                  4 1692 0049 82

                  cvethand ch frac14 cvTHORN

                  frac14 0275

                  Ur frac14 073 (from Figure 730)

                  58 Consolidation theory

                  Using Equation 740

                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                  U frac14 080

                  710

                  Without sand drains

                  Uv frac14 090

                  Tv frac14 0848

                  t frac14 Tvd2

                  cvfrac14 0848 102

                  96frac14 88 years

                  With sand drains

                  R frac14 0564S frac14 0564 4 frac14 226m

                  n frac14 Rrfrac14 226

                  015frac14 15

                  Tr

                  Tvfrac14 chcv

                  d2

                  4R2ethsame tTHORN

                  Tr

                  Tvfrac14 140

                  96 102

                  4 2262frac14 714 eth1THORN

                  Using Equation 740

                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                  An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                  Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                  040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                  Thus

                  Uv frac14 0295 and Ur frac14 086

                  t frac14 88 00683

                  0848frac14 07 years

                  Consolidation theory 59

                  Chapter 8

                  Bearing capacity

                  81

                  (a) The ultimate bearing capacity is given by Equation 83

                  qf frac14 cNc thorn DNq thorn 1

                  2BN

                  For u frac14 0

                  Nc frac14 514 Nq frac14 1 N frac14 0

                  qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                  The net ultimate bearing capacity is

                  qnf frac14 qf D frac14 540 kN=m2

                  The net foundation pressure is

                  qn frac14 q D frac14 425

                  2 eth21 1THORN frac14 192 kN=m2

                  The factor of safety (Equation 86) is

                  F frac14 qnfqnfrac14 540

                  192frac14 28

                  (b) For 0 frac14 28

                  Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                  qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                  2 112 2 13

                  frac14 260thorn 168thorn 146 frac14 574 kN=m2

                  qnf frac14 574 112 frac14 563 kN=m2

                  F frac14 563

                  192frac14 29

                  (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                  82

                  For 0 frac14 38

                  Nq frac14 49 N frac14 67

                  qnf frac14 DethNq 1THORN thorn 1

                  2BN ethfrom Equation 83THORN

                  frac14 eth18 075 48THORN thorn 1

                  2 18 15 67

                  frac14 648thorn 905 frac14 1553 kN=m2

                  qn frac14 500

                  15 eth18 075THORN frac14 320 kN=m2

                  F frac14 qnfqnfrac14 1553

                  320frac14 48

                  0d frac14 tan1tan 38

                  125

                  frac14 32 therefore Nq frac14 23 and N frac14 25

                  Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                  2 18 15 25

                  frac14 15eth310thorn 337THORNfrac14 970 kN=m

                  Design load (action) Vd frac14 500 kN=m

                  The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                  83

                  D

                  Bfrac14 350

                  225frac14 155

                  From Figure 85 for a square foundation

                  Nc frac14 81

                  Bearing capacity 61

                  For a rectangular foundation (L frac14 450m B frac14 225m)

                  Nc frac14 084thorn 016B

                  L

                  81 frac14 745

                  Using Equation 810

                  qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                  For F frac14 3

                  qn frac14 1006

                  3frac14 335 kN=m2

                  q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                  Design load frac14 405 450 225 frac14 4100 kN

                  Design undrained strength cud frac14 135

                  14frac14 96 kN=m2

                  Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                  frac14 7241 kN

                  Design load Vd frac14 4100 kN

                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                  84

                  For 0 frac14 40

                  Nq frac14 64 N frac14 95

                  qnf frac14 DethNq 1THORN thorn 04BN

                  (a) Water table 5m below ground level

                  qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                  qn frac14 400 17 frac14 383 kN=m2

                  F frac14 2686

                  383frac14 70

                  (b) Water table 1m below ground level (ie at foundation level)

                  0 frac14 20 98 frac14 102 kN=m3

                  62 Bearing capacity

                  qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                  F frac14 2040

                  383frac14 53

                  (c) Water table at ground level with upward hydraulic gradient 02

                  eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                  qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                  F frac14 1296

                  392frac14 33

                  85

                  The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                  Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                  Design value of 0 frac14 tan1tan 39

                  125

                  frac14 33

                  For 0 frac14 33 Nq frac14 26 and N frac14 29

                  Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                  86

                  (a) Undrained shear for u frac14 0

                  Nc frac14 514 Nq frac14 1 N frac14 0

                  qnf frac14 12cuNc

                  frac14 12 100 514 frac14 617 kN=m2

                  qn frac14 qnfFfrac14 617

                  3frac14 206 kN=m2

                  q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                  Bearing capacity 63

                  Drained shear for 0 frac14 32

                  Nq frac14 23 N frac14 25

                  0 frac14 21 98 frac14 112 kN=m3

                  qnf frac14 0DethNq 1THORN thorn 040BN

                  frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                  frac14 694 kN=m2

                  q frac14 694

                  3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                  Design load frac14 42 227 frac14 3632 kN

                  (b) Design undrained strength cud frac14 100

                  14frac14 71 kNm2

                  Design bearing resistance Rd frac14 12cudNe area

                  frac14 12 71 514 42

                  frac14 7007 kN

                  For drained shear 0d frac14 tan1tan 32

                  125

                  frac14 26

                  Nq frac14 12 N frac14 10

                  Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                  (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                  Layer z (m) m n Ir 0 (kNm2) sod (mm)

                  1 2 100 0175 0700qn 0182qn

                  2 6 033 0044 0176qn 0046qn

                  3 10 020 0017 0068qn 0018qn

                  0246qn

                  Diameter of equivalent circle B frac14 45m

                  H

                  Bfrac14 12

                  45frac14 27 and A frac14 042

                  13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                  64 Bearing capacity

                  For sc frac14 30mm

                  qn frac14 30

                  0147frac14 204 kN=m2

                  q frac14 204thorn 21 frac14 225 kN=m2

                  Design load frac14 42 225 frac14 3600 kN

                  The design load is 3600 kN settlement being the limiting criterion

                  87

                  D

                  Bfrac14 8

                  4frac14 20

                  From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                  F frac14 cuNc

                  Dfrac14 40 71

                  20 8frac14 18

                  88

                  Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                  Design value of 0 frac14 tan1tan 38

                  125

                  frac14 32

                  Figure Q86

                  Bearing capacity 65

                  For 0 frac14 32 Nq frac14 23 and N frac14 25

                  Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                  The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                  Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                  For B frac14 250m qn frac14 3750

                  2502 17 frac14 583 kN=m2

                  From Figure 510 m frac14 n frac14 126

                  6frac14 021

                  Ir frac14 0019

                  Stress increment frac14 4 0019 583 frac14 44 kN=m2

                  Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                  The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                  89

                  Depth (m) N 0v (kNm2) CN N1

                  070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                  Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                  (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                  Cw frac14 05thorn 0530

                  47

                  frac14 082

                  66 Bearing capacity

                  Thus

                  qa frac14 150 082 frac14 120 kN=m2

                  (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                  Thus

                  qa frac14 90 15 frac14 135 kN=m2

                  (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                  Ic frac14 171

                  1014frac14 0068

                  From Equation 819(a) with s frac14 25mm

                  q frac14 25

                  3507 0068frac14 150 kN=m2

                  810

                  Peak value of strain influence factor occurs at a depth of 27m and is given by

                  Izp frac14 05thorn 01130

                  16 27

                  05

                  frac14 067

                  Refer to Figure Q810

                  E frac14 25qc

                  Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                  Ez (mm3MN)

                  1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                  0203

                  C1 frac14 1 0500qnfrac14 1 05 12 16

                  130frac14 093

                  C2 frac14 1 ethsayTHORN

                  s frac14 C1C2qnX Iz

                  Ez frac14 093 1 130 0203 frac14 25mm

                  Bearing capacity 67

                  811

                  At pile base level

                  cu frac14 220 kN=m2

                  qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                  Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                  00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                  qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                  Then

                  Qf frac14 Abqb thorn Asqs

                  frac14

                  4 32 1980

                  thorn eth 105 139 86THORN

                  frac14 13 996thorn 3941 frac14 17 937 kN

                  0 01 02 03 04 05 06 07

                  0 2 4 6 8 10 12 14

                  1

                  2

                  3

                  4

                  5

                  6

                  7

                  8

                  (1)

                  (2)

                  (3)

                  (4)

                  (5)

                  qc

                  qc

                  Iz

                  Iz

                  (MNm2)

                  z (m)

                  Figure Q810

                  68 Bearing capacity

                  Allowable load

                  ethaTHORN Qf

                  2frac14 17 937

                  2frac14 8968 kN

                  ethbTHORN Abqb

                  3thorn Asqs frac14 13 996

                  3thorn 3941 frac14 8606 kN

                  ie allowable load frac14 8600 kN

                  Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                  According to the limit state method

                  Characteristic undrained strength at base level cuk frac14 220

                  150kN=m2

                  Characteristic base resistance qbk frac14 9cuk frac14 9 220

                  150frac14 1320 kN=m2

                  Characteristic shaft resistance qsk frac14 00150

                  frac14 86

                  150frac14 57 kN=m2

                  Characteristic base and shaft resistances

                  Rbk frac14

                  4 32 1320 frac14 9330 kN

                  Rsk frac14 105 139 86

                  150frac14 2629 kN

                  For a bored pile the partial factors are b frac14 160 and s frac14 130

                  Design bearing resistance Rcd frac14 9330

                  160thorn 2629

                  130

                  frac14 5831thorn 2022

                  frac14 7850 kN

                  Adding ethDAb W) the design bearing resistance becomes 9650 kN

                  812

                  ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                  qs frac14 cu frac14 040 105 frac14 42 kN=m2

                  For a single pile

                  Qf frac14 Abqb thorn Asqs

                  frac14

                  4 062 1305

                  thorn eth 06 15 42THORN

                  frac14 369thorn 1187 frac14 1556 kN

                  Bearing capacity 69

                  Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                  eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                  (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                  (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                  qbkfrac14 9cuk frac14 9 220

                  150frac14 1320 kN=m2

                  qskfrac14cuk frac14 040 105

                  150frac14 28 kN=m2

                  Rbkfrac14

                  4 0602 1320 frac14 373 kN

                  Rskfrac14 060 15 28 frac14 791 kN

                  Rcdfrac14 373

                  160thorn 791

                  130frac14 233thorn 608 frac14 841 kN

                  Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                  Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                  (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                  q frac14 21 000

                  1762frac14 68 kN=m2

                  Immediate settlement

                  H

                  Bfrac14 15

                  176frac14 085

                  D

                  Bfrac14 13

                  176frac14 074

                  L

                  Bfrac14 1

                  Hence from Figure 515

                  130 frac14 078 and 131 frac14 041

                  70 Bearing capacity

                  Thus using Equation 528

                  si frac14 078 041 68 176

                  65frac14 6mm

                  Consolidation settlement

                  Layer z (m) Area (m2) (kNm2) mvH (mm)

                  1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                  434 (sod)

                  Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                  sc frac14 056 434 frac14 24mm

                  The total settlement is (6thorn 24) frac14 30mm

                  813

                  At base level N frac14 26 Then using Equation 830

                  qb frac14 40NDb

                  Bfrac14 40 26 2

                  025frac14 8320 kN=m2

                  ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                  Figure Q812

                  Bearing capacity 71

                  Over the length embedded in sand

                  N frac14 21 ie18thorn 24

                  2

                  Using Equation 831

                  qs frac14 2N frac14 2 21 frac14 42 kN=m2

                  For a single pile

                  Qf frac14 Abqb thorn Asqs

                  frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                  For the pile group assuming a group efficiency of 12

                  XQf frac14 12 9 604 frac14 6523 kN

                  Then the load factor is

                  F frac14 6523

                  2000thorn 1000frac14 21

                  (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                  Characteristic base resistance per unit area qbk frac14 8320

                  150frac14 5547 kNm2

                  Characteristic shaft resistance per unit area qsk frac14 42

                  150frac14 28 kNm2

                  Characteristic base and shaft resistances for a single pile

                  Rbk frac14 0252 5547 frac14 347 kN

                  Rsk frac14 4 025 2 28 frac14 56 kN

                  For a driven pile the partial factors are b frac14 s frac14 130

                  Design bearing resistance Rcd frac14 347

                  130thorn 56

                  130frac14 310 kN

                  For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                  Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                  (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                  From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                  72 Bearing capacity

                  N frac14 24thorn 26thorn 34

                  3frac14 28

                  Ic frac14 171

                  2814frac14 0016 ethEquation 818THORN

                  s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                  814

                  Using Equation 841

                  Tf frac14 DLcu thorn

                  4ethD2 d2THORNcuNc

                  frac14 eth 02 5 06 110THORN thorn

                  4eth022 012THORN110 9

                  frac14 207thorn 23 frac14 230 kN

                  Figure Q813

                  Bearing capacity 73

                  Chapter 9

                  Stability of slopes

                  91

                  Referring to Figure Q91

                  W frac14 417 19 frac14 792 kN=m

                  Q frac14 20 28 frac14 56 kN=m

                  Arc lengthAB frac14

                  180 73 90 frac14 115m

                  Arc length BC frac14

                  180 28 90 frac14 44m

                  The factor of safety is given by

                  F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                  90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                  Depth of tension crack z0 frac14 2cu

                  frac14 2 20

                  19frac14 21m

                  Arc length BD frac14

                  180 13

                  1

                  2 90 frac14 21m

                  F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                  The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                  Design resisting moment frac14 rXethcudLaTHORN frac14 90

                  14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                  Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                  The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                  92

                  u frac14 0

                  Depth factor D frac14 11

                  9frac14 122

                  Using Equation 92 with F frac14 10

                  Ns frac14 cu

                  FHfrac14 30

                  10 19 9frac14 0175

                  Hence from Figure 93

                  frac14 50

                  For F frac14 12

                  Ns frac14 30

                  12 19 9frac14 0146

                  frac14 27

                  93

                  Refer to Figure Q93

                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                  1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                  74 m

                  214 1deg

                  213 1deg

                  39 m

                  WB

                  D

                  C

                  28 m

                  21 m

                  A

                  Q

                  Soil (1)Soil (2)

                  73deg

                  Figure Q91

                  Stability of slopes 75

                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                  9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                  599 256 328 1372

                  Figure Q93

                  76 Stability of slopes

                  XW cos frac14 b

                  Xh cos frac14 21 2 599 frac14 2516 kN=mX

                  W sin frac14 bX

                  h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                  Arc length La frac14

                  180 57

                  1

                  2 326 frac14 327m

                  The factor of safety is given by

                  F frac14 c0La thorn tan0ethW cos ulTHORN

                  W sin

                  frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                  frac14 091

                  According to the limit state method

                  0d frac14 tan1tan 32

                  125

                  frac14 265

                  c0 frac14 8

                  160frac14 5 kN=m2

                  Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                  Design disturbing moment frac14 1075 kN=m

                  The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                  94

                  F frac14 1

                  W sin

                  Xfc0bthorn ethW ubTHORN tan0g sec

                  1thorn ethtan tan0=FTHORN

                  c0 frac14 8 kN=m2

                  0 frac14 32

                  c0b frac14 8 2 frac14 16 kN=m

                  W frac14 bh frac14 21 2 h frac14 42h kN=m

                  Try F frac14 100

                  tan0

                  Ffrac14 0625

                  Stability of slopes 77

                  Values of u are as obtained in Figure Q93

                  SliceNo

                  h(m)

                  W frac14 bh(kNm)

                  W sin(kNm)

                  ub(kNm)

                  c0bthorn (W ub) tan0(kNm)

                  sec

                  1thorn (tan tan0)FProduct(kNm)

                  1 05 21 6 2 8 24 1078 262 13 55 31

                  23 33 30 1042 31

                  3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                  224 92 72 0931 67

                  6 50 210 11 40 100 85 0907 777 55 231 14

                  12 58 112 90 0889 80

                  8 60 252 1812

                  80 114 102 0874 899 63 265 22 99 116 109 0861 94

                  10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                  2154 88 116 0853 99

                  14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                  1074 1091

                  F frac14 1091

                  1074frac14 102 (assumed value 100)

                  Thus

                  F frac14 101

                  95

                  F frac14 1

                  W sin

                  XfWeth1 ruTHORN tan0g sec

                  1thorn ethtan tan0THORN=F

                  0 frac14 33

                  ru frac14 020

                  W frac14 bh frac14 20 5 h frac14 100h kN=m

                  eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                  Try F frac14 110

                  tan 0

                  Ffrac14 tan 33

                  110frac14 0590

                  78 Stability of slopes

                  Referring to Figure Q95

                  SliceNo

                  h(m)

                  W frac14 bh(kNm)

                  W sin(kNm)

                  W(1 ru) tan0(kNm)

                  sec

                  1thorn ( tan tan0)FProduct(kNm)

                  1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                  2120 234 0892 209

                  4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                  1185 1271

                  Figure Q95

                  Stability of slopes 79

                  F frac14 1271

                  1185frac14 107

                  The trial value was 110 therefore take F to be 108

                  96

                  (a) Water table at surface the factor of safety is given by Equation 912

                  F frac14 0

                  sat

                  tan0

                  tan

                  ptie 15 frac14 92

                  19

                  tan 36

                  tan

                  tan frac14 0234

                  frac14 13

                  Water table well below surface the factor of safety is given by Equation 911

                  F frac14 tan0

                  tan

                  frac14 tan 36

                  tan 13

                  frac14 31

                  (b) 0d frac14 tan1tan 36

                  125

                  frac14 30

                  Depth of potential failure surface frac14 z

                  Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                  frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                  frac14 504z kN

                  Design disturbing moment per unit area Sd frac14 sat sin cos

                  frac14 19 z sin 13 cos 13

                  frac14 416z kN

                  Rd gtSd therefore the limit state for overall stability is satisfied

                  80 Stability of slopes

                  • Book Cover
                  • Title
                  • Contents
                  • Basic characteristics of soils
                  • Seepage
                  • Effective stress
                  • Shear strength
                  • Stresses and displacements
                  • Lateral earth pressure
                  • Consolidation theory
                  • Bearing capacity
                  • Stability of slopes

                    Using Equation 115

                    Afrac14 e wGs

                    1thorn e frac14038 0318

                    138frac14 0045 eth45THORN

                    The zero air voids dry density is given by Equation 125

                    d frac14 Gs

                    1thorn wGsw frac14 265

                    1thorn eth0135 265THORN 100 frac14 195Mg=m3

                    ie a dry density of 200Mgm3 would not be possible

                    16

                    Mass (g) (Mgm3) w d (Mgm3) d0(Mgm3) d5

                    (Mgm3) d10(Mgm3)

                    2010 2010 0128 1782 1990 1890 17912092 2092 0145 1827 1925 1829 17332114 2114 0156 1829 1884 1790 16962100 2100 0168 1798 1843 1751 16582055 2055 0192 1724 1765 1676 1588

                    In each case the bulk density () is equal to the mass of compacted soil divided by thevolume of the mould The corresponding value of dry density (d) is obtained fromEquation 124 The dry densityndashwater content curve is plotted from which

                    wopt frac14 15 and dmaxfrac14 183Mg=m3

                    Figure Q16

                    4 Basic characteristics of soils

                    Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

                    d5 d10

                    respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

                    17

                    From Equation 120

                    e frac14 Gswd 1

                    The maximum and minimum values of void ratio are given by

                    emax frac14 Gsw

                    dmin

                    1

                    emin frac14 Gswdmax

                    1

                    From Equation 123

                    ID frac14 Gsweth1=dmin 1=dTHORN

                    Gsweth1=dmin 1=dmax

                    THORN

                    frac14 frac121 ethdmin=dTHORN1=dmin

                    frac121 ethdmin=dmax

                    THORN1=dmin

                    frac14 d dmin

                    dmax dmin

                    dmax

                    d

                    frac14 172 154

                    181 154

                    181

                    172

                    frac14 070 eth70THORN

                    Basic characteristics of soils 5

                    Chapter 2

                    Seepage

                    21

                    The coefficient of permeability is determined from the equation

                    k frac14 23al

                    At1log

                    h0

                    h1

                    where

                    a frac14

                    4 00052 m2 l frac14 02m

                    A frac14

                    4 012 m2 t1 frac14 3 602 s

                    logh0

                    h1frac14 log

                    100

                    035frac14 0456

                    k frac14 23 00052 02 0456

                    012 3 602frac14 49 108 m=s

                    22

                    The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

                    q frac14 kh Nf

                    Ndfrac14 106 400 37

                    11frac14 13 106 m3=s per m

                    Figure Q22

                    23

                    The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

                    q frac14 kh Nf

                    Ndfrac14 5 105 300 35

                    9frac14 58 105 m3=s per m

                    The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

                    Point h (m) h z (m) u frac14 w(h z)(kNm2)

                    1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

                    eg for Point 1

                    h1 frac14 7

                    9 300 frac14 233m

                    h1 z1 frac14 233 eth250THORN frac14 483m

                    Figure Q23

                    Seepage 7

                    u1 frac14 98 483 frac14 47 kN=m2

                    The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

                    24

                    The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

                    q frac14 kh Nf

                    Ndfrac14 40 107 550 10

                    11frac14 20 106 m3=s per m

                    25

                    The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

                    q frac14 kh Nf

                    Ndfrac14 20 106 500 42

                    9frac14 47 106 m3=s per m

                    Figure Q24

                    8 Seepage

                    26

                    The scale transformation factor in the x direction is given by Equation 221 ie

                    xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                    ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

                    Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

                    k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                    pfrac14

                    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

                    p 107 frac14 30 107 m=s

                    Thus the quantity of seepage is given by

                    q frac14 k0h Nf

                    Ndfrac14 30 107 1400 325

                    12frac14 11 106 m3=s per m

                    Figure Q25

                    Seepage 9

                    27

                    The scale transformation factor in the x direction is

                    xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                    ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

                    Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

                    k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                    pfrac14

                    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

                    p 106 frac14 45 106 m=s

                    The focus of the basic parabola is at point A The parabola passes through point Gsuch that

                    GC frac14 03HC frac14 03 30 frac14 90m

                    Thus the coordinates of G are

                    x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

                    480 frac14 x0 2002

                    4x0

                    Figure Q26

                    10 Seepage

                    Hence

                    x0 frac14 20m

                    Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

                    x frac14 20 z2

                    80

                    x 20 0 50 100 200 300z 0 400 748 980 1327 1600

                    The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

                    No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

                    q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

                    28

                    The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

                    Figure Q27

                    Seepage 11

                    q frac14 kh Nf

                    Ndfrac14 45 105 28 33

                    7

                    frac14 59 105 m3=s per m

                    29

                    The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

                    kx frac14 H1k1 thornH2k2

                    H1 thornH2frac14 106

                    10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

                    kz frac14 H1 thornH2

                    H1

                    k1thornH2

                    k2

                    frac14 10

                    5

                    eth2 106THORN thorn5

                    eth16 106THORNfrac14 36 106 m=s

                    Then the scale transformation factor is given by

                    xt frac14 xffiffiffiffiffikz

                    pffiffiffiffiffikx

                    p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

                    Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

                    Figure Q28

                    12 Seepage

                    k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                    qfrac14

                    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

                    p 106 frac14 57 106 m=s

                    Then the quantity of seepage is given by

                    q frac14 k0h Nf

                    Ndfrac14 57 106 350 56

                    11

                    frac14 10 105 m3=s per m

                    Figure Q29

                    Seepage 13

                    Chapter 3

                    Effective stress

                    31

                    Buoyant unit weight

                    0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                    Effective vertical stress

                    0v frac14 5 102 frac14 51 kN=m2 or

                    Total vertical stress

                    v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

                    Pore water pressure

                    u frac14 7 98 frac14 686 kN=m2

                    Effective vertical stress

                    0v frac14 v u frac14 1196 686 frac14 51 kN=m2

                    32

                    Buoyant unit weight

                    0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                    Effective vertical stress

                    0v frac14 5 102 frac14 51 kN=m2 or

                    Total vertical stress

                    v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

                    Pore water pressure

                    u frac14 205 98 frac14 2009 kN=m2

                    Effective vertical stress

                    0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

                    33

                    At top of the clay

                    v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

                    u frac14 2 98 frac14 196 kN=m2

                    0v frac14 v u frac14 710 196 frac14 514 kN=m2

                    Alternatively

                    0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                    0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

                    At bottom of the clay

                    v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

                    u frac14 12 98 frac14 1176 kN=m2

                    0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

                    NB The alternative method of calculation is not applicable because of the artesiancondition

                    Figure Q3132

                    Effective stress 15

                    34

                    0 frac14 20 98 frac14 102 kN=m3

                    At 8m depth

                    0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                    35

                    0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                    0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                    Figure Q33

                    Figure Q34

                    16 Effective stress

                    (a) Immediately after WT rise

                    At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                    0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                    At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                    0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                    (b) Several years after WT rise

                    At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                    At 8m depth

                    0v frac14 940 kN=m2 (as above)

                    At 12m depth

                    0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                    Figure Q35

                    Effective stress 17

                    36

                    Total weight

                    ab frac14 210 kN

                    Effective weight

                    ac frac14 112 kN

                    Resultant boundary water force

                    be frac14 119 kN

                    Seepage force

                    ce frac14 34 kN

                    Resultant body force

                    ae frac14 99 kN eth73 to horizontalTHORN

                    (Refer to Figure Q36)

                    Figure Q36

                    18 Effective stress

                    37

                    Situation (1)(a)

                    frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                    u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                    0 frac14 u frac14 694 392 frac14 302 kN=m2

                    (b)

                    i frac14 2

                    4frac14 05

                    j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                    Situation (2)(a)

                    frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                    u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                    0 frac14 u frac14 498 392 frac14 106 kN=m2

                    (b)

                    i frac14 2

                    4frac14 05

                    j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                    38

                    The flow net is drawn in Figure Q24

                    Loss in total head between adjacent equipotentials

                    h frac14 550

                    Ndfrac14 550

                    11frac14 050m

                    Exit hydraulic gradient

                    ie frac14 h

                    sfrac14 050

                    070frac14 071

                    Effective stress 19

                    The critical hydraulic gradient is given by Equation 39

                    ic frac14 0

                    wfrac14 102

                    98frac14 104

                    Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                    F frac14 iciefrac14 104

                    071frac14 15

                    Total head at C

                    hC frac14 nd

                    Ndh frac14 24

                    11 550 frac14 120m

                    Elevation head at C

                    zC frac14 250m

                    Pore water pressure at C

                    uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                    Therefore effective vertical stress at C

                    0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                    For point D

                    hD frac14 73

                    11 550 frac14 365m

                    zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                    0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                    39

                    The flow net is drawn in Figure Q25

                    For a soil prism 150 300m adjacent to the piling

                    hm frac14 26

                    9 500 frac14 145m

                    20 Effective stress

                    Factor of safety against lsquoheavingrsquo (Equation 310)

                    F frac14 ic

                    imfrac14 0d

                    whmfrac14 97 300

                    98 145frac14 20

                    With a filter

                    F frac14 0d thorn wwhm

                    3 frac14 eth97 300THORN thorn w98 145

                    w frac14 135 kN=m2

                    Depth of filterfrac14 13521frac14 065m (if above water level)

                    Effective stress 21

                    Chapter 4

                    Shear strength

                    41

                    frac14 295 kN=m2

                    u frac14 120 kN=m2

                    0 frac14 u frac14 295 120 frac14 175 kN=m2

                    f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                    42

                    03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                    100 452 552200 908 1108400 1810 2210800 3624 4424

                    The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                    Figure Q42

                    43

                    3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                    200 222 422400 218 618600 220 820

                    The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                    44

                    The modified shear strength parameters are

                    0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                    a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                    The coordinates of the stress point representing failure conditions in the test are

                    1

                    2eth1 2THORN frac14 1

                    2 170 frac14 85 kN=m2

                    1

                    2eth1 thorn 3THORN frac14 1

                    2eth270thorn 100THORN frac14 185 kN=m2

                    The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                    uf frac14 36 kN=m2

                    Figure Q43

                    Figure Q44

                    Shear strength 23

                    45

                    3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                    150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                    The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                    The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                    46

                    03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                    200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                    The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                    Figure Q45

                    24 Shear strength

                    47

                    The torque required to produce shear failure is given by

                    T frac14 dh cud

                    2thorn 2

                    Z d=2

                    0

                    2r drcur

                    frac14 cud2h

                    2thorn 4cu

                    Z d=2

                    0

                    r2dr

                    frac14 cud2h

                    2thorn d

                    3

                    6

                    Then

                    35 frac14 cu52 10

                    2thorn 53

                    6

                    103

                    cu frac14 76 kN=m3

                    400

                    0 400 800 1200 1600

                    τ (k

                    Nm

                    2 )

                    σprime (kNm2)

                    34deg

                    315deg29deg

                    (a)

                    (b)

                    0 400

                    400

                    800 1200 1600

                    Failure envelope

                    300 500

                    σprime (kNm2)

                    τ (k

                    Nm

                    2 )

                    20 (kNm2)

                    31deg

                    Figure Q46

                    Shear strength 25

                    48

                    The relevant stress values are calculated as follows

                    3 frac14 600 kN=m2

                    1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                    2(1 3) 0 40 79 107 139 159

                    1

                    2(01 thorn 03) 400 411 402 389 351 326

                    1

                    2(1 thorn 3) 600 640 679 707 739 759

                    The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                    Af frac14 433 200

                    319frac14 073

                    49

                    B frac14 u33

                    frac14 144

                    350 200frac14 096

                    a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                    0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                    10 283 65 023

                    Figure Q48

                    26 Shear strength

                    The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                    Figure Q49

                    Shear strength 27

                    Chapter 5

                    Stresses and displacements

                    51

                    Vertical stress is given by

                    z frac14 Qz2Ip frac14 5000

                    52Ip

                    Values of Ip are obtained from Table 51

                    r (m) rz Ip z (kNm2)

                    0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                    10 20 0009 2

                    The variation of z with radial distance (r) is plotted in Figure Q51

                    Figure Q51

                    52

                    Below the centre load (Figure Q52)

                    r

                    zfrac14 0 for the 7500-kN load

                    Ip frac14 0478

                    r

                    zfrac14 5

                    4frac14 125 for the 10 000- and 9000-kN loads

                    Ip frac14 0045

                    Then

                    z frac14X Q

                    z2Ip

                    frac14 7500 0478

                    42thorn 10 000 0045

                    42thorn 9000 0045

                    42

                    frac14 224thorn 28thorn 25 frac14 277 kN=m2

                    53

                    The vertical stress under a corner of a rectangular area is given by

                    z frac14 qIr

                    where values of Ir are obtained from Figure 510 In this case

                    z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                    z

                    Figure Q52

                    Stresses and displacements 29

                    z (m) m n Ir z (kNm2)

                    0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                    10 010 0005 5

                    z is plotted against z in Figure Q53

                    54

                    (a)

                    m frac14 125

                    12frac14 104

                    n frac14 18

                    12frac14 150

                    From Figure 510 Irfrac14 0196

                    z frac14 2 175 0196 frac14 68 kN=m2

                    Figure Q53

                    30 Stresses and displacements

                    (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                    z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                    55

                    Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                    Px frac14 2Q

                    1

                    m2 thorn 1frac14 2 150

                    125frac14 76 kN=m

                    Equation 517 is used to obtain the pressure distribution

                    px frac14 4Q

                    h

                    m2n

                    ethm2 thorn n2THORN2 frac14150

                    m2n

                    ethm2 thorn n2THORN2 ethkN=m2THORN

                    Figure Q54

                    Stresses and displacements 31

                    n m2n

                    (m2 thorn n2)2

                    px(kNm2)

                    0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                    The pressure distribution is plotted in Figure Q55

                    56

                    H

                    Bfrac14 10

                    2frac14 5

                    L

                    Bfrac14 4

                    2frac14 2

                    D

                    Bfrac14 1

                    2frac14 05

                    Hence from Figure 515

                    131 frac14 082

                    130 frac14 094

                    Figure Q55

                    32 Stresses and displacements

                    The immediate settlement is given by Equation 528

                    si frac14 130131qB

                    Eu

                    frac14 094 082 200 2

                    45frac14 7mm

                    Stresses and displacements 33

                    Chapter 6

                    Lateral earth pressure

                    61

                    For 0 frac14 37 the active pressure coefficient is given by

                    Ka frac14 1 sin 37

                    1thorn sin 37frac14 025

                    The total active thrust (Equation 66a with c0 frac14 0) is

                    Pa frac14 1

                    2KaH

                    2 frac14 1

                    2 025 17 62 frac14 765 kN=m

                    If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                    K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                    and the thrust on the wall is

                    P0 frac14 1

                    2K0H

                    2 frac14 1

                    2 040 17 62 frac14 122 kN=m

                    62

                    The active pressure coefficients for the three soil types are as follows

                    Ka1 frac141 sin 35

                    1thorn sin 35frac14 0271

                    Ka2 frac141 sin 27

                    1thorn sin 27frac14 0375

                    ffiffiffiffiffiffiffiKa2

                    p frac14 0613

                    Ka3 frac141 sin 42

                    1thorn sin 42frac14 0198

                    Distribution of active pressure (plotted in Figure Q62)

                    Depth (m) Soil Active pressure (kNm2)

                    3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                    12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                    At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                    Total thrust frac14 571 kNm

                    Point of application is (4893571) m from the top of the wall ie 857m

                    Force (kN) Arm (m) Moment (kN m)

                    (1)1

                    2 0271 16 32 frac14 195 20 390

                    (2) 0271 16 3 2 frac14 260 40 1040

                    (3)1

                    2 0271 92 22 frac14 50 433 217

                    (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                    (5)1

                    2 0375 102 32 frac14 172 70 1204

                    (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                    (7)1

                    2 0198 112 42 frac14 177 1067 1889

                    (8)1

                    2 98 92 frac14 3969 90 35721

                    5713 48934

                    Figure Q62

                    Lateral earth pressure 35

                    63

                    (a) For u frac14 0 Ka frac14 Kp frac14 1

                    Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                    frac14 245

                    At the lower end of the piling

                    pa frac14 Kaqthorn Kasatz Kaccu

                    frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                    frac14 115 kN=m2

                    pp frac14 Kpsatzthorn Kpccu

                    frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                    frac14 202 kN=m2

                    (b) For 0 frac14 26 and frac14 1

                    20

                    Ka frac14 035

                    Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                    pfrac14 145 ethEquation 619THORN

                    Kp frac14 37

                    Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                    pfrac14 47 ethEquation 624THORN

                    At the lower end of the piling

                    pa frac14 Kaqthorn Ka0z Kacc

                    0

                    frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                    frac14 187 kN=m2

                    pp frac14 Kp0zthorn Kpcc

                    0

                    frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                    frac14 198 kN=m2

                    36 Lateral earth pressure

                    64

                    (a) For 0 frac14 38 Ka frac14 024

                    0 frac14 20 98 frac14 102 kN=m3

                    The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                    Force (kN) Arm (m) Moment (kN m)

                    (1) 024 10 66 frac14 159 33 525

                    (2)1

                    2 024 17 392 frac14 310 400 1240

                    (3) 024 17 39 27 frac14 430 135 580

                    (4)1

                    2 024 102 272 frac14 89 090 80

                    (5)1

                    2 98 272 frac14 357 090 321

                    Hfrac14 1345 MH frac14 2746

                    (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                    (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                    XM frac14MV MH frac14 7790 kNm

                    Lever arm of base resultant

                    M

                    Vfrac14 779

                    488frac14 160

                    Eccentricity of base resultant

                    e frac14 200 160 frac14 040m

                    39 m

                    27 m

                    40 m

                    04 m

                    04 m

                    26 m

                    (7)

                    (9)

                    (1)(2)

                    (3)

                    (4)

                    (5)

                    (8)(6)

                    (10)

                    WT

                    10 kNm2

                    Hydrostatic

                    Figure Q64

                    Lateral earth pressure 37

                    Base pressures (Equation 627)

                    p frac14 VB

                    1 6e

                    B

                    frac14 488

                    4eth1 060THORN

                    frac14 195 kN=m2 and 49 kN=m2

                    Factor of safety against sliding (Equation 628)

                    F frac14 V tan

                    Hfrac14 488 tan 25

                    1345frac14 17

                    (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                    Hfrac14 1633 kN

                    V frac14 4879 kN

                    MH frac14 3453 kNm

                    MV frac14 10536 kNm

                    The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                    65

                    For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                    Kp

                    Ffrac14 385

                    2

                    0 frac14 20 98 frac14 102 kN=m3

                    The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                    Force (kN) Arm (m) Moment (kN m)

                    (1)1

                    2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                    (2) 026 17 45 d frac14 199d d2 995d2

                    (3)1

                    2 026 102 d2 frac14 133d2 d3 044d3

                    (4)1

                    2 385

                    2 17 152 frac14 368 dthorn 05 368d 184

                    (5)385

                    2 17 15 d frac14 491d d2 2455d2

                    (6)1

                    2 385

                    2 102 d2 frac14 982d2 d3 327d3

                    38 Lateral earth pressure

                    XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                    d3 thorn 516d2 283d 1724 frac14 0

                    d frac14 179m

                    Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                    Over additional 20 embedded depth

                    pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                    Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                    66

                    The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                    Ka frac14 sin 69=sin 105

                    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                    pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                    26664

                    37775

                    2

                    frac14 050

                    The total active thrust (acting at 25 above the normal) is given by Equation 616

                    Pa frac14 1

                    2 050 19 7502 frac14 267 kN=m

                    Figure Q65

                    Lateral earth pressure 39

                    Horizontal component

                    Ph frac14 267 cos 40 frac14 205 kN=m

                    Vertical component

                    Pv frac14 267 sin 40 frac14 172 kN=m

                    Consider moments about the toe of the wall (Figure Q66) (per m)

                    Force (kN) Arm (m) Moment (kN m)

                    (1)1

                    2 175 650 235 frac14 1337 258 345

                    (2) 050 650 235 frac14 764 175 134

                    (3)1

                    2 070 650 235 frac14 535 127 68

                    (4) 100 400 235 frac14 940 200 188

                    (5) 1

                    2 080 050 235 frac14 47 027 1

                    Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                    Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                    Lever arm of base resultant

                    M

                    Vfrac14 795

                    525frac14 151m

                    Eccentricity of base resultant

                    e frac14 200 151 frac14 049m

                    Figure Q66

                    40 Lateral earth pressure

                    Base pressures (Equation 627)

                    p frac14 525

                    41 6 049

                    4

                    frac14 228 kN=m2 and 35 kN=m2

                    The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                    The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                    The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                    67

                    For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                    Force (kN) Arm (m) Moment (kNm)

                    (1)1

                    2 027 17 52 frac14 574 183 1050

                    (2) 027 17 5 3 frac14 689 500 3445

                    (3)1

                    2 027 102 32 frac14 124 550 682

                    (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                    (5)1

                    2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                    (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                    (7) 1

                    2 267

                    2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                    (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                    2 d frac14 163d d2thorn 650 82d2 1060d

                    Tie rod force per m frac14 T 0 0

                    XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                    d3 thorn 77d2 269d 1438 frac14 0

                    d frac14 467m

                    Depth of penetration frac14 12d frac14 560m

                    Lateral earth pressure 41

                    Algebraic sum of forces for d frac14 467m isX

                    F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                    T frac14 905 kN=m

                    Force in each tie rod frac14 25T frac14 226 kN

                    68

                    (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                    0 frac14 21 98 frac14 112 kN=m3

                    The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                    uC frac14 150

                    165 15 98 frac14 134 kN=m2

                    The average seepage pressure is

                    j frac14 15

                    165 98 frac14 09 kN=m3

                    Hence

                    0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                    0 j frac14 112 09 frac14 103 kN=m3

                    Figure Q67

                    42 Lateral earth pressure

                    Consider moments about the anchor point A (per m)

                    Force (kN) Arm (m) Moment (kN m)

                    (1) 10 026 150 frac14 390 60 2340

                    (2)1

                    2 026 18 452 frac14 474 15 711

                    (3) 026 18 45 105 frac14 2211 825 18240

                    (4)1

                    2 026 121 1052 frac14 1734 100 17340

                    (5)1

                    2 134 15 frac14 101 40 404

                    (6) 134 30 frac14 402 60 2412

                    (7)1

                    2 134 60 frac14 402 95 3819

                    571 4527(8) Ppm

                    115 115PPm

                    XM frac14 0

                    Ppm frac144527

                    115frac14 394 kN=m

                    Available passive resistance

                    Pp frac14 1

                    2 385 103 62 frac14 714 kN=m

                    Factor of safety

                    Fp frac14 Pp

                    Ppm

                    frac14 714

                    394frac14 18

                    Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                    Figure Q68

                    Lateral earth pressure 43

                    (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                    Consider moments (per m) about the tie point A

                    Force (kN) Arm (m)

                    (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                    (2)1

                    2 033 18 452 frac14 601 15

                    (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                    (4)1

                    2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                    (5)1

                    2 134 15 frac14 101 40

                    (6) 134 30 frac14 402 60

                    (7)1

                    2 134 d frac14 67d d3thorn 75

                    (8) 1

                    2 30 103 d2 frac141545d2 2d3thorn 75

                    Moment (kN m)

                    (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                    XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                    d3 thorn 827d2 466d 1518 frac14 0

                    By trial

                    d frac14 544m

                    The minimum depth of embedment required is 544m

                    69

                    For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                    0 frac14 20 98 frac14 102 kN=m3

                    The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                    44 Lateral earth pressure

                    uC frac14 147

                    173 26 98 frac14 216 kN=m2

                    and the average seepage pressure around the wall is

                    j frac14 26

                    173 98 frac14 15 kN=m3

                    Consider moments about the prop (A) (per m)

                    Force (kN) Arm (m) Moment (kN m)

                    (1)1

                    2 03 17 272 frac14 186 020 37

                    (2) 03 17 27 53 frac14 730 335 2445

                    (3)1

                    2 03 (102thorn 15) 532 frac14 493 423 2085

                    (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                    (5)1

                    2 216 26 frac14 281 243 684

                    (6) 216 27 frac14 583 465 2712

                    (7)1

                    2 216 60 frac14 648 800 5184

                    3055(8)

                    1

                    2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                    Factor of safety

                    Fr frac14 6885

                    3055frac14 225

                    Figure Q69

                    Lateral earth pressure 45

                    610

                    For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                    p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                    Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                    Using the recommendations of Twine and Roscoe

                    p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                    Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                    611

                    frac14 18 kN=m3 0 frac14 34

                    H frac14 350m nH frac14 335m mH frac14 185m

                    Consider a trial value of F frac14 20 Refer to Figure 635

                    0m frac14 tan1tan 34

                    20

                    frac14 186

                    Then

                    frac14 45 thorn 0m2frac14 543

                    W frac14 1

                    2 18 3502 cot 543 frac14 792 kN=m

                    Figure Q610

                    46 Lateral earth pressure

                    P frac14 1

                    2 s 3352 frac14 561s kN=m

                    U frac14 1

                    2 98 1852 cosec 543 frac14 206 kN=m

                    Equations 630 and 631 then become

                    561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                    792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                    ie

                    561s 0616N 405 frac14 0

                    792 0857N thorn 563 frac14 0

                    N frac14 848

                    0857frac14 989 kN=m

                    Then

                    561s 609 405 frac14 0

                    s frac14 649

                    561frac14 116 kN=m3

                    The calculations for trial values of F of 20 15 and 10 are summarized below

                    F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                    20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                    s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                    Figure Q611

                    Lateral earth pressure 47

                    612

                    For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                    45 thorn 0

                    2frac14 63

                    For the retained material between the surface and a depth of 36m

                    Pa frac14 1

                    2 030 18 362 frac14 350 kN=m

                    Weight of reinforced fill between the surface and a depth of 36m is

                    Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                    eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                    Eccentricity of Rv

                    e frac14 263 250 frac14 013m

                    The average vertical stress at a depth of 36m is

                    z frac14 Rv

                    L 2efrac14 324

                    474frac14 68 kN=m2

                    (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                    Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                    Tensile stress in the element frac14 138 103

                    65 3frac14 71N=mm2

                    Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                    Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                    Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                    The weight of ABC is

                    W frac14 1

                    2 18 52 265 frac14 124 kN=m

                    From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                    48 Lateral earth pressure

                    (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                    Tp frac14 032 68 120 065 frac14 170 kN

                    Tr frac14 213 420

                    418frac14 214 kN

                    Again the tensile failure and slipping limit states are satisfied for this element

                    Figure Q612

                    Lateral earth pressure 49

                    Chapter 7

                    Consolidation theory

                    71

                    Total change in thickness

                    H frac14 782 602 frac14 180mm

                    Average thickness frac14 1530thorn 180

                    2frac14 1620mm

                    Length of drainage path d frac14 1620

                    2frac14 810mm

                    Root time plot (Figure Q71a)

                    ffiffiffiffiffiffit90p frac14 33

                    t90 frac14 109min

                    cv frac14 0848d2

                    t90frac14 0848 8102

                    109 1440 365

                    106frac14 27m2=year

                    r0 frac14 782 764

                    782 602frac14 018

                    180frac14 0100

                    rp frac14 10eth764 645THORN9eth782 602THORN frac14

                    10 119

                    9 180frac14 0735

                    rs frac14 1 eth0100thorn 0735THORN frac14 0165

                    Log time plot (Figure Q71b)

                    t50 frac14 26min

                    cv frac14 0196d2

                    t50frac14 0196 8102

                    26 1440 365

                    106frac14 26m2=year

                    r0 frac14 782 763

                    782 602frac14 019

                    180frac14 0106

                    rp frac14 763 623

                    782 602frac14 140

                    180frac14 0778

                    rs frac14 1 eth0106thorn 0778THORN frac14 0116

                    Figure Q71(a)

                    Figure Q71(b)

                    Final void ratio

                    e1 frac14 w1Gs frac14 0232 272 frac14 0631

                    e

                    Hfrac14 1thorn e0

                    H0frac14 1thorn e1 thorne

                    H0

                    ie

                    e

                    180frac14 1631thorne

                    1710

                    e frac14 2936

                    1530frac14 0192

                    Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                    mv frac14 1

                    1thorn e0 e0 e101 00

                    frac14 1

                    1823 0192

                    0107frac14 098m2=MN

                    k frac14 cvmvw frac14 265 098 98

                    60 1440 365 103frac14 81 1010 m=s

                    72

                    Using Equation 77 (one-dimensional method)

                    sc frac14 e0 e11thorn e0 H

                    Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                    Figure Q72

                    52 Consolidation theory

                    Settlement

                    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                    1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                    318

                    Notes 5 92y 460thorn 84

                    Heave

                    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                    1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                    38

                    73

                    U frac14 f ethTvTHORN frac14 f cvt

                    d2

                    Hence if cv is constant

                    t1

                    t2frac14 d

                    21

                    d22

                    where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                    d1 frac14 95mm and d2 frac14 2500mm

                    for U frac14 050 t2 frac14 t1 d22

                    d21

                    frac14 20

                    60 24 365 25002

                    952frac14 263 years

                    for U lt 060 Tv frac14

                    4U2 (Equation 724(a))

                    t030 frac14 t050 0302

                    0502

                    frac14 263 036 frac14 095 years

                    Consolidation theory 53

                    74

                    The layer is open

                    d frac14 8

                    2frac14 4m

                    Tv frac14 cvtd2frac14 24 3

                    42frac14 0450

                    ui frac14 frac14 84 kN=m2

                    The excess pore water pressure is given by Equation 721

                    ue frac14Xmfrac141mfrac140

                    2ui

                    Msin

                    Mz

                    d

                    expethM2TvTHORN

                    In this case z frac14 d

                    sinMz

                    d

                    frac14 sinM

                    where

                    M frac14

                    23

                    25

                    2

                    M sin M M2Tv exp (M2Tv)

                    2thorn1 1110 0329

                    3

                    21 9993 457 105

                    ue frac14 2 84 2

                    1 0329 ethother terms negligibleTHORN

                    frac14 352 kN=m2

                    75

                    The layer is open

                    d frac14 6

                    2frac14 3m

                    Tv frac14 cvtd2frac14 10 3

                    32frac14 0333

                    The layer thickness will be divided into six equal parts ie m frac14 6

                    54 Consolidation theory

                    For an open layer

                    Tv frac14 4n

                    m2

                    n frac14 0333 62

                    4frac14 300

                    The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                    ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                    i j

                    0 1 2 3 4 5 6 7 8 9 10 11 12

                    0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                    The initial and 3-year isochrones are plotted in Figure Q75

                    Area under initial isochrone frac14 180 units

                    Area under 3-year isochrone frac14 63 units

                    The average degree of consolidation is given by Equation 725Thus

                    U frac14 1 63

                    180frac14 065

                    Figure Q75

                    Consolidation theory 55

                    76

                    At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                    0 frac14 2w frac14 2 98 frac14 196 kN=m2

                    The final consolidation settlement (one-dimensional method) is

                    sc frac14 mv0H frac14 083 196 8 frac14 130mm

                    Corrected time t frac14 2 1

                    2

                    40

                    52

                    frac14 1615 years

                    Tv frac14 cvtd2frac14 44 1615

                    42frac14 0444

                    From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                    77

                    The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                    Figure Q77

                    56 Consolidation theory

                    Point m n Ir (kNm2) sc (mm)

                    13020frac14 15 20

                    20frac14 10 0194 (4) 113 124

                    260

                    20frac14 30

                    20

                    20frac14 10 0204 (2) 59 65

                    360

                    20frac14 30

                    40

                    20frac14 20 0238 (1) 35 38

                    430

                    20frac14 15

                    40

                    20frac14 20 0224 (2) 65 72

                    Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                    78

                    Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                    (a) Immediate settlement

                    H

                    Bfrac14 30

                    35frac14 086

                    D

                    Bfrac14 2

                    35frac14 006

                    Figure Q78

                    Consolidation theory 57

                    From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                    si frac14 130131qB

                    Eufrac14 10 032 105 35

                    40frac14 30mm

                    (b) Consolidation settlement

                    Layer z (m) Dz Ic (kNm2) syod (mm)

                    1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                    3150

                    Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                    Now

                    H

                    Bfrac14 30

                    35frac14 086 and A frac14 065

                    from Figure 712 13 frac14 079

                    sc frac14 13sod frac14 079 315 frac14 250mm

                    Total settlement

                    s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                    79

                    Without sand drains

                    Uv frac14 025

                    Tv frac14 0049 ethfrom Figure 718THORN

                    t frac14 Tvd2

                    cvfrac14 0049 82

                    cvWith sand drains

                    R frac14 0564S frac14 0564 3 frac14 169m

                    n frac14 Rrfrac14 169

                    015frac14 113

                    Tr frac14 cht

                    4R2frac14 ch

                    4 1692 0049 82

                    cvethand ch frac14 cvTHORN

                    frac14 0275

                    Ur frac14 073 (from Figure 730)

                    58 Consolidation theory

                    Using Equation 740

                    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                    U frac14 080

                    710

                    Without sand drains

                    Uv frac14 090

                    Tv frac14 0848

                    t frac14 Tvd2

                    cvfrac14 0848 102

                    96frac14 88 years

                    With sand drains

                    R frac14 0564S frac14 0564 4 frac14 226m

                    n frac14 Rrfrac14 226

                    015frac14 15

                    Tr

                    Tvfrac14 chcv

                    d2

                    4R2ethsame tTHORN

                    Tr

                    Tvfrac14 140

                    96 102

                    4 2262frac14 714 eth1THORN

                    Using Equation 740

                    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                    An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                    Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                    040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                    Thus

                    Uv frac14 0295 and Ur frac14 086

                    t frac14 88 00683

                    0848frac14 07 years

                    Consolidation theory 59

                    Chapter 8

                    Bearing capacity

                    81

                    (a) The ultimate bearing capacity is given by Equation 83

                    qf frac14 cNc thorn DNq thorn 1

                    2BN

                    For u frac14 0

                    Nc frac14 514 Nq frac14 1 N frac14 0

                    qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                    The net ultimate bearing capacity is

                    qnf frac14 qf D frac14 540 kN=m2

                    The net foundation pressure is

                    qn frac14 q D frac14 425

                    2 eth21 1THORN frac14 192 kN=m2

                    The factor of safety (Equation 86) is

                    F frac14 qnfqnfrac14 540

                    192frac14 28

                    (b) For 0 frac14 28

                    Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                    qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                    2 112 2 13

                    frac14 260thorn 168thorn 146 frac14 574 kN=m2

                    qnf frac14 574 112 frac14 563 kN=m2

                    F frac14 563

                    192frac14 29

                    (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                    82

                    For 0 frac14 38

                    Nq frac14 49 N frac14 67

                    qnf frac14 DethNq 1THORN thorn 1

                    2BN ethfrom Equation 83THORN

                    frac14 eth18 075 48THORN thorn 1

                    2 18 15 67

                    frac14 648thorn 905 frac14 1553 kN=m2

                    qn frac14 500

                    15 eth18 075THORN frac14 320 kN=m2

                    F frac14 qnfqnfrac14 1553

                    320frac14 48

                    0d frac14 tan1tan 38

                    125

                    frac14 32 therefore Nq frac14 23 and N frac14 25

                    Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                    2 18 15 25

                    frac14 15eth310thorn 337THORNfrac14 970 kN=m

                    Design load (action) Vd frac14 500 kN=m

                    The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                    83

                    D

                    Bfrac14 350

                    225frac14 155

                    From Figure 85 for a square foundation

                    Nc frac14 81

                    Bearing capacity 61

                    For a rectangular foundation (L frac14 450m B frac14 225m)

                    Nc frac14 084thorn 016B

                    L

                    81 frac14 745

                    Using Equation 810

                    qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                    For F frac14 3

                    qn frac14 1006

                    3frac14 335 kN=m2

                    q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                    Design load frac14 405 450 225 frac14 4100 kN

                    Design undrained strength cud frac14 135

                    14frac14 96 kN=m2

                    Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                    frac14 7241 kN

                    Design load Vd frac14 4100 kN

                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                    84

                    For 0 frac14 40

                    Nq frac14 64 N frac14 95

                    qnf frac14 DethNq 1THORN thorn 04BN

                    (a) Water table 5m below ground level

                    qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                    qn frac14 400 17 frac14 383 kN=m2

                    F frac14 2686

                    383frac14 70

                    (b) Water table 1m below ground level (ie at foundation level)

                    0 frac14 20 98 frac14 102 kN=m3

                    62 Bearing capacity

                    qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                    F frac14 2040

                    383frac14 53

                    (c) Water table at ground level with upward hydraulic gradient 02

                    eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                    qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                    F frac14 1296

                    392frac14 33

                    85

                    The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                    Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                    Design value of 0 frac14 tan1tan 39

                    125

                    frac14 33

                    For 0 frac14 33 Nq frac14 26 and N frac14 29

                    Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                    86

                    (a) Undrained shear for u frac14 0

                    Nc frac14 514 Nq frac14 1 N frac14 0

                    qnf frac14 12cuNc

                    frac14 12 100 514 frac14 617 kN=m2

                    qn frac14 qnfFfrac14 617

                    3frac14 206 kN=m2

                    q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                    Bearing capacity 63

                    Drained shear for 0 frac14 32

                    Nq frac14 23 N frac14 25

                    0 frac14 21 98 frac14 112 kN=m3

                    qnf frac14 0DethNq 1THORN thorn 040BN

                    frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                    frac14 694 kN=m2

                    q frac14 694

                    3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                    Design load frac14 42 227 frac14 3632 kN

                    (b) Design undrained strength cud frac14 100

                    14frac14 71 kNm2

                    Design bearing resistance Rd frac14 12cudNe area

                    frac14 12 71 514 42

                    frac14 7007 kN

                    For drained shear 0d frac14 tan1tan 32

                    125

                    frac14 26

                    Nq frac14 12 N frac14 10

                    Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                    (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                    Layer z (m) m n Ir 0 (kNm2) sod (mm)

                    1 2 100 0175 0700qn 0182qn

                    2 6 033 0044 0176qn 0046qn

                    3 10 020 0017 0068qn 0018qn

                    0246qn

                    Diameter of equivalent circle B frac14 45m

                    H

                    Bfrac14 12

                    45frac14 27 and A frac14 042

                    13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                    64 Bearing capacity

                    For sc frac14 30mm

                    qn frac14 30

                    0147frac14 204 kN=m2

                    q frac14 204thorn 21 frac14 225 kN=m2

                    Design load frac14 42 225 frac14 3600 kN

                    The design load is 3600 kN settlement being the limiting criterion

                    87

                    D

                    Bfrac14 8

                    4frac14 20

                    From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                    F frac14 cuNc

                    Dfrac14 40 71

                    20 8frac14 18

                    88

                    Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                    Design value of 0 frac14 tan1tan 38

                    125

                    frac14 32

                    Figure Q86

                    Bearing capacity 65

                    For 0 frac14 32 Nq frac14 23 and N frac14 25

                    Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                    The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                    Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                    For B frac14 250m qn frac14 3750

                    2502 17 frac14 583 kN=m2

                    From Figure 510 m frac14 n frac14 126

                    6frac14 021

                    Ir frac14 0019

                    Stress increment frac14 4 0019 583 frac14 44 kN=m2

                    Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                    The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                    89

                    Depth (m) N 0v (kNm2) CN N1

                    070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                    Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                    (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                    Cw frac14 05thorn 0530

                    47

                    frac14 082

                    66 Bearing capacity

                    Thus

                    qa frac14 150 082 frac14 120 kN=m2

                    (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                    Thus

                    qa frac14 90 15 frac14 135 kN=m2

                    (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                    Ic frac14 171

                    1014frac14 0068

                    From Equation 819(a) with s frac14 25mm

                    q frac14 25

                    3507 0068frac14 150 kN=m2

                    810

                    Peak value of strain influence factor occurs at a depth of 27m and is given by

                    Izp frac14 05thorn 01130

                    16 27

                    05

                    frac14 067

                    Refer to Figure Q810

                    E frac14 25qc

                    Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                    Ez (mm3MN)

                    1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                    0203

                    C1 frac14 1 0500qnfrac14 1 05 12 16

                    130frac14 093

                    C2 frac14 1 ethsayTHORN

                    s frac14 C1C2qnX Iz

                    Ez frac14 093 1 130 0203 frac14 25mm

                    Bearing capacity 67

                    811

                    At pile base level

                    cu frac14 220 kN=m2

                    qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                    Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                    00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                    qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                    Then

                    Qf frac14 Abqb thorn Asqs

                    frac14

                    4 32 1980

                    thorn eth 105 139 86THORN

                    frac14 13 996thorn 3941 frac14 17 937 kN

                    0 01 02 03 04 05 06 07

                    0 2 4 6 8 10 12 14

                    1

                    2

                    3

                    4

                    5

                    6

                    7

                    8

                    (1)

                    (2)

                    (3)

                    (4)

                    (5)

                    qc

                    qc

                    Iz

                    Iz

                    (MNm2)

                    z (m)

                    Figure Q810

                    68 Bearing capacity

                    Allowable load

                    ethaTHORN Qf

                    2frac14 17 937

                    2frac14 8968 kN

                    ethbTHORN Abqb

                    3thorn Asqs frac14 13 996

                    3thorn 3941 frac14 8606 kN

                    ie allowable load frac14 8600 kN

                    Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                    According to the limit state method

                    Characteristic undrained strength at base level cuk frac14 220

                    150kN=m2

                    Characteristic base resistance qbk frac14 9cuk frac14 9 220

                    150frac14 1320 kN=m2

                    Characteristic shaft resistance qsk frac14 00150

                    frac14 86

                    150frac14 57 kN=m2

                    Characteristic base and shaft resistances

                    Rbk frac14

                    4 32 1320 frac14 9330 kN

                    Rsk frac14 105 139 86

                    150frac14 2629 kN

                    For a bored pile the partial factors are b frac14 160 and s frac14 130

                    Design bearing resistance Rcd frac14 9330

                    160thorn 2629

                    130

                    frac14 5831thorn 2022

                    frac14 7850 kN

                    Adding ethDAb W) the design bearing resistance becomes 9650 kN

                    812

                    ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                    qs frac14 cu frac14 040 105 frac14 42 kN=m2

                    For a single pile

                    Qf frac14 Abqb thorn Asqs

                    frac14

                    4 062 1305

                    thorn eth 06 15 42THORN

                    frac14 369thorn 1187 frac14 1556 kN

                    Bearing capacity 69

                    Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                    eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                    (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                    (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                    qbkfrac14 9cuk frac14 9 220

                    150frac14 1320 kN=m2

                    qskfrac14cuk frac14 040 105

                    150frac14 28 kN=m2

                    Rbkfrac14

                    4 0602 1320 frac14 373 kN

                    Rskfrac14 060 15 28 frac14 791 kN

                    Rcdfrac14 373

                    160thorn 791

                    130frac14 233thorn 608 frac14 841 kN

                    Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                    Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                    (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                    q frac14 21 000

                    1762frac14 68 kN=m2

                    Immediate settlement

                    H

                    Bfrac14 15

                    176frac14 085

                    D

                    Bfrac14 13

                    176frac14 074

                    L

                    Bfrac14 1

                    Hence from Figure 515

                    130 frac14 078 and 131 frac14 041

                    70 Bearing capacity

                    Thus using Equation 528

                    si frac14 078 041 68 176

                    65frac14 6mm

                    Consolidation settlement

                    Layer z (m) Area (m2) (kNm2) mvH (mm)

                    1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                    434 (sod)

                    Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                    sc frac14 056 434 frac14 24mm

                    The total settlement is (6thorn 24) frac14 30mm

                    813

                    At base level N frac14 26 Then using Equation 830

                    qb frac14 40NDb

                    Bfrac14 40 26 2

                    025frac14 8320 kN=m2

                    ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                    Figure Q812

                    Bearing capacity 71

                    Over the length embedded in sand

                    N frac14 21 ie18thorn 24

                    2

                    Using Equation 831

                    qs frac14 2N frac14 2 21 frac14 42 kN=m2

                    For a single pile

                    Qf frac14 Abqb thorn Asqs

                    frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                    For the pile group assuming a group efficiency of 12

                    XQf frac14 12 9 604 frac14 6523 kN

                    Then the load factor is

                    F frac14 6523

                    2000thorn 1000frac14 21

                    (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                    Characteristic base resistance per unit area qbk frac14 8320

                    150frac14 5547 kNm2

                    Characteristic shaft resistance per unit area qsk frac14 42

                    150frac14 28 kNm2

                    Characteristic base and shaft resistances for a single pile

                    Rbk frac14 0252 5547 frac14 347 kN

                    Rsk frac14 4 025 2 28 frac14 56 kN

                    For a driven pile the partial factors are b frac14 s frac14 130

                    Design bearing resistance Rcd frac14 347

                    130thorn 56

                    130frac14 310 kN

                    For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                    Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                    (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                    From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                    72 Bearing capacity

                    N frac14 24thorn 26thorn 34

                    3frac14 28

                    Ic frac14 171

                    2814frac14 0016 ethEquation 818THORN

                    s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                    814

                    Using Equation 841

                    Tf frac14 DLcu thorn

                    4ethD2 d2THORNcuNc

                    frac14 eth 02 5 06 110THORN thorn

                    4eth022 012THORN110 9

                    frac14 207thorn 23 frac14 230 kN

                    Figure Q813

                    Bearing capacity 73

                    Chapter 9

                    Stability of slopes

                    91

                    Referring to Figure Q91

                    W frac14 417 19 frac14 792 kN=m

                    Q frac14 20 28 frac14 56 kN=m

                    Arc lengthAB frac14

                    180 73 90 frac14 115m

                    Arc length BC frac14

                    180 28 90 frac14 44m

                    The factor of safety is given by

                    F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                    90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                    Depth of tension crack z0 frac14 2cu

                    frac14 2 20

                    19frac14 21m

                    Arc length BD frac14

                    180 13

                    1

                    2 90 frac14 21m

                    F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                    The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                    Design resisting moment frac14 rXethcudLaTHORN frac14 90

                    14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                    Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                    The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                    92

                    u frac14 0

                    Depth factor D frac14 11

                    9frac14 122

                    Using Equation 92 with F frac14 10

                    Ns frac14 cu

                    FHfrac14 30

                    10 19 9frac14 0175

                    Hence from Figure 93

                    frac14 50

                    For F frac14 12

                    Ns frac14 30

                    12 19 9frac14 0146

                    frac14 27

                    93

                    Refer to Figure Q93

                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                    1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                    74 m

                    214 1deg

                    213 1deg

                    39 m

                    WB

                    D

                    C

                    28 m

                    21 m

                    A

                    Q

                    Soil (1)Soil (2)

                    73deg

                    Figure Q91

                    Stability of slopes 75

                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                    9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                    599 256 328 1372

                    Figure Q93

                    76 Stability of slopes

                    XW cos frac14 b

                    Xh cos frac14 21 2 599 frac14 2516 kN=mX

                    W sin frac14 bX

                    h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                    Arc length La frac14

                    180 57

                    1

                    2 326 frac14 327m

                    The factor of safety is given by

                    F frac14 c0La thorn tan0ethW cos ulTHORN

                    W sin

                    frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                    frac14 091

                    According to the limit state method

                    0d frac14 tan1tan 32

                    125

                    frac14 265

                    c0 frac14 8

                    160frac14 5 kN=m2

                    Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                    Design disturbing moment frac14 1075 kN=m

                    The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                    94

                    F frac14 1

                    W sin

                    Xfc0bthorn ethW ubTHORN tan0g sec

                    1thorn ethtan tan0=FTHORN

                    c0 frac14 8 kN=m2

                    0 frac14 32

                    c0b frac14 8 2 frac14 16 kN=m

                    W frac14 bh frac14 21 2 h frac14 42h kN=m

                    Try F frac14 100

                    tan0

                    Ffrac14 0625

                    Stability of slopes 77

                    Values of u are as obtained in Figure Q93

                    SliceNo

                    h(m)

                    W frac14 bh(kNm)

                    W sin(kNm)

                    ub(kNm)

                    c0bthorn (W ub) tan0(kNm)

                    sec

                    1thorn (tan tan0)FProduct(kNm)

                    1 05 21 6 2 8 24 1078 262 13 55 31

                    23 33 30 1042 31

                    3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                    224 92 72 0931 67

                    6 50 210 11 40 100 85 0907 777 55 231 14

                    12 58 112 90 0889 80

                    8 60 252 1812

                    80 114 102 0874 899 63 265 22 99 116 109 0861 94

                    10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                    2154 88 116 0853 99

                    14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                    1074 1091

                    F frac14 1091

                    1074frac14 102 (assumed value 100)

                    Thus

                    F frac14 101

                    95

                    F frac14 1

                    W sin

                    XfWeth1 ruTHORN tan0g sec

                    1thorn ethtan tan0THORN=F

                    0 frac14 33

                    ru frac14 020

                    W frac14 bh frac14 20 5 h frac14 100h kN=m

                    eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                    Try F frac14 110

                    tan 0

                    Ffrac14 tan 33

                    110frac14 0590

                    78 Stability of slopes

                    Referring to Figure Q95

                    SliceNo

                    h(m)

                    W frac14 bh(kNm)

                    W sin(kNm)

                    W(1 ru) tan0(kNm)

                    sec

                    1thorn ( tan tan0)FProduct(kNm)

                    1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                    2120 234 0892 209

                    4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                    1185 1271

                    Figure Q95

                    Stability of slopes 79

                    F frac14 1271

                    1185frac14 107

                    The trial value was 110 therefore take F to be 108

                    96

                    (a) Water table at surface the factor of safety is given by Equation 912

                    F frac14 0

                    sat

                    tan0

                    tan

                    ptie 15 frac14 92

                    19

                    tan 36

                    tan

                    tan frac14 0234

                    frac14 13

                    Water table well below surface the factor of safety is given by Equation 911

                    F frac14 tan0

                    tan

                    frac14 tan 36

                    tan 13

                    frac14 31

                    (b) 0d frac14 tan1tan 36

                    125

                    frac14 30

                    Depth of potential failure surface frac14 z

                    Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                    frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                    frac14 504z kN

                    Design disturbing moment per unit area Sd frac14 sat sin cos

                    frac14 19 z sin 13 cos 13

                    frac14 416z kN

                    Rd gtSd therefore the limit state for overall stability is satisfied

                    80 Stability of slopes

                    • Book Cover
                    • Title
                    • Contents
                    • Basic characteristics of soils
                    • Seepage
                    • Effective stress
                    • Shear strength
                    • Stresses and displacements
                    • Lateral earth pressure
                    • Consolidation theory
                    • Bearing capacity
                    • Stability of slopes

                      Equation 126 with A equal in turn to 0 005 and 010 is used to calculate values ofdry density (d0

                      d5 d10

                      respectively) for use in plotting the air content curves Theexperimental values of w have been used in these calculations however any series ofw values within the relevant range could be used By inspection the value of aircontent at maximum dry density is 35

                      17

                      From Equation 120

                      e frac14 Gswd 1

                      The maximum and minimum values of void ratio are given by

                      emax frac14 Gsw

                      dmin

                      1

                      emin frac14 Gswdmax

                      1

                      From Equation 123

                      ID frac14 Gsweth1=dmin 1=dTHORN

                      Gsweth1=dmin 1=dmax

                      THORN

                      frac14 frac121 ethdmin=dTHORN1=dmin

                      frac121 ethdmin=dmax

                      THORN1=dmin

                      frac14 d dmin

                      dmax dmin

                      dmax

                      d

                      frac14 172 154

                      181 154

                      181

                      172

                      frac14 070 eth70THORN

                      Basic characteristics of soils 5

                      Chapter 2

                      Seepage

                      21

                      The coefficient of permeability is determined from the equation

                      k frac14 23al

                      At1log

                      h0

                      h1

                      where

                      a frac14

                      4 00052 m2 l frac14 02m

                      A frac14

                      4 012 m2 t1 frac14 3 602 s

                      logh0

                      h1frac14 log

                      100

                      035frac14 0456

                      k frac14 23 00052 02 0456

                      012 3 602frac14 49 108 m=s

                      22

                      The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

                      q frac14 kh Nf

                      Ndfrac14 106 400 37

                      11frac14 13 106 m3=s per m

                      Figure Q22

                      23

                      The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

                      q frac14 kh Nf

                      Ndfrac14 5 105 300 35

                      9frac14 58 105 m3=s per m

                      The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

                      Point h (m) h z (m) u frac14 w(h z)(kNm2)

                      1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

                      eg for Point 1

                      h1 frac14 7

                      9 300 frac14 233m

                      h1 z1 frac14 233 eth250THORN frac14 483m

                      Figure Q23

                      Seepage 7

                      u1 frac14 98 483 frac14 47 kN=m2

                      The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

                      24

                      The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

                      q frac14 kh Nf

                      Ndfrac14 40 107 550 10

                      11frac14 20 106 m3=s per m

                      25

                      The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

                      q frac14 kh Nf

                      Ndfrac14 20 106 500 42

                      9frac14 47 106 m3=s per m

                      Figure Q24

                      8 Seepage

                      26

                      The scale transformation factor in the x direction is given by Equation 221 ie

                      xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                      ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

                      Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

                      k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                      pfrac14

                      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

                      p 107 frac14 30 107 m=s

                      Thus the quantity of seepage is given by

                      q frac14 k0h Nf

                      Ndfrac14 30 107 1400 325

                      12frac14 11 106 m3=s per m

                      Figure Q25

                      Seepage 9

                      27

                      The scale transformation factor in the x direction is

                      xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                      ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

                      Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

                      k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                      pfrac14

                      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

                      p 106 frac14 45 106 m=s

                      The focus of the basic parabola is at point A The parabola passes through point Gsuch that

                      GC frac14 03HC frac14 03 30 frac14 90m

                      Thus the coordinates of G are

                      x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

                      480 frac14 x0 2002

                      4x0

                      Figure Q26

                      10 Seepage

                      Hence

                      x0 frac14 20m

                      Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

                      x frac14 20 z2

                      80

                      x 20 0 50 100 200 300z 0 400 748 980 1327 1600

                      The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

                      No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

                      q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

                      28

                      The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

                      Figure Q27

                      Seepage 11

                      q frac14 kh Nf

                      Ndfrac14 45 105 28 33

                      7

                      frac14 59 105 m3=s per m

                      29

                      The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

                      kx frac14 H1k1 thornH2k2

                      H1 thornH2frac14 106

                      10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

                      kz frac14 H1 thornH2

                      H1

                      k1thornH2

                      k2

                      frac14 10

                      5

                      eth2 106THORN thorn5

                      eth16 106THORNfrac14 36 106 m=s

                      Then the scale transformation factor is given by

                      xt frac14 xffiffiffiffiffikz

                      pffiffiffiffiffikx

                      p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

                      Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

                      Figure Q28

                      12 Seepage

                      k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                      qfrac14

                      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

                      p 106 frac14 57 106 m=s

                      Then the quantity of seepage is given by

                      q frac14 k0h Nf

                      Ndfrac14 57 106 350 56

                      11

                      frac14 10 105 m3=s per m

                      Figure Q29

                      Seepage 13

                      Chapter 3

                      Effective stress

                      31

                      Buoyant unit weight

                      0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                      Effective vertical stress

                      0v frac14 5 102 frac14 51 kN=m2 or

                      Total vertical stress

                      v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

                      Pore water pressure

                      u frac14 7 98 frac14 686 kN=m2

                      Effective vertical stress

                      0v frac14 v u frac14 1196 686 frac14 51 kN=m2

                      32

                      Buoyant unit weight

                      0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                      Effective vertical stress

                      0v frac14 5 102 frac14 51 kN=m2 or

                      Total vertical stress

                      v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

                      Pore water pressure

                      u frac14 205 98 frac14 2009 kN=m2

                      Effective vertical stress

                      0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

                      33

                      At top of the clay

                      v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

                      u frac14 2 98 frac14 196 kN=m2

                      0v frac14 v u frac14 710 196 frac14 514 kN=m2

                      Alternatively

                      0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                      0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

                      At bottom of the clay

                      v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

                      u frac14 12 98 frac14 1176 kN=m2

                      0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

                      NB The alternative method of calculation is not applicable because of the artesiancondition

                      Figure Q3132

                      Effective stress 15

                      34

                      0 frac14 20 98 frac14 102 kN=m3

                      At 8m depth

                      0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                      35

                      0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                      0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                      Figure Q33

                      Figure Q34

                      16 Effective stress

                      (a) Immediately after WT rise

                      At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                      0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                      At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                      0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                      (b) Several years after WT rise

                      At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                      At 8m depth

                      0v frac14 940 kN=m2 (as above)

                      At 12m depth

                      0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                      Figure Q35

                      Effective stress 17

                      36

                      Total weight

                      ab frac14 210 kN

                      Effective weight

                      ac frac14 112 kN

                      Resultant boundary water force

                      be frac14 119 kN

                      Seepage force

                      ce frac14 34 kN

                      Resultant body force

                      ae frac14 99 kN eth73 to horizontalTHORN

                      (Refer to Figure Q36)

                      Figure Q36

                      18 Effective stress

                      37

                      Situation (1)(a)

                      frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                      u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                      0 frac14 u frac14 694 392 frac14 302 kN=m2

                      (b)

                      i frac14 2

                      4frac14 05

                      j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                      Situation (2)(a)

                      frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                      u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                      0 frac14 u frac14 498 392 frac14 106 kN=m2

                      (b)

                      i frac14 2

                      4frac14 05

                      j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                      38

                      The flow net is drawn in Figure Q24

                      Loss in total head between adjacent equipotentials

                      h frac14 550

                      Ndfrac14 550

                      11frac14 050m

                      Exit hydraulic gradient

                      ie frac14 h

                      sfrac14 050

                      070frac14 071

                      Effective stress 19

                      The critical hydraulic gradient is given by Equation 39

                      ic frac14 0

                      wfrac14 102

                      98frac14 104

                      Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                      F frac14 iciefrac14 104

                      071frac14 15

                      Total head at C

                      hC frac14 nd

                      Ndh frac14 24

                      11 550 frac14 120m

                      Elevation head at C

                      zC frac14 250m

                      Pore water pressure at C

                      uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                      Therefore effective vertical stress at C

                      0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                      For point D

                      hD frac14 73

                      11 550 frac14 365m

                      zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                      0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                      39

                      The flow net is drawn in Figure Q25

                      For a soil prism 150 300m adjacent to the piling

                      hm frac14 26

                      9 500 frac14 145m

                      20 Effective stress

                      Factor of safety against lsquoheavingrsquo (Equation 310)

                      F frac14 ic

                      imfrac14 0d

                      whmfrac14 97 300

                      98 145frac14 20

                      With a filter

                      F frac14 0d thorn wwhm

                      3 frac14 eth97 300THORN thorn w98 145

                      w frac14 135 kN=m2

                      Depth of filterfrac14 13521frac14 065m (if above water level)

                      Effective stress 21

                      Chapter 4

                      Shear strength

                      41

                      frac14 295 kN=m2

                      u frac14 120 kN=m2

                      0 frac14 u frac14 295 120 frac14 175 kN=m2

                      f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                      42

                      03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                      100 452 552200 908 1108400 1810 2210800 3624 4424

                      The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                      Figure Q42

                      43

                      3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                      200 222 422400 218 618600 220 820

                      The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                      44

                      The modified shear strength parameters are

                      0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                      a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                      The coordinates of the stress point representing failure conditions in the test are

                      1

                      2eth1 2THORN frac14 1

                      2 170 frac14 85 kN=m2

                      1

                      2eth1 thorn 3THORN frac14 1

                      2eth270thorn 100THORN frac14 185 kN=m2

                      The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                      uf frac14 36 kN=m2

                      Figure Q43

                      Figure Q44

                      Shear strength 23

                      45

                      3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                      150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                      The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                      The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                      46

                      03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                      200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                      The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                      Figure Q45

                      24 Shear strength

                      47

                      The torque required to produce shear failure is given by

                      T frac14 dh cud

                      2thorn 2

                      Z d=2

                      0

                      2r drcur

                      frac14 cud2h

                      2thorn 4cu

                      Z d=2

                      0

                      r2dr

                      frac14 cud2h

                      2thorn d

                      3

                      6

                      Then

                      35 frac14 cu52 10

                      2thorn 53

                      6

                      103

                      cu frac14 76 kN=m3

                      400

                      0 400 800 1200 1600

                      τ (k

                      Nm

                      2 )

                      σprime (kNm2)

                      34deg

                      315deg29deg

                      (a)

                      (b)

                      0 400

                      400

                      800 1200 1600

                      Failure envelope

                      300 500

                      σprime (kNm2)

                      τ (k

                      Nm

                      2 )

                      20 (kNm2)

                      31deg

                      Figure Q46

                      Shear strength 25

                      48

                      The relevant stress values are calculated as follows

                      3 frac14 600 kN=m2

                      1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                      2(1 3) 0 40 79 107 139 159

                      1

                      2(01 thorn 03) 400 411 402 389 351 326

                      1

                      2(1 thorn 3) 600 640 679 707 739 759

                      The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                      Af frac14 433 200

                      319frac14 073

                      49

                      B frac14 u33

                      frac14 144

                      350 200frac14 096

                      a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                      0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                      10 283 65 023

                      Figure Q48

                      26 Shear strength

                      The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                      Figure Q49

                      Shear strength 27

                      Chapter 5

                      Stresses and displacements

                      51

                      Vertical stress is given by

                      z frac14 Qz2Ip frac14 5000

                      52Ip

                      Values of Ip are obtained from Table 51

                      r (m) rz Ip z (kNm2)

                      0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                      10 20 0009 2

                      The variation of z with radial distance (r) is plotted in Figure Q51

                      Figure Q51

                      52

                      Below the centre load (Figure Q52)

                      r

                      zfrac14 0 for the 7500-kN load

                      Ip frac14 0478

                      r

                      zfrac14 5

                      4frac14 125 for the 10 000- and 9000-kN loads

                      Ip frac14 0045

                      Then

                      z frac14X Q

                      z2Ip

                      frac14 7500 0478

                      42thorn 10 000 0045

                      42thorn 9000 0045

                      42

                      frac14 224thorn 28thorn 25 frac14 277 kN=m2

                      53

                      The vertical stress under a corner of a rectangular area is given by

                      z frac14 qIr

                      where values of Ir are obtained from Figure 510 In this case

                      z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                      z

                      Figure Q52

                      Stresses and displacements 29

                      z (m) m n Ir z (kNm2)

                      0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                      10 010 0005 5

                      z is plotted against z in Figure Q53

                      54

                      (a)

                      m frac14 125

                      12frac14 104

                      n frac14 18

                      12frac14 150

                      From Figure 510 Irfrac14 0196

                      z frac14 2 175 0196 frac14 68 kN=m2

                      Figure Q53

                      30 Stresses and displacements

                      (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                      z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                      55

                      Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                      Px frac14 2Q

                      1

                      m2 thorn 1frac14 2 150

                      125frac14 76 kN=m

                      Equation 517 is used to obtain the pressure distribution

                      px frac14 4Q

                      h

                      m2n

                      ethm2 thorn n2THORN2 frac14150

                      m2n

                      ethm2 thorn n2THORN2 ethkN=m2THORN

                      Figure Q54

                      Stresses and displacements 31

                      n m2n

                      (m2 thorn n2)2

                      px(kNm2)

                      0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                      The pressure distribution is plotted in Figure Q55

                      56

                      H

                      Bfrac14 10

                      2frac14 5

                      L

                      Bfrac14 4

                      2frac14 2

                      D

                      Bfrac14 1

                      2frac14 05

                      Hence from Figure 515

                      131 frac14 082

                      130 frac14 094

                      Figure Q55

                      32 Stresses and displacements

                      The immediate settlement is given by Equation 528

                      si frac14 130131qB

                      Eu

                      frac14 094 082 200 2

                      45frac14 7mm

                      Stresses and displacements 33

                      Chapter 6

                      Lateral earth pressure

                      61

                      For 0 frac14 37 the active pressure coefficient is given by

                      Ka frac14 1 sin 37

                      1thorn sin 37frac14 025

                      The total active thrust (Equation 66a with c0 frac14 0) is

                      Pa frac14 1

                      2KaH

                      2 frac14 1

                      2 025 17 62 frac14 765 kN=m

                      If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                      K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                      and the thrust on the wall is

                      P0 frac14 1

                      2K0H

                      2 frac14 1

                      2 040 17 62 frac14 122 kN=m

                      62

                      The active pressure coefficients for the three soil types are as follows

                      Ka1 frac141 sin 35

                      1thorn sin 35frac14 0271

                      Ka2 frac141 sin 27

                      1thorn sin 27frac14 0375

                      ffiffiffiffiffiffiffiKa2

                      p frac14 0613

                      Ka3 frac141 sin 42

                      1thorn sin 42frac14 0198

                      Distribution of active pressure (plotted in Figure Q62)

                      Depth (m) Soil Active pressure (kNm2)

                      3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                      12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                      At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                      Total thrust frac14 571 kNm

                      Point of application is (4893571) m from the top of the wall ie 857m

                      Force (kN) Arm (m) Moment (kN m)

                      (1)1

                      2 0271 16 32 frac14 195 20 390

                      (2) 0271 16 3 2 frac14 260 40 1040

                      (3)1

                      2 0271 92 22 frac14 50 433 217

                      (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                      (5)1

                      2 0375 102 32 frac14 172 70 1204

                      (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                      (7)1

                      2 0198 112 42 frac14 177 1067 1889

                      (8)1

                      2 98 92 frac14 3969 90 35721

                      5713 48934

                      Figure Q62

                      Lateral earth pressure 35

                      63

                      (a) For u frac14 0 Ka frac14 Kp frac14 1

                      Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                      frac14 245

                      At the lower end of the piling

                      pa frac14 Kaqthorn Kasatz Kaccu

                      frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                      frac14 115 kN=m2

                      pp frac14 Kpsatzthorn Kpccu

                      frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                      frac14 202 kN=m2

                      (b) For 0 frac14 26 and frac14 1

                      20

                      Ka frac14 035

                      Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                      pfrac14 145 ethEquation 619THORN

                      Kp frac14 37

                      Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                      pfrac14 47 ethEquation 624THORN

                      At the lower end of the piling

                      pa frac14 Kaqthorn Ka0z Kacc

                      0

                      frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                      frac14 187 kN=m2

                      pp frac14 Kp0zthorn Kpcc

                      0

                      frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                      frac14 198 kN=m2

                      36 Lateral earth pressure

                      64

                      (a) For 0 frac14 38 Ka frac14 024

                      0 frac14 20 98 frac14 102 kN=m3

                      The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                      Force (kN) Arm (m) Moment (kN m)

                      (1) 024 10 66 frac14 159 33 525

                      (2)1

                      2 024 17 392 frac14 310 400 1240

                      (3) 024 17 39 27 frac14 430 135 580

                      (4)1

                      2 024 102 272 frac14 89 090 80

                      (5)1

                      2 98 272 frac14 357 090 321

                      Hfrac14 1345 MH frac14 2746

                      (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                      (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                      XM frac14MV MH frac14 7790 kNm

                      Lever arm of base resultant

                      M

                      Vfrac14 779

                      488frac14 160

                      Eccentricity of base resultant

                      e frac14 200 160 frac14 040m

                      39 m

                      27 m

                      40 m

                      04 m

                      04 m

                      26 m

                      (7)

                      (9)

                      (1)(2)

                      (3)

                      (4)

                      (5)

                      (8)(6)

                      (10)

                      WT

                      10 kNm2

                      Hydrostatic

                      Figure Q64

                      Lateral earth pressure 37

                      Base pressures (Equation 627)

                      p frac14 VB

                      1 6e

                      B

                      frac14 488

                      4eth1 060THORN

                      frac14 195 kN=m2 and 49 kN=m2

                      Factor of safety against sliding (Equation 628)

                      F frac14 V tan

                      Hfrac14 488 tan 25

                      1345frac14 17

                      (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                      Hfrac14 1633 kN

                      V frac14 4879 kN

                      MH frac14 3453 kNm

                      MV frac14 10536 kNm

                      The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                      65

                      For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                      Kp

                      Ffrac14 385

                      2

                      0 frac14 20 98 frac14 102 kN=m3

                      The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                      Force (kN) Arm (m) Moment (kN m)

                      (1)1

                      2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                      (2) 026 17 45 d frac14 199d d2 995d2

                      (3)1

                      2 026 102 d2 frac14 133d2 d3 044d3

                      (4)1

                      2 385

                      2 17 152 frac14 368 dthorn 05 368d 184

                      (5)385

                      2 17 15 d frac14 491d d2 2455d2

                      (6)1

                      2 385

                      2 102 d2 frac14 982d2 d3 327d3

                      38 Lateral earth pressure

                      XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                      d3 thorn 516d2 283d 1724 frac14 0

                      d frac14 179m

                      Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                      Over additional 20 embedded depth

                      pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                      Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                      66

                      The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                      Ka frac14 sin 69=sin 105

                      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                      pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                      26664

                      37775

                      2

                      frac14 050

                      The total active thrust (acting at 25 above the normal) is given by Equation 616

                      Pa frac14 1

                      2 050 19 7502 frac14 267 kN=m

                      Figure Q65

                      Lateral earth pressure 39

                      Horizontal component

                      Ph frac14 267 cos 40 frac14 205 kN=m

                      Vertical component

                      Pv frac14 267 sin 40 frac14 172 kN=m

                      Consider moments about the toe of the wall (Figure Q66) (per m)

                      Force (kN) Arm (m) Moment (kN m)

                      (1)1

                      2 175 650 235 frac14 1337 258 345

                      (2) 050 650 235 frac14 764 175 134

                      (3)1

                      2 070 650 235 frac14 535 127 68

                      (4) 100 400 235 frac14 940 200 188

                      (5) 1

                      2 080 050 235 frac14 47 027 1

                      Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                      Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                      Lever arm of base resultant

                      M

                      Vfrac14 795

                      525frac14 151m

                      Eccentricity of base resultant

                      e frac14 200 151 frac14 049m

                      Figure Q66

                      40 Lateral earth pressure

                      Base pressures (Equation 627)

                      p frac14 525

                      41 6 049

                      4

                      frac14 228 kN=m2 and 35 kN=m2

                      The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                      The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                      The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                      67

                      For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                      Force (kN) Arm (m) Moment (kNm)

                      (1)1

                      2 027 17 52 frac14 574 183 1050

                      (2) 027 17 5 3 frac14 689 500 3445

                      (3)1

                      2 027 102 32 frac14 124 550 682

                      (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                      (5)1

                      2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                      (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                      (7) 1

                      2 267

                      2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                      (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                      2 d frac14 163d d2thorn 650 82d2 1060d

                      Tie rod force per m frac14 T 0 0

                      XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                      d3 thorn 77d2 269d 1438 frac14 0

                      d frac14 467m

                      Depth of penetration frac14 12d frac14 560m

                      Lateral earth pressure 41

                      Algebraic sum of forces for d frac14 467m isX

                      F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                      T frac14 905 kN=m

                      Force in each tie rod frac14 25T frac14 226 kN

                      68

                      (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                      0 frac14 21 98 frac14 112 kN=m3

                      The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                      uC frac14 150

                      165 15 98 frac14 134 kN=m2

                      The average seepage pressure is

                      j frac14 15

                      165 98 frac14 09 kN=m3

                      Hence

                      0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                      0 j frac14 112 09 frac14 103 kN=m3

                      Figure Q67

                      42 Lateral earth pressure

                      Consider moments about the anchor point A (per m)

                      Force (kN) Arm (m) Moment (kN m)

                      (1) 10 026 150 frac14 390 60 2340

                      (2)1

                      2 026 18 452 frac14 474 15 711

                      (3) 026 18 45 105 frac14 2211 825 18240

                      (4)1

                      2 026 121 1052 frac14 1734 100 17340

                      (5)1

                      2 134 15 frac14 101 40 404

                      (6) 134 30 frac14 402 60 2412

                      (7)1

                      2 134 60 frac14 402 95 3819

                      571 4527(8) Ppm

                      115 115PPm

                      XM frac14 0

                      Ppm frac144527

                      115frac14 394 kN=m

                      Available passive resistance

                      Pp frac14 1

                      2 385 103 62 frac14 714 kN=m

                      Factor of safety

                      Fp frac14 Pp

                      Ppm

                      frac14 714

                      394frac14 18

                      Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                      Figure Q68

                      Lateral earth pressure 43

                      (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                      Consider moments (per m) about the tie point A

                      Force (kN) Arm (m)

                      (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                      (2)1

                      2 033 18 452 frac14 601 15

                      (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                      (4)1

                      2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                      (5)1

                      2 134 15 frac14 101 40

                      (6) 134 30 frac14 402 60

                      (7)1

                      2 134 d frac14 67d d3thorn 75

                      (8) 1

                      2 30 103 d2 frac141545d2 2d3thorn 75

                      Moment (kN m)

                      (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                      XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                      d3 thorn 827d2 466d 1518 frac14 0

                      By trial

                      d frac14 544m

                      The minimum depth of embedment required is 544m

                      69

                      For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                      0 frac14 20 98 frac14 102 kN=m3

                      The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                      44 Lateral earth pressure

                      uC frac14 147

                      173 26 98 frac14 216 kN=m2

                      and the average seepage pressure around the wall is

                      j frac14 26

                      173 98 frac14 15 kN=m3

                      Consider moments about the prop (A) (per m)

                      Force (kN) Arm (m) Moment (kN m)

                      (1)1

                      2 03 17 272 frac14 186 020 37

                      (2) 03 17 27 53 frac14 730 335 2445

                      (3)1

                      2 03 (102thorn 15) 532 frac14 493 423 2085

                      (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                      (5)1

                      2 216 26 frac14 281 243 684

                      (6) 216 27 frac14 583 465 2712

                      (7)1

                      2 216 60 frac14 648 800 5184

                      3055(8)

                      1

                      2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                      Factor of safety

                      Fr frac14 6885

                      3055frac14 225

                      Figure Q69

                      Lateral earth pressure 45

                      610

                      For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                      p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                      Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                      Using the recommendations of Twine and Roscoe

                      p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                      Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                      611

                      frac14 18 kN=m3 0 frac14 34

                      H frac14 350m nH frac14 335m mH frac14 185m

                      Consider a trial value of F frac14 20 Refer to Figure 635

                      0m frac14 tan1tan 34

                      20

                      frac14 186

                      Then

                      frac14 45 thorn 0m2frac14 543

                      W frac14 1

                      2 18 3502 cot 543 frac14 792 kN=m

                      Figure Q610

                      46 Lateral earth pressure

                      P frac14 1

                      2 s 3352 frac14 561s kN=m

                      U frac14 1

                      2 98 1852 cosec 543 frac14 206 kN=m

                      Equations 630 and 631 then become

                      561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                      792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                      ie

                      561s 0616N 405 frac14 0

                      792 0857N thorn 563 frac14 0

                      N frac14 848

                      0857frac14 989 kN=m

                      Then

                      561s 609 405 frac14 0

                      s frac14 649

                      561frac14 116 kN=m3

                      The calculations for trial values of F of 20 15 and 10 are summarized below

                      F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                      20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                      s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                      Figure Q611

                      Lateral earth pressure 47

                      612

                      For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                      45 thorn 0

                      2frac14 63

                      For the retained material between the surface and a depth of 36m

                      Pa frac14 1

                      2 030 18 362 frac14 350 kN=m

                      Weight of reinforced fill between the surface and a depth of 36m is

                      Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                      eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                      Eccentricity of Rv

                      e frac14 263 250 frac14 013m

                      The average vertical stress at a depth of 36m is

                      z frac14 Rv

                      L 2efrac14 324

                      474frac14 68 kN=m2

                      (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                      Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                      Tensile stress in the element frac14 138 103

                      65 3frac14 71N=mm2

                      Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                      Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                      Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                      The weight of ABC is

                      W frac14 1

                      2 18 52 265 frac14 124 kN=m

                      From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                      48 Lateral earth pressure

                      (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                      Tp frac14 032 68 120 065 frac14 170 kN

                      Tr frac14 213 420

                      418frac14 214 kN

                      Again the tensile failure and slipping limit states are satisfied for this element

                      Figure Q612

                      Lateral earth pressure 49

                      Chapter 7

                      Consolidation theory

                      71

                      Total change in thickness

                      H frac14 782 602 frac14 180mm

                      Average thickness frac14 1530thorn 180

                      2frac14 1620mm

                      Length of drainage path d frac14 1620

                      2frac14 810mm

                      Root time plot (Figure Q71a)

                      ffiffiffiffiffiffit90p frac14 33

                      t90 frac14 109min

                      cv frac14 0848d2

                      t90frac14 0848 8102

                      109 1440 365

                      106frac14 27m2=year

                      r0 frac14 782 764

                      782 602frac14 018

                      180frac14 0100

                      rp frac14 10eth764 645THORN9eth782 602THORN frac14

                      10 119

                      9 180frac14 0735

                      rs frac14 1 eth0100thorn 0735THORN frac14 0165

                      Log time plot (Figure Q71b)

                      t50 frac14 26min

                      cv frac14 0196d2

                      t50frac14 0196 8102

                      26 1440 365

                      106frac14 26m2=year

                      r0 frac14 782 763

                      782 602frac14 019

                      180frac14 0106

                      rp frac14 763 623

                      782 602frac14 140

                      180frac14 0778

                      rs frac14 1 eth0106thorn 0778THORN frac14 0116

                      Figure Q71(a)

                      Figure Q71(b)

                      Final void ratio

                      e1 frac14 w1Gs frac14 0232 272 frac14 0631

                      e

                      Hfrac14 1thorn e0

                      H0frac14 1thorn e1 thorne

                      H0

                      ie

                      e

                      180frac14 1631thorne

                      1710

                      e frac14 2936

                      1530frac14 0192

                      Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                      mv frac14 1

                      1thorn e0 e0 e101 00

                      frac14 1

                      1823 0192

                      0107frac14 098m2=MN

                      k frac14 cvmvw frac14 265 098 98

                      60 1440 365 103frac14 81 1010 m=s

                      72

                      Using Equation 77 (one-dimensional method)

                      sc frac14 e0 e11thorn e0 H

                      Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                      Figure Q72

                      52 Consolidation theory

                      Settlement

                      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                      1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                      318

                      Notes 5 92y 460thorn 84

                      Heave

                      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                      1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                      38

                      73

                      U frac14 f ethTvTHORN frac14 f cvt

                      d2

                      Hence if cv is constant

                      t1

                      t2frac14 d

                      21

                      d22

                      where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                      d1 frac14 95mm and d2 frac14 2500mm

                      for U frac14 050 t2 frac14 t1 d22

                      d21

                      frac14 20

                      60 24 365 25002

                      952frac14 263 years

                      for U lt 060 Tv frac14

                      4U2 (Equation 724(a))

                      t030 frac14 t050 0302

                      0502

                      frac14 263 036 frac14 095 years

                      Consolidation theory 53

                      74

                      The layer is open

                      d frac14 8

                      2frac14 4m

                      Tv frac14 cvtd2frac14 24 3

                      42frac14 0450

                      ui frac14 frac14 84 kN=m2

                      The excess pore water pressure is given by Equation 721

                      ue frac14Xmfrac141mfrac140

                      2ui

                      Msin

                      Mz

                      d

                      expethM2TvTHORN

                      In this case z frac14 d

                      sinMz

                      d

                      frac14 sinM

                      where

                      M frac14

                      23

                      25

                      2

                      M sin M M2Tv exp (M2Tv)

                      2thorn1 1110 0329

                      3

                      21 9993 457 105

                      ue frac14 2 84 2

                      1 0329 ethother terms negligibleTHORN

                      frac14 352 kN=m2

                      75

                      The layer is open

                      d frac14 6

                      2frac14 3m

                      Tv frac14 cvtd2frac14 10 3

                      32frac14 0333

                      The layer thickness will be divided into six equal parts ie m frac14 6

                      54 Consolidation theory

                      For an open layer

                      Tv frac14 4n

                      m2

                      n frac14 0333 62

                      4frac14 300

                      The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                      ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                      i j

                      0 1 2 3 4 5 6 7 8 9 10 11 12

                      0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                      The initial and 3-year isochrones are plotted in Figure Q75

                      Area under initial isochrone frac14 180 units

                      Area under 3-year isochrone frac14 63 units

                      The average degree of consolidation is given by Equation 725Thus

                      U frac14 1 63

                      180frac14 065

                      Figure Q75

                      Consolidation theory 55

                      76

                      At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                      0 frac14 2w frac14 2 98 frac14 196 kN=m2

                      The final consolidation settlement (one-dimensional method) is

                      sc frac14 mv0H frac14 083 196 8 frac14 130mm

                      Corrected time t frac14 2 1

                      2

                      40

                      52

                      frac14 1615 years

                      Tv frac14 cvtd2frac14 44 1615

                      42frac14 0444

                      From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                      77

                      The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                      Figure Q77

                      56 Consolidation theory

                      Point m n Ir (kNm2) sc (mm)

                      13020frac14 15 20

                      20frac14 10 0194 (4) 113 124

                      260

                      20frac14 30

                      20

                      20frac14 10 0204 (2) 59 65

                      360

                      20frac14 30

                      40

                      20frac14 20 0238 (1) 35 38

                      430

                      20frac14 15

                      40

                      20frac14 20 0224 (2) 65 72

                      Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                      78

                      Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                      (a) Immediate settlement

                      H

                      Bfrac14 30

                      35frac14 086

                      D

                      Bfrac14 2

                      35frac14 006

                      Figure Q78

                      Consolidation theory 57

                      From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                      si frac14 130131qB

                      Eufrac14 10 032 105 35

                      40frac14 30mm

                      (b) Consolidation settlement

                      Layer z (m) Dz Ic (kNm2) syod (mm)

                      1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                      3150

                      Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                      Now

                      H

                      Bfrac14 30

                      35frac14 086 and A frac14 065

                      from Figure 712 13 frac14 079

                      sc frac14 13sod frac14 079 315 frac14 250mm

                      Total settlement

                      s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                      79

                      Without sand drains

                      Uv frac14 025

                      Tv frac14 0049 ethfrom Figure 718THORN

                      t frac14 Tvd2

                      cvfrac14 0049 82

                      cvWith sand drains

                      R frac14 0564S frac14 0564 3 frac14 169m

                      n frac14 Rrfrac14 169

                      015frac14 113

                      Tr frac14 cht

                      4R2frac14 ch

                      4 1692 0049 82

                      cvethand ch frac14 cvTHORN

                      frac14 0275

                      Ur frac14 073 (from Figure 730)

                      58 Consolidation theory

                      Using Equation 740

                      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                      U frac14 080

                      710

                      Without sand drains

                      Uv frac14 090

                      Tv frac14 0848

                      t frac14 Tvd2

                      cvfrac14 0848 102

                      96frac14 88 years

                      With sand drains

                      R frac14 0564S frac14 0564 4 frac14 226m

                      n frac14 Rrfrac14 226

                      015frac14 15

                      Tr

                      Tvfrac14 chcv

                      d2

                      4R2ethsame tTHORN

                      Tr

                      Tvfrac14 140

                      96 102

                      4 2262frac14 714 eth1THORN

                      Using Equation 740

                      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                      An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                      Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                      040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                      Thus

                      Uv frac14 0295 and Ur frac14 086

                      t frac14 88 00683

                      0848frac14 07 years

                      Consolidation theory 59

                      Chapter 8

                      Bearing capacity

                      81

                      (a) The ultimate bearing capacity is given by Equation 83

                      qf frac14 cNc thorn DNq thorn 1

                      2BN

                      For u frac14 0

                      Nc frac14 514 Nq frac14 1 N frac14 0

                      qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                      The net ultimate bearing capacity is

                      qnf frac14 qf D frac14 540 kN=m2

                      The net foundation pressure is

                      qn frac14 q D frac14 425

                      2 eth21 1THORN frac14 192 kN=m2

                      The factor of safety (Equation 86) is

                      F frac14 qnfqnfrac14 540

                      192frac14 28

                      (b) For 0 frac14 28

                      Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                      qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                      2 112 2 13

                      frac14 260thorn 168thorn 146 frac14 574 kN=m2

                      qnf frac14 574 112 frac14 563 kN=m2

                      F frac14 563

                      192frac14 29

                      (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                      82

                      For 0 frac14 38

                      Nq frac14 49 N frac14 67

                      qnf frac14 DethNq 1THORN thorn 1

                      2BN ethfrom Equation 83THORN

                      frac14 eth18 075 48THORN thorn 1

                      2 18 15 67

                      frac14 648thorn 905 frac14 1553 kN=m2

                      qn frac14 500

                      15 eth18 075THORN frac14 320 kN=m2

                      F frac14 qnfqnfrac14 1553

                      320frac14 48

                      0d frac14 tan1tan 38

                      125

                      frac14 32 therefore Nq frac14 23 and N frac14 25

                      Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                      2 18 15 25

                      frac14 15eth310thorn 337THORNfrac14 970 kN=m

                      Design load (action) Vd frac14 500 kN=m

                      The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                      83

                      D

                      Bfrac14 350

                      225frac14 155

                      From Figure 85 for a square foundation

                      Nc frac14 81

                      Bearing capacity 61

                      For a rectangular foundation (L frac14 450m B frac14 225m)

                      Nc frac14 084thorn 016B

                      L

                      81 frac14 745

                      Using Equation 810

                      qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                      For F frac14 3

                      qn frac14 1006

                      3frac14 335 kN=m2

                      q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                      Design load frac14 405 450 225 frac14 4100 kN

                      Design undrained strength cud frac14 135

                      14frac14 96 kN=m2

                      Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                      frac14 7241 kN

                      Design load Vd frac14 4100 kN

                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                      84

                      For 0 frac14 40

                      Nq frac14 64 N frac14 95

                      qnf frac14 DethNq 1THORN thorn 04BN

                      (a) Water table 5m below ground level

                      qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                      qn frac14 400 17 frac14 383 kN=m2

                      F frac14 2686

                      383frac14 70

                      (b) Water table 1m below ground level (ie at foundation level)

                      0 frac14 20 98 frac14 102 kN=m3

                      62 Bearing capacity

                      qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                      F frac14 2040

                      383frac14 53

                      (c) Water table at ground level with upward hydraulic gradient 02

                      eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                      qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                      F frac14 1296

                      392frac14 33

                      85

                      The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                      Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                      Design value of 0 frac14 tan1tan 39

                      125

                      frac14 33

                      For 0 frac14 33 Nq frac14 26 and N frac14 29

                      Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                      86

                      (a) Undrained shear for u frac14 0

                      Nc frac14 514 Nq frac14 1 N frac14 0

                      qnf frac14 12cuNc

                      frac14 12 100 514 frac14 617 kN=m2

                      qn frac14 qnfFfrac14 617

                      3frac14 206 kN=m2

                      q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                      Bearing capacity 63

                      Drained shear for 0 frac14 32

                      Nq frac14 23 N frac14 25

                      0 frac14 21 98 frac14 112 kN=m3

                      qnf frac14 0DethNq 1THORN thorn 040BN

                      frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                      frac14 694 kN=m2

                      q frac14 694

                      3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                      Design load frac14 42 227 frac14 3632 kN

                      (b) Design undrained strength cud frac14 100

                      14frac14 71 kNm2

                      Design bearing resistance Rd frac14 12cudNe area

                      frac14 12 71 514 42

                      frac14 7007 kN

                      For drained shear 0d frac14 tan1tan 32

                      125

                      frac14 26

                      Nq frac14 12 N frac14 10

                      Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                      (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                      Layer z (m) m n Ir 0 (kNm2) sod (mm)

                      1 2 100 0175 0700qn 0182qn

                      2 6 033 0044 0176qn 0046qn

                      3 10 020 0017 0068qn 0018qn

                      0246qn

                      Diameter of equivalent circle B frac14 45m

                      H

                      Bfrac14 12

                      45frac14 27 and A frac14 042

                      13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                      64 Bearing capacity

                      For sc frac14 30mm

                      qn frac14 30

                      0147frac14 204 kN=m2

                      q frac14 204thorn 21 frac14 225 kN=m2

                      Design load frac14 42 225 frac14 3600 kN

                      The design load is 3600 kN settlement being the limiting criterion

                      87

                      D

                      Bfrac14 8

                      4frac14 20

                      From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                      F frac14 cuNc

                      Dfrac14 40 71

                      20 8frac14 18

                      88

                      Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                      Design value of 0 frac14 tan1tan 38

                      125

                      frac14 32

                      Figure Q86

                      Bearing capacity 65

                      For 0 frac14 32 Nq frac14 23 and N frac14 25

                      Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                      The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                      Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                      For B frac14 250m qn frac14 3750

                      2502 17 frac14 583 kN=m2

                      From Figure 510 m frac14 n frac14 126

                      6frac14 021

                      Ir frac14 0019

                      Stress increment frac14 4 0019 583 frac14 44 kN=m2

                      Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                      The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                      89

                      Depth (m) N 0v (kNm2) CN N1

                      070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                      Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                      (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                      Cw frac14 05thorn 0530

                      47

                      frac14 082

                      66 Bearing capacity

                      Thus

                      qa frac14 150 082 frac14 120 kN=m2

                      (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                      Thus

                      qa frac14 90 15 frac14 135 kN=m2

                      (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                      Ic frac14 171

                      1014frac14 0068

                      From Equation 819(a) with s frac14 25mm

                      q frac14 25

                      3507 0068frac14 150 kN=m2

                      810

                      Peak value of strain influence factor occurs at a depth of 27m and is given by

                      Izp frac14 05thorn 01130

                      16 27

                      05

                      frac14 067

                      Refer to Figure Q810

                      E frac14 25qc

                      Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                      Ez (mm3MN)

                      1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                      0203

                      C1 frac14 1 0500qnfrac14 1 05 12 16

                      130frac14 093

                      C2 frac14 1 ethsayTHORN

                      s frac14 C1C2qnX Iz

                      Ez frac14 093 1 130 0203 frac14 25mm

                      Bearing capacity 67

                      811

                      At pile base level

                      cu frac14 220 kN=m2

                      qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                      Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                      00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                      qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                      Then

                      Qf frac14 Abqb thorn Asqs

                      frac14

                      4 32 1980

                      thorn eth 105 139 86THORN

                      frac14 13 996thorn 3941 frac14 17 937 kN

                      0 01 02 03 04 05 06 07

                      0 2 4 6 8 10 12 14

                      1

                      2

                      3

                      4

                      5

                      6

                      7

                      8

                      (1)

                      (2)

                      (3)

                      (4)

                      (5)

                      qc

                      qc

                      Iz

                      Iz

                      (MNm2)

                      z (m)

                      Figure Q810

                      68 Bearing capacity

                      Allowable load

                      ethaTHORN Qf

                      2frac14 17 937

                      2frac14 8968 kN

                      ethbTHORN Abqb

                      3thorn Asqs frac14 13 996

                      3thorn 3941 frac14 8606 kN

                      ie allowable load frac14 8600 kN

                      Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                      According to the limit state method

                      Characteristic undrained strength at base level cuk frac14 220

                      150kN=m2

                      Characteristic base resistance qbk frac14 9cuk frac14 9 220

                      150frac14 1320 kN=m2

                      Characteristic shaft resistance qsk frac14 00150

                      frac14 86

                      150frac14 57 kN=m2

                      Characteristic base and shaft resistances

                      Rbk frac14

                      4 32 1320 frac14 9330 kN

                      Rsk frac14 105 139 86

                      150frac14 2629 kN

                      For a bored pile the partial factors are b frac14 160 and s frac14 130

                      Design bearing resistance Rcd frac14 9330

                      160thorn 2629

                      130

                      frac14 5831thorn 2022

                      frac14 7850 kN

                      Adding ethDAb W) the design bearing resistance becomes 9650 kN

                      812

                      ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                      qs frac14 cu frac14 040 105 frac14 42 kN=m2

                      For a single pile

                      Qf frac14 Abqb thorn Asqs

                      frac14

                      4 062 1305

                      thorn eth 06 15 42THORN

                      frac14 369thorn 1187 frac14 1556 kN

                      Bearing capacity 69

                      Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                      eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                      (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                      (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                      qbkfrac14 9cuk frac14 9 220

                      150frac14 1320 kN=m2

                      qskfrac14cuk frac14 040 105

                      150frac14 28 kN=m2

                      Rbkfrac14

                      4 0602 1320 frac14 373 kN

                      Rskfrac14 060 15 28 frac14 791 kN

                      Rcdfrac14 373

                      160thorn 791

                      130frac14 233thorn 608 frac14 841 kN

                      Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                      Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                      (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                      q frac14 21 000

                      1762frac14 68 kN=m2

                      Immediate settlement

                      H

                      Bfrac14 15

                      176frac14 085

                      D

                      Bfrac14 13

                      176frac14 074

                      L

                      Bfrac14 1

                      Hence from Figure 515

                      130 frac14 078 and 131 frac14 041

                      70 Bearing capacity

                      Thus using Equation 528

                      si frac14 078 041 68 176

                      65frac14 6mm

                      Consolidation settlement

                      Layer z (m) Area (m2) (kNm2) mvH (mm)

                      1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                      434 (sod)

                      Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                      sc frac14 056 434 frac14 24mm

                      The total settlement is (6thorn 24) frac14 30mm

                      813

                      At base level N frac14 26 Then using Equation 830

                      qb frac14 40NDb

                      Bfrac14 40 26 2

                      025frac14 8320 kN=m2

                      ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                      Figure Q812

                      Bearing capacity 71

                      Over the length embedded in sand

                      N frac14 21 ie18thorn 24

                      2

                      Using Equation 831

                      qs frac14 2N frac14 2 21 frac14 42 kN=m2

                      For a single pile

                      Qf frac14 Abqb thorn Asqs

                      frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                      For the pile group assuming a group efficiency of 12

                      XQf frac14 12 9 604 frac14 6523 kN

                      Then the load factor is

                      F frac14 6523

                      2000thorn 1000frac14 21

                      (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                      Characteristic base resistance per unit area qbk frac14 8320

                      150frac14 5547 kNm2

                      Characteristic shaft resistance per unit area qsk frac14 42

                      150frac14 28 kNm2

                      Characteristic base and shaft resistances for a single pile

                      Rbk frac14 0252 5547 frac14 347 kN

                      Rsk frac14 4 025 2 28 frac14 56 kN

                      For a driven pile the partial factors are b frac14 s frac14 130

                      Design bearing resistance Rcd frac14 347

                      130thorn 56

                      130frac14 310 kN

                      For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                      Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                      (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                      From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                      72 Bearing capacity

                      N frac14 24thorn 26thorn 34

                      3frac14 28

                      Ic frac14 171

                      2814frac14 0016 ethEquation 818THORN

                      s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                      814

                      Using Equation 841

                      Tf frac14 DLcu thorn

                      4ethD2 d2THORNcuNc

                      frac14 eth 02 5 06 110THORN thorn

                      4eth022 012THORN110 9

                      frac14 207thorn 23 frac14 230 kN

                      Figure Q813

                      Bearing capacity 73

                      Chapter 9

                      Stability of slopes

                      91

                      Referring to Figure Q91

                      W frac14 417 19 frac14 792 kN=m

                      Q frac14 20 28 frac14 56 kN=m

                      Arc lengthAB frac14

                      180 73 90 frac14 115m

                      Arc length BC frac14

                      180 28 90 frac14 44m

                      The factor of safety is given by

                      F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                      90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                      Depth of tension crack z0 frac14 2cu

                      frac14 2 20

                      19frac14 21m

                      Arc length BD frac14

                      180 13

                      1

                      2 90 frac14 21m

                      F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                      The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                      Design resisting moment frac14 rXethcudLaTHORN frac14 90

                      14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                      Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                      The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                      92

                      u frac14 0

                      Depth factor D frac14 11

                      9frac14 122

                      Using Equation 92 with F frac14 10

                      Ns frac14 cu

                      FHfrac14 30

                      10 19 9frac14 0175

                      Hence from Figure 93

                      frac14 50

                      For F frac14 12

                      Ns frac14 30

                      12 19 9frac14 0146

                      frac14 27

                      93

                      Refer to Figure Q93

                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                      1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                      74 m

                      214 1deg

                      213 1deg

                      39 m

                      WB

                      D

                      C

                      28 m

                      21 m

                      A

                      Q

                      Soil (1)Soil (2)

                      73deg

                      Figure Q91

                      Stability of slopes 75

                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                      9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                      599 256 328 1372

                      Figure Q93

                      76 Stability of slopes

                      XW cos frac14 b

                      Xh cos frac14 21 2 599 frac14 2516 kN=mX

                      W sin frac14 bX

                      h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                      Arc length La frac14

                      180 57

                      1

                      2 326 frac14 327m

                      The factor of safety is given by

                      F frac14 c0La thorn tan0ethW cos ulTHORN

                      W sin

                      frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                      frac14 091

                      According to the limit state method

                      0d frac14 tan1tan 32

                      125

                      frac14 265

                      c0 frac14 8

                      160frac14 5 kN=m2

                      Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                      Design disturbing moment frac14 1075 kN=m

                      The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                      94

                      F frac14 1

                      W sin

                      Xfc0bthorn ethW ubTHORN tan0g sec

                      1thorn ethtan tan0=FTHORN

                      c0 frac14 8 kN=m2

                      0 frac14 32

                      c0b frac14 8 2 frac14 16 kN=m

                      W frac14 bh frac14 21 2 h frac14 42h kN=m

                      Try F frac14 100

                      tan0

                      Ffrac14 0625

                      Stability of slopes 77

                      Values of u are as obtained in Figure Q93

                      SliceNo

                      h(m)

                      W frac14 bh(kNm)

                      W sin(kNm)

                      ub(kNm)

                      c0bthorn (W ub) tan0(kNm)

                      sec

                      1thorn (tan tan0)FProduct(kNm)

                      1 05 21 6 2 8 24 1078 262 13 55 31

                      23 33 30 1042 31

                      3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                      224 92 72 0931 67

                      6 50 210 11 40 100 85 0907 777 55 231 14

                      12 58 112 90 0889 80

                      8 60 252 1812

                      80 114 102 0874 899 63 265 22 99 116 109 0861 94

                      10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                      2154 88 116 0853 99

                      14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                      1074 1091

                      F frac14 1091

                      1074frac14 102 (assumed value 100)

                      Thus

                      F frac14 101

                      95

                      F frac14 1

                      W sin

                      XfWeth1 ruTHORN tan0g sec

                      1thorn ethtan tan0THORN=F

                      0 frac14 33

                      ru frac14 020

                      W frac14 bh frac14 20 5 h frac14 100h kN=m

                      eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                      Try F frac14 110

                      tan 0

                      Ffrac14 tan 33

                      110frac14 0590

                      78 Stability of slopes

                      Referring to Figure Q95

                      SliceNo

                      h(m)

                      W frac14 bh(kNm)

                      W sin(kNm)

                      W(1 ru) tan0(kNm)

                      sec

                      1thorn ( tan tan0)FProduct(kNm)

                      1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                      2120 234 0892 209

                      4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                      1185 1271

                      Figure Q95

                      Stability of slopes 79

                      F frac14 1271

                      1185frac14 107

                      The trial value was 110 therefore take F to be 108

                      96

                      (a) Water table at surface the factor of safety is given by Equation 912

                      F frac14 0

                      sat

                      tan0

                      tan

                      ptie 15 frac14 92

                      19

                      tan 36

                      tan

                      tan frac14 0234

                      frac14 13

                      Water table well below surface the factor of safety is given by Equation 911

                      F frac14 tan0

                      tan

                      frac14 tan 36

                      tan 13

                      frac14 31

                      (b) 0d frac14 tan1tan 36

                      125

                      frac14 30

                      Depth of potential failure surface frac14 z

                      Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                      frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                      frac14 504z kN

                      Design disturbing moment per unit area Sd frac14 sat sin cos

                      frac14 19 z sin 13 cos 13

                      frac14 416z kN

                      Rd gtSd therefore the limit state for overall stability is satisfied

                      80 Stability of slopes

                      • Book Cover
                      • Title
                      • Contents
                      • Basic characteristics of soils
                      • Seepage
                      • Effective stress
                      • Shear strength
                      • Stresses and displacements
                      • Lateral earth pressure
                      • Consolidation theory
                      • Bearing capacity
                      • Stability of slopes

                        Chapter 2

                        Seepage

                        21

                        The coefficient of permeability is determined from the equation

                        k frac14 23al

                        At1log

                        h0

                        h1

                        where

                        a frac14

                        4 00052 m2 l frac14 02m

                        A frac14

                        4 012 m2 t1 frac14 3 602 s

                        logh0

                        h1frac14 log

                        100

                        035frac14 0456

                        k frac14 23 00052 02 0456

                        012 3 602frac14 49 108 m=s

                        22

                        The flow net is drawn in Figure Q22 In the flow net there are 37 flow channels and 11equipotential drops ie Nffrac14 37 and Ndfrac14 11 The overall loss in total head is 400mThe quantity of seepage is calculated by using Equation 216

                        q frac14 kh Nf

                        Ndfrac14 106 400 37

                        11frac14 13 106 m3=s per m

                        Figure Q22

                        23

                        The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

                        q frac14 kh Nf

                        Ndfrac14 5 105 300 35

                        9frac14 58 105 m3=s per m

                        The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

                        Point h (m) h z (m) u frac14 w(h z)(kNm2)

                        1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

                        eg for Point 1

                        h1 frac14 7

                        9 300 frac14 233m

                        h1 z1 frac14 233 eth250THORN frac14 483m

                        Figure Q23

                        Seepage 7

                        u1 frac14 98 483 frac14 47 kN=m2

                        The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

                        24

                        The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

                        q frac14 kh Nf

                        Ndfrac14 40 107 550 10

                        11frac14 20 106 m3=s per m

                        25

                        The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

                        q frac14 kh Nf

                        Ndfrac14 20 106 500 42

                        9frac14 47 106 m3=s per m

                        Figure Q24

                        8 Seepage

                        26

                        The scale transformation factor in the x direction is given by Equation 221 ie

                        xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                        ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

                        Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

                        k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                        pfrac14

                        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

                        p 107 frac14 30 107 m=s

                        Thus the quantity of seepage is given by

                        q frac14 k0h Nf

                        Ndfrac14 30 107 1400 325

                        12frac14 11 106 m3=s per m

                        Figure Q25

                        Seepage 9

                        27

                        The scale transformation factor in the x direction is

                        xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                        ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

                        Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

                        k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                        pfrac14

                        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

                        p 106 frac14 45 106 m=s

                        The focus of the basic parabola is at point A The parabola passes through point Gsuch that

                        GC frac14 03HC frac14 03 30 frac14 90m

                        Thus the coordinates of G are

                        x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

                        480 frac14 x0 2002

                        4x0

                        Figure Q26

                        10 Seepage

                        Hence

                        x0 frac14 20m

                        Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

                        x frac14 20 z2

                        80

                        x 20 0 50 100 200 300z 0 400 748 980 1327 1600

                        The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

                        No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

                        q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

                        28

                        The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

                        Figure Q27

                        Seepage 11

                        q frac14 kh Nf

                        Ndfrac14 45 105 28 33

                        7

                        frac14 59 105 m3=s per m

                        29

                        The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

                        kx frac14 H1k1 thornH2k2

                        H1 thornH2frac14 106

                        10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

                        kz frac14 H1 thornH2

                        H1

                        k1thornH2

                        k2

                        frac14 10

                        5

                        eth2 106THORN thorn5

                        eth16 106THORNfrac14 36 106 m=s

                        Then the scale transformation factor is given by

                        xt frac14 xffiffiffiffiffikz

                        pffiffiffiffiffikx

                        p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

                        Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

                        Figure Q28

                        12 Seepage

                        k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                        qfrac14

                        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

                        p 106 frac14 57 106 m=s

                        Then the quantity of seepage is given by

                        q frac14 k0h Nf

                        Ndfrac14 57 106 350 56

                        11

                        frac14 10 105 m3=s per m

                        Figure Q29

                        Seepage 13

                        Chapter 3

                        Effective stress

                        31

                        Buoyant unit weight

                        0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                        Effective vertical stress

                        0v frac14 5 102 frac14 51 kN=m2 or

                        Total vertical stress

                        v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

                        Pore water pressure

                        u frac14 7 98 frac14 686 kN=m2

                        Effective vertical stress

                        0v frac14 v u frac14 1196 686 frac14 51 kN=m2

                        32

                        Buoyant unit weight

                        0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                        Effective vertical stress

                        0v frac14 5 102 frac14 51 kN=m2 or

                        Total vertical stress

                        v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

                        Pore water pressure

                        u frac14 205 98 frac14 2009 kN=m2

                        Effective vertical stress

                        0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

                        33

                        At top of the clay

                        v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

                        u frac14 2 98 frac14 196 kN=m2

                        0v frac14 v u frac14 710 196 frac14 514 kN=m2

                        Alternatively

                        0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                        0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

                        At bottom of the clay

                        v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

                        u frac14 12 98 frac14 1176 kN=m2

                        0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

                        NB The alternative method of calculation is not applicable because of the artesiancondition

                        Figure Q3132

                        Effective stress 15

                        34

                        0 frac14 20 98 frac14 102 kN=m3

                        At 8m depth

                        0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                        35

                        0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                        0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                        Figure Q33

                        Figure Q34

                        16 Effective stress

                        (a) Immediately after WT rise

                        At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                        0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                        At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                        0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                        (b) Several years after WT rise

                        At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                        At 8m depth

                        0v frac14 940 kN=m2 (as above)

                        At 12m depth

                        0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                        Figure Q35

                        Effective stress 17

                        36

                        Total weight

                        ab frac14 210 kN

                        Effective weight

                        ac frac14 112 kN

                        Resultant boundary water force

                        be frac14 119 kN

                        Seepage force

                        ce frac14 34 kN

                        Resultant body force

                        ae frac14 99 kN eth73 to horizontalTHORN

                        (Refer to Figure Q36)

                        Figure Q36

                        18 Effective stress

                        37

                        Situation (1)(a)

                        frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                        u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                        0 frac14 u frac14 694 392 frac14 302 kN=m2

                        (b)

                        i frac14 2

                        4frac14 05

                        j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                        Situation (2)(a)

                        frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                        u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                        0 frac14 u frac14 498 392 frac14 106 kN=m2

                        (b)

                        i frac14 2

                        4frac14 05

                        j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                        38

                        The flow net is drawn in Figure Q24

                        Loss in total head between adjacent equipotentials

                        h frac14 550

                        Ndfrac14 550

                        11frac14 050m

                        Exit hydraulic gradient

                        ie frac14 h

                        sfrac14 050

                        070frac14 071

                        Effective stress 19

                        The critical hydraulic gradient is given by Equation 39

                        ic frac14 0

                        wfrac14 102

                        98frac14 104

                        Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                        F frac14 iciefrac14 104

                        071frac14 15

                        Total head at C

                        hC frac14 nd

                        Ndh frac14 24

                        11 550 frac14 120m

                        Elevation head at C

                        zC frac14 250m

                        Pore water pressure at C

                        uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                        Therefore effective vertical stress at C

                        0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                        For point D

                        hD frac14 73

                        11 550 frac14 365m

                        zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                        0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                        39

                        The flow net is drawn in Figure Q25

                        For a soil prism 150 300m adjacent to the piling

                        hm frac14 26

                        9 500 frac14 145m

                        20 Effective stress

                        Factor of safety against lsquoheavingrsquo (Equation 310)

                        F frac14 ic

                        imfrac14 0d

                        whmfrac14 97 300

                        98 145frac14 20

                        With a filter

                        F frac14 0d thorn wwhm

                        3 frac14 eth97 300THORN thorn w98 145

                        w frac14 135 kN=m2

                        Depth of filterfrac14 13521frac14 065m (if above water level)

                        Effective stress 21

                        Chapter 4

                        Shear strength

                        41

                        frac14 295 kN=m2

                        u frac14 120 kN=m2

                        0 frac14 u frac14 295 120 frac14 175 kN=m2

                        f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                        42

                        03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                        100 452 552200 908 1108400 1810 2210800 3624 4424

                        The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                        Figure Q42

                        43

                        3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                        200 222 422400 218 618600 220 820

                        The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                        44

                        The modified shear strength parameters are

                        0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                        a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                        The coordinates of the stress point representing failure conditions in the test are

                        1

                        2eth1 2THORN frac14 1

                        2 170 frac14 85 kN=m2

                        1

                        2eth1 thorn 3THORN frac14 1

                        2eth270thorn 100THORN frac14 185 kN=m2

                        The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                        uf frac14 36 kN=m2

                        Figure Q43

                        Figure Q44

                        Shear strength 23

                        45

                        3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                        150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                        The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                        The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                        46

                        03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                        200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                        The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                        Figure Q45

                        24 Shear strength

                        47

                        The torque required to produce shear failure is given by

                        T frac14 dh cud

                        2thorn 2

                        Z d=2

                        0

                        2r drcur

                        frac14 cud2h

                        2thorn 4cu

                        Z d=2

                        0

                        r2dr

                        frac14 cud2h

                        2thorn d

                        3

                        6

                        Then

                        35 frac14 cu52 10

                        2thorn 53

                        6

                        103

                        cu frac14 76 kN=m3

                        400

                        0 400 800 1200 1600

                        τ (k

                        Nm

                        2 )

                        σprime (kNm2)

                        34deg

                        315deg29deg

                        (a)

                        (b)

                        0 400

                        400

                        800 1200 1600

                        Failure envelope

                        300 500

                        σprime (kNm2)

                        τ (k

                        Nm

                        2 )

                        20 (kNm2)

                        31deg

                        Figure Q46

                        Shear strength 25

                        48

                        The relevant stress values are calculated as follows

                        3 frac14 600 kN=m2

                        1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                        2(1 3) 0 40 79 107 139 159

                        1

                        2(01 thorn 03) 400 411 402 389 351 326

                        1

                        2(1 thorn 3) 600 640 679 707 739 759

                        The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                        Af frac14 433 200

                        319frac14 073

                        49

                        B frac14 u33

                        frac14 144

                        350 200frac14 096

                        a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                        0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                        10 283 65 023

                        Figure Q48

                        26 Shear strength

                        The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                        Figure Q49

                        Shear strength 27

                        Chapter 5

                        Stresses and displacements

                        51

                        Vertical stress is given by

                        z frac14 Qz2Ip frac14 5000

                        52Ip

                        Values of Ip are obtained from Table 51

                        r (m) rz Ip z (kNm2)

                        0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                        10 20 0009 2

                        The variation of z with radial distance (r) is plotted in Figure Q51

                        Figure Q51

                        52

                        Below the centre load (Figure Q52)

                        r

                        zfrac14 0 for the 7500-kN load

                        Ip frac14 0478

                        r

                        zfrac14 5

                        4frac14 125 for the 10 000- and 9000-kN loads

                        Ip frac14 0045

                        Then

                        z frac14X Q

                        z2Ip

                        frac14 7500 0478

                        42thorn 10 000 0045

                        42thorn 9000 0045

                        42

                        frac14 224thorn 28thorn 25 frac14 277 kN=m2

                        53

                        The vertical stress under a corner of a rectangular area is given by

                        z frac14 qIr

                        where values of Ir are obtained from Figure 510 In this case

                        z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                        z

                        Figure Q52

                        Stresses and displacements 29

                        z (m) m n Ir z (kNm2)

                        0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                        10 010 0005 5

                        z is plotted against z in Figure Q53

                        54

                        (a)

                        m frac14 125

                        12frac14 104

                        n frac14 18

                        12frac14 150

                        From Figure 510 Irfrac14 0196

                        z frac14 2 175 0196 frac14 68 kN=m2

                        Figure Q53

                        30 Stresses and displacements

                        (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                        z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                        55

                        Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                        Px frac14 2Q

                        1

                        m2 thorn 1frac14 2 150

                        125frac14 76 kN=m

                        Equation 517 is used to obtain the pressure distribution

                        px frac14 4Q

                        h

                        m2n

                        ethm2 thorn n2THORN2 frac14150

                        m2n

                        ethm2 thorn n2THORN2 ethkN=m2THORN

                        Figure Q54

                        Stresses and displacements 31

                        n m2n

                        (m2 thorn n2)2

                        px(kNm2)

                        0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                        The pressure distribution is plotted in Figure Q55

                        56

                        H

                        Bfrac14 10

                        2frac14 5

                        L

                        Bfrac14 4

                        2frac14 2

                        D

                        Bfrac14 1

                        2frac14 05

                        Hence from Figure 515

                        131 frac14 082

                        130 frac14 094

                        Figure Q55

                        32 Stresses and displacements

                        The immediate settlement is given by Equation 528

                        si frac14 130131qB

                        Eu

                        frac14 094 082 200 2

                        45frac14 7mm

                        Stresses and displacements 33

                        Chapter 6

                        Lateral earth pressure

                        61

                        For 0 frac14 37 the active pressure coefficient is given by

                        Ka frac14 1 sin 37

                        1thorn sin 37frac14 025

                        The total active thrust (Equation 66a with c0 frac14 0) is

                        Pa frac14 1

                        2KaH

                        2 frac14 1

                        2 025 17 62 frac14 765 kN=m

                        If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                        K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                        and the thrust on the wall is

                        P0 frac14 1

                        2K0H

                        2 frac14 1

                        2 040 17 62 frac14 122 kN=m

                        62

                        The active pressure coefficients for the three soil types are as follows

                        Ka1 frac141 sin 35

                        1thorn sin 35frac14 0271

                        Ka2 frac141 sin 27

                        1thorn sin 27frac14 0375

                        ffiffiffiffiffiffiffiKa2

                        p frac14 0613

                        Ka3 frac141 sin 42

                        1thorn sin 42frac14 0198

                        Distribution of active pressure (plotted in Figure Q62)

                        Depth (m) Soil Active pressure (kNm2)

                        3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                        12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                        At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                        Total thrust frac14 571 kNm

                        Point of application is (4893571) m from the top of the wall ie 857m

                        Force (kN) Arm (m) Moment (kN m)

                        (1)1

                        2 0271 16 32 frac14 195 20 390

                        (2) 0271 16 3 2 frac14 260 40 1040

                        (3)1

                        2 0271 92 22 frac14 50 433 217

                        (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                        (5)1

                        2 0375 102 32 frac14 172 70 1204

                        (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                        (7)1

                        2 0198 112 42 frac14 177 1067 1889

                        (8)1

                        2 98 92 frac14 3969 90 35721

                        5713 48934

                        Figure Q62

                        Lateral earth pressure 35

                        63

                        (a) For u frac14 0 Ka frac14 Kp frac14 1

                        Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                        frac14 245

                        At the lower end of the piling

                        pa frac14 Kaqthorn Kasatz Kaccu

                        frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                        frac14 115 kN=m2

                        pp frac14 Kpsatzthorn Kpccu

                        frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                        frac14 202 kN=m2

                        (b) For 0 frac14 26 and frac14 1

                        20

                        Ka frac14 035

                        Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                        pfrac14 145 ethEquation 619THORN

                        Kp frac14 37

                        Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                        pfrac14 47 ethEquation 624THORN

                        At the lower end of the piling

                        pa frac14 Kaqthorn Ka0z Kacc

                        0

                        frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                        frac14 187 kN=m2

                        pp frac14 Kp0zthorn Kpcc

                        0

                        frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                        frac14 198 kN=m2

                        36 Lateral earth pressure

                        64

                        (a) For 0 frac14 38 Ka frac14 024

                        0 frac14 20 98 frac14 102 kN=m3

                        The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                        Force (kN) Arm (m) Moment (kN m)

                        (1) 024 10 66 frac14 159 33 525

                        (2)1

                        2 024 17 392 frac14 310 400 1240

                        (3) 024 17 39 27 frac14 430 135 580

                        (4)1

                        2 024 102 272 frac14 89 090 80

                        (5)1

                        2 98 272 frac14 357 090 321

                        Hfrac14 1345 MH frac14 2746

                        (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                        (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                        XM frac14MV MH frac14 7790 kNm

                        Lever arm of base resultant

                        M

                        Vfrac14 779

                        488frac14 160

                        Eccentricity of base resultant

                        e frac14 200 160 frac14 040m

                        39 m

                        27 m

                        40 m

                        04 m

                        04 m

                        26 m

                        (7)

                        (9)

                        (1)(2)

                        (3)

                        (4)

                        (5)

                        (8)(6)

                        (10)

                        WT

                        10 kNm2

                        Hydrostatic

                        Figure Q64

                        Lateral earth pressure 37

                        Base pressures (Equation 627)

                        p frac14 VB

                        1 6e

                        B

                        frac14 488

                        4eth1 060THORN

                        frac14 195 kN=m2 and 49 kN=m2

                        Factor of safety against sliding (Equation 628)

                        F frac14 V tan

                        Hfrac14 488 tan 25

                        1345frac14 17

                        (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                        Hfrac14 1633 kN

                        V frac14 4879 kN

                        MH frac14 3453 kNm

                        MV frac14 10536 kNm

                        The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                        65

                        For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                        Kp

                        Ffrac14 385

                        2

                        0 frac14 20 98 frac14 102 kN=m3

                        The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                        Force (kN) Arm (m) Moment (kN m)

                        (1)1

                        2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                        (2) 026 17 45 d frac14 199d d2 995d2

                        (3)1

                        2 026 102 d2 frac14 133d2 d3 044d3

                        (4)1

                        2 385

                        2 17 152 frac14 368 dthorn 05 368d 184

                        (5)385

                        2 17 15 d frac14 491d d2 2455d2

                        (6)1

                        2 385

                        2 102 d2 frac14 982d2 d3 327d3

                        38 Lateral earth pressure

                        XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                        d3 thorn 516d2 283d 1724 frac14 0

                        d frac14 179m

                        Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                        Over additional 20 embedded depth

                        pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                        Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                        66

                        The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                        Ka frac14 sin 69=sin 105

                        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                        pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                        26664

                        37775

                        2

                        frac14 050

                        The total active thrust (acting at 25 above the normal) is given by Equation 616

                        Pa frac14 1

                        2 050 19 7502 frac14 267 kN=m

                        Figure Q65

                        Lateral earth pressure 39

                        Horizontal component

                        Ph frac14 267 cos 40 frac14 205 kN=m

                        Vertical component

                        Pv frac14 267 sin 40 frac14 172 kN=m

                        Consider moments about the toe of the wall (Figure Q66) (per m)

                        Force (kN) Arm (m) Moment (kN m)

                        (1)1

                        2 175 650 235 frac14 1337 258 345

                        (2) 050 650 235 frac14 764 175 134

                        (3)1

                        2 070 650 235 frac14 535 127 68

                        (4) 100 400 235 frac14 940 200 188

                        (5) 1

                        2 080 050 235 frac14 47 027 1

                        Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                        Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                        Lever arm of base resultant

                        M

                        Vfrac14 795

                        525frac14 151m

                        Eccentricity of base resultant

                        e frac14 200 151 frac14 049m

                        Figure Q66

                        40 Lateral earth pressure

                        Base pressures (Equation 627)

                        p frac14 525

                        41 6 049

                        4

                        frac14 228 kN=m2 and 35 kN=m2

                        The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                        The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                        The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                        67

                        For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                        Force (kN) Arm (m) Moment (kNm)

                        (1)1

                        2 027 17 52 frac14 574 183 1050

                        (2) 027 17 5 3 frac14 689 500 3445

                        (3)1

                        2 027 102 32 frac14 124 550 682

                        (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                        (5)1

                        2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                        (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                        (7) 1

                        2 267

                        2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                        (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                        2 d frac14 163d d2thorn 650 82d2 1060d

                        Tie rod force per m frac14 T 0 0

                        XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                        d3 thorn 77d2 269d 1438 frac14 0

                        d frac14 467m

                        Depth of penetration frac14 12d frac14 560m

                        Lateral earth pressure 41

                        Algebraic sum of forces for d frac14 467m isX

                        F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                        T frac14 905 kN=m

                        Force in each tie rod frac14 25T frac14 226 kN

                        68

                        (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                        0 frac14 21 98 frac14 112 kN=m3

                        The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                        uC frac14 150

                        165 15 98 frac14 134 kN=m2

                        The average seepage pressure is

                        j frac14 15

                        165 98 frac14 09 kN=m3

                        Hence

                        0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                        0 j frac14 112 09 frac14 103 kN=m3

                        Figure Q67

                        42 Lateral earth pressure

                        Consider moments about the anchor point A (per m)

                        Force (kN) Arm (m) Moment (kN m)

                        (1) 10 026 150 frac14 390 60 2340

                        (2)1

                        2 026 18 452 frac14 474 15 711

                        (3) 026 18 45 105 frac14 2211 825 18240

                        (4)1

                        2 026 121 1052 frac14 1734 100 17340

                        (5)1

                        2 134 15 frac14 101 40 404

                        (6) 134 30 frac14 402 60 2412

                        (7)1

                        2 134 60 frac14 402 95 3819

                        571 4527(8) Ppm

                        115 115PPm

                        XM frac14 0

                        Ppm frac144527

                        115frac14 394 kN=m

                        Available passive resistance

                        Pp frac14 1

                        2 385 103 62 frac14 714 kN=m

                        Factor of safety

                        Fp frac14 Pp

                        Ppm

                        frac14 714

                        394frac14 18

                        Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                        Figure Q68

                        Lateral earth pressure 43

                        (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                        Consider moments (per m) about the tie point A

                        Force (kN) Arm (m)

                        (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                        (2)1

                        2 033 18 452 frac14 601 15

                        (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                        (4)1

                        2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                        (5)1

                        2 134 15 frac14 101 40

                        (6) 134 30 frac14 402 60

                        (7)1

                        2 134 d frac14 67d d3thorn 75

                        (8) 1

                        2 30 103 d2 frac141545d2 2d3thorn 75

                        Moment (kN m)

                        (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                        XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                        d3 thorn 827d2 466d 1518 frac14 0

                        By trial

                        d frac14 544m

                        The minimum depth of embedment required is 544m

                        69

                        For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                        0 frac14 20 98 frac14 102 kN=m3

                        The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                        44 Lateral earth pressure

                        uC frac14 147

                        173 26 98 frac14 216 kN=m2

                        and the average seepage pressure around the wall is

                        j frac14 26

                        173 98 frac14 15 kN=m3

                        Consider moments about the prop (A) (per m)

                        Force (kN) Arm (m) Moment (kN m)

                        (1)1

                        2 03 17 272 frac14 186 020 37

                        (2) 03 17 27 53 frac14 730 335 2445

                        (3)1

                        2 03 (102thorn 15) 532 frac14 493 423 2085

                        (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                        (5)1

                        2 216 26 frac14 281 243 684

                        (6) 216 27 frac14 583 465 2712

                        (7)1

                        2 216 60 frac14 648 800 5184

                        3055(8)

                        1

                        2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                        Factor of safety

                        Fr frac14 6885

                        3055frac14 225

                        Figure Q69

                        Lateral earth pressure 45

                        610

                        For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                        p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                        Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                        Using the recommendations of Twine and Roscoe

                        p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                        Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                        611

                        frac14 18 kN=m3 0 frac14 34

                        H frac14 350m nH frac14 335m mH frac14 185m

                        Consider a trial value of F frac14 20 Refer to Figure 635

                        0m frac14 tan1tan 34

                        20

                        frac14 186

                        Then

                        frac14 45 thorn 0m2frac14 543

                        W frac14 1

                        2 18 3502 cot 543 frac14 792 kN=m

                        Figure Q610

                        46 Lateral earth pressure

                        P frac14 1

                        2 s 3352 frac14 561s kN=m

                        U frac14 1

                        2 98 1852 cosec 543 frac14 206 kN=m

                        Equations 630 and 631 then become

                        561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                        792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                        ie

                        561s 0616N 405 frac14 0

                        792 0857N thorn 563 frac14 0

                        N frac14 848

                        0857frac14 989 kN=m

                        Then

                        561s 609 405 frac14 0

                        s frac14 649

                        561frac14 116 kN=m3

                        The calculations for trial values of F of 20 15 and 10 are summarized below

                        F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                        20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                        s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                        Figure Q611

                        Lateral earth pressure 47

                        612

                        For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                        45 thorn 0

                        2frac14 63

                        For the retained material between the surface and a depth of 36m

                        Pa frac14 1

                        2 030 18 362 frac14 350 kN=m

                        Weight of reinforced fill between the surface and a depth of 36m is

                        Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                        eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                        Eccentricity of Rv

                        e frac14 263 250 frac14 013m

                        The average vertical stress at a depth of 36m is

                        z frac14 Rv

                        L 2efrac14 324

                        474frac14 68 kN=m2

                        (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                        Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                        Tensile stress in the element frac14 138 103

                        65 3frac14 71N=mm2

                        Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                        Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                        Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                        The weight of ABC is

                        W frac14 1

                        2 18 52 265 frac14 124 kN=m

                        From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                        48 Lateral earth pressure

                        (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                        Tp frac14 032 68 120 065 frac14 170 kN

                        Tr frac14 213 420

                        418frac14 214 kN

                        Again the tensile failure and slipping limit states are satisfied for this element

                        Figure Q612

                        Lateral earth pressure 49

                        Chapter 7

                        Consolidation theory

                        71

                        Total change in thickness

                        H frac14 782 602 frac14 180mm

                        Average thickness frac14 1530thorn 180

                        2frac14 1620mm

                        Length of drainage path d frac14 1620

                        2frac14 810mm

                        Root time plot (Figure Q71a)

                        ffiffiffiffiffiffit90p frac14 33

                        t90 frac14 109min

                        cv frac14 0848d2

                        t90frac14 0848 8102

                        109 1440 365

                        106frac14 27m2=year

                        r0 frac14 782 764

                        782 602frac14 018

                        180frac14 0100

                        rp frac14 10eth764 645THORN9eth782 602THORN frac14

                        10 119

                        9 180frac14 0735

                        rs frac14 1 eth0100thorn 0735THORN frac14 0165

                        Log time plot (Figure Q71b)

                        t50 frac14 26min

                        cv frac14 0196d2

                        t50frac14 0196 8102

                        26 1440 365

                        106frac14 26m2=year

                        r0 frac14 782 763

                        782 602frac14 019

                        180frac14 0106

                        rp frac14 763 623

                        782 602frac14 140

                        180frac14 0778

                        rs frac14 1 eth0106thorn 0778THORN frac14 0116

                        Figure Q71(a)

                        Figure Q71(b)

                        Final void ratio

                        e1 frac14 w1Gs frac14 0232 272 frac14 0631

                        e

                        Hfrac14 1thorn e0

                        H0frac14 1thorn e1 thorne

                        H0

                        ie

                        e

                        180frac14 1631thorne

                        1710

                        e frac14 2936

                        1530frac14 0192

                        Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                        mv frac14 1

                        1thorn e0 e0 e101 00

                        frac14 1

                        1823 0192

                        0107frac14 098m2=MN

                        k frac14 cvmvw frac14 265 098 98

                        60 1440 365 103frac14 81 1010 m=s

                        72

                        Using Equation 77 (one-dimensional method)

                        sc frac14 e0 e11thorn e0 H

                        Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                        Figure Q72

                        52 Consolidation theory

                        Settlement

                        Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                        1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                        318

                        Notes 5 92y 460thorn 84

                        Heave

                        Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                        1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                        38

                        73

                        U frac14 f ethTvTHORN frac14 f cvt

                        d2

                        Hence if cv is constant

                        t1

                        t2frac14 d

                        21

                        d22

                        where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                        d1 frac14 95mm and d2 frac14 2500mm

                        for U frac14 050 t2 frac14 t1 d22

                        d21

                        frac14 20

                        60 24 365 25002

                        952frac14 263 years

                        for U lt 060 Tv frac14

                        4U2 (Equation 724(a))

                        t030 frac14 t050 0302

                        0502

                        frac14 263 036 frac14 095 years

                        Consolidation theory 53

                        74

                        The layer is open

                        d frac14 8

                        2frac14 4m

                        Tv frac14 cvtd2frac14 24 3

                        42frac14 0450

                        ui frac14 frac14 84 kN=m2

                        The excess pore water pressure is given by Equation 721

                        ue frac14Xmfrac141mfrac140

                        2ui

                        Msin

                        Mz

                        d

                        expethM2TvTHORN

                        In this case z frac14 d

                        sinMz

                        d

                        frac14 sinM

                        where

                        M frac14

                        23

                        25

                        2

                        M sin M M2Tv exp (M2Tv)

                        2thorn1 1110 0329

                        3

                        21 9993 457 105

                        ue frac14 2 84 2

                        1 0329 ethother terms negligibleTHORN

                        frac14 352 kN=m2

                        75

                        The layer is open

                        d frac14 6

                        2frac14 3m

                        Tv frac14 cvtd2frac14 10 3

                        32frac14 0333

                        The layer thickness will be divided into six equal parts ie m frac14 6

                        54 Consolidation theory

                        For an open layer

                        Tv frac14 4n

                        m2

                        n frac14 0333 62

                        4frac14 300

                        The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                        ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                        i j

                        0 1 2 3 4 5 6 7 8 9 10 11 12

                        0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                        The initial and 3-year isochrones are plotted in Figure Q75

                        Area under initial isochrone frac14 180 units

                        Area under 3-year isochrone frac14 63 units

                        The average degree of consolidation is given by Equation 725Thus

                        U frac14 1 63

                        180frac14 065

                        Figure Q75

                        Consolidation theory 55

                        76

                        At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                        0 frac14 2w frac14 2 98 frac14 196 kN=m2

                        The final consolidation settlement (one-dimensional method) is

                        sc frac14 mv0H frac14 083 196 8 frac14 130mm

                        Corrected time t frac14 2 1

                        2

                        40

                        52

                        frac14 1615 years

                        Tv frac14 cvtd2frac14 44 1615

                        42frac14 0444

                        From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                        77

                        The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                        Figure Q77

                        56 Consolidation theory

                        Point m n Ir (kNm2) sc (mm)

                        13020frac14 15 20

                        20frac14 10 0194 (4) 113 124

                        260

                        20frac14 30

                        20

                        20frac14 10 0204 (2) 59 65

                        360

                        20frac14 30

                        40

                        20frac14 20 0238 (1) 35 38

                        430

                        20frac14 15

                        40

                        20frac14 20 0224 (2) 65 72

                        Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                        78

                        Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                        (a) Immediate settlement

                        H

                        Bfrac14 30

                        35frac14 086

                        D

                        Bfrac14 2

                        35frac14 006

                        Figure Q78

                        Consolidation theory 57

                        From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                        si frac14 130131qB

                        Eufrac14 10 032 105 35

                        40frac14 30mm

                        (b) Consolidation settlement

                        Layer z (m) Dz Ic (kNm2) syod (mm)

                        1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                        3150

                        Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                        Now

                        H

                        Bfrac14 30

                        35frac14 086 and A frac14 065

                        from Figure 712 13 frac14 079

                        sc frac14 13sod frac14 079 315 frac14 250mm

                        Total settlement

                        s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                        79

                        Without sand drains

                        Uv frac14 025

                        Tv frac14 0049 ethfrom Figure 718THORN

                        t frac14 Tvd2

                        cvfrac14 0049 82

                        cvWith sand drains

                        R frac14 0564S frac14 0564 3 frac14 169m

                        n frac14 Rrfrac14 169

                        015frac14 113

                        Tr frac14 cht

                        4R2frac14 ch

                        4 1692 0049 82

                        cvethand ch frac14 cvTHORN

                        frac14 0275

                        Ur frac14 073 (from Figure 730)

                        58 Consolidation theory

                        Using Equation 740

                        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                        U frac14 080

                        710

                        Without sand drains

                        Uv frac14 090

                        Tv frac14 0848

                        t frac14 Tvd2

                        cvfrac14 0848 102

                        96frac14 88 years

                        With sand drains

                        R frac14 0564S frac14 0564 4 frac14 226m

                        n frac14 Rrfrac14 226

                        015frac14 15

                        Tr

                        Tvfrac14 chcv

                        d2

                        4R2ethsame tTHORN

                        Tr

                        Tvfrac14 140

                        96 102

                        4 2262frac14 714 eth1THORN

                        Using Equation 740

                        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                        An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                        Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                        040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                        Thus

                        Uv frac14 0295 and Ur frac14 086

                        t frac14 88 00683

                        0848frac14 07 years

                        Consolidation theory 59

                        Chapter 8

                        Bearing capacity

                        81

                        (a) The ultimate bearing capacity is given by Equation 83

                        qf frac14 cNc thorn DNq thorn 1

                        2BN

                        For u frac14 0

                        Nc frac14 514 Nq frac14 1 N frac14 0

                        qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                        The net ultimate bearing capacity is

                        qnf frac14 qf D frac14 540 kN=m2

                        The net foundation pressure is

                        qn frac14 q D frac14 425

                        2 eth21 1THORN frac14 192 kN=m2

                        The factor of safety (Equation 86) is

                        F frac14 qnfqnfrac14 540

                        192frac14 28

                        (b) For 0 frac14 28

                        Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                        qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                        2 112 2 13

                        frac14 260thorn 168thorn 146 frac14 574 kN=m2

                        qnf frac14 574 112 frac14 563 kN=m2

                        F frac14 563

                        192frac14 29

                        (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                        82

                        For 0 frac14 38

                        Nq frac14 49 N frac14 67

                        qnf frac14 DethNq 1THORN thorn 1

                        2BN ethfrom Equation 83THORN

                        frac14 eth18 075 48THORN thorn 1

                        2 18 15 67

                        frac14 648thorn 905 frac14 1553 kN=m2

                        qn frac14 500

                        15 eth18 075THORN frac14 320 kN=m2

                        F frac14 qnfqnfrac14 1553

                        320frac14 48

                        0d frac14 tan1tan 38

                        125

                        frac14 32 therefore Nq frac14 23 and N frac14 25

                        Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                        2 18 15 25

                        frac14 15eth310thorn 337THORNfrac14 970 kN=m

                        Design load (action) Vd frac14 500 kN=m

                        The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                        83

                        D

                        Bfrac14 350

                        225frac14 155

                        From Figure 85 for a square foundation

                        Nc frac14 81

                        Bearing capacity 61

                        For a rectangular foundation (L frac14 450m B frac14 225m)

                        Nc frac14 084thorn 016B

                        L

                        81 frac14 745

                        Using Equation 810

                        qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                        For F frac14 3

                        qn frac14 1006

                        3frac14 335 kN=m2

                        q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                        Design load frac14 405 450 225 frac14 4100 kN

                        Design undrained strength cud frac14 135

                        14frac14 96 kN=m2

                        Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                        frac14 7241 kN

                        Design load Vd frac14 4100 kN

                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                        84

                        For 0 frac14 40

                        Nq frac14 64 N frac14 95

                        qnf frac14 DethNq 1THORN thorn 04BN

                        (a) Water table 5m below ground level

                        qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                        qn frac14 400 17 frac14 383 kN=m2

                        F frac14 2686

                        383frac14 70

                        (b) Water table 1m below ground level (ie at foundation level)

                        0 frac14 20 98 frac14 102 kN=m3

                        62 Bearing capacity

                        qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                        F frac14 2040

                        383frac14 53

                        (c) Water table at ground level with upward hydraulic gradient 02

                        eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                        qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                        F frac14 1296

                        392frac14 33

                        85

                        The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                        Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                        Design value of 0 frac14 tan1tan 39

                        125

                        frac14 33

                        For 0 frac14 33 Nq frac14 26 and N frac14 29

                        Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                        86

                        (a) Undrained shear for u frac14 0

                        Nc frac14 514 Nq frac14 1 N frac14 0

                        qnf frac14 12cuNc

                        frac14 12 100 514 frac14 617 kN=m2

                        qn frac14 qnfFfrac14 617

                        3frac14 206 kN=m2

                        q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                        Bearing capacity 63

                        Drained shear for 0 frac14 32

                        Nq frac14 23 N frac14 25

                        0 frac14 21 98 frac14 112 kN=m3

                        qnf frac14 0DethNq 1THORN thorn 040BN

                        frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                        frac14 694 kN=m2

                        q frac14 694

                        3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                        Design load frac14 42 227 frac14 3632 kN

                        (b) Design undrained strength cud frac14 100

                        14frac14 71 kNm2

                        Design bearing resistance Rd frac14 12cudNe area

                        frac14 12 71 514 42

                        frac14 7007 kN

                        For drained shear 0d frac14 tan1tan 32

                        125

                        frac14 26

                        Nq frac14 12 N frac14 10

                        Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                        (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                        Layer z (m) m n Ir 0 (kNm2) sod (mm)

                        1 2 100 0175 0700qn 0182qn

                        2 6 033 0044 0176qn 0046qn

                        3 10 020 0017 0068qn 0018qn

                        0246qn

                        Diameter of equivalent circle B frac14 45m

                        H

                        Bfrac14 12

                        45frac14 27 and A frac14 042

                        13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                        64 Bearing capacity

                        For sc frac14 30mm

                        qn frac14 30

                        0147frac14 204 kN=m2

                        q frac14 204thorn 21 frac14 225 kN=m2

                        Design load frac14 42 225 frac14 3600 kN

                        The design load is 3600 kN settlement being the limiting criterion

                        87

                        D

                        Bfrac14 8

                        4frac14 20

                        From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                        F frac14 cuNc

                        Dfrac14 40 71

                        20 8frac14 18

                        88

                        Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                        Design value of 0 frac14 tan1tan 38

                        125

                        frac14 32

                        Figure Q86

                        Bearing capacity 65

                        For 0 frac14 32 Nq frac14 23 and N frac14 25

                        Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                        The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                        Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                        For B frac14 250m qn frac14 3750

                        2502 17 frac14 583 kN=m2

                        From Figure 510 m frac14 n frac14 126

                        6frac14 021

                        Ir frac14 0019

                        Stress increment frac14 4 0019 583 frac14 44 kN=m2

                        Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                        The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                        89

                        Depth (m) N 0v (kNm2) CN N1

                        070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                        Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                        (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                        Cw frac14 05thorn 0530

                        47

                        frac14 082

                        66 Bearing capacity

                        Thus

                        qa frac14 150 082 frac14 120 kN=m2

                        (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                        Thus

                        qa frac14 90 15 frac14 135 kN=m2

                        (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                        Ic frac14 171

                        1014frac14 0068

                        From Equation 819(a) with s frac14 25mm

                        q frac14 25

                        3507 0068frac14 150 kN=m2

                        810

                        Peak value of strain influence factor occurs at a depth of 27m and is given by

                        Izp frac14 05thorn 01130

                        16 27

                        05

                        frac14 067

                        Refer to Figure Q810

                        E frac14 25qc

                        Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                        Ez (mm3MN)

                        1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                        0203

                        C1 frac14 1 0500qnfrac14 1 05 12 16

                        130frac14 093

                        C2 frac14 1 ethsayTHORN

                        s frac14 C1C2qnX Iz

                        Ez frac14 093 1 130 0203 frac14 25mm

                        Bearing capacity 67

                        811

                        At pile base level

                        cu frac14 220 kN=m2

                        qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                        Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                        00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                        qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                        Then

                        Qf frac14 Abqb thorn Asqs

                        frac14

                        4 32 1980

                        thorn eth 105 139 86THORN

                        frac14 13 996thorn 3941 frac14 17 937 kN

                        0 01 02 03 04 05 06 07

                        0 2 4 6 8 10 12 14

                        1

                        2

                        3

                        4

                        5

                        6

                        7

                        8

                        (1)

                        (2)

                        (3)

                        (4)

                        (5)

                        qc

                        qc

                        Iz

                        Iz

                        (MNm2)

                        z (m)

                        Figure Q810

                        68 Bearing capacity

                        Allowable load

                        ethaTHORN Qf

                        2frac14 17 937

                        2frac14 8968 kN

                        ethbTHORN Abqb

                        3thorn Asqs frac14 13 996

                        3thorn 3941 frac14 8606 kN

                        ie allowable load frac14 8600 kN

                        Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                        According to the limit state method

                        Characteristic undrained strength at base level cuk frac14 220

                        150kN=m2

                        Characteristic base resistance qbk frac14 9cuk frac14 9 220

                        150frac14 1320 kN=m2

                        Characteristic shaft resistance qsk frac14 00150

                        frac14 86

                        150frac14 57 kN=m2

                        Characteristic base and shaft resistances

                        Rbk frac14

                        4 32 1320 frac14 9330 kN

                        Rsk frac14 105 139 86

                        150frac14 2629 kN

                        For a bored pile the partial factors are b frac14 160 and s frac14 130

                        Design bearing resistance Rcd frac14 9330

                        160thorn 2629

                        130

                        frac14 5831thorn 2022

                        frac14 7850 kN

                        Adding ethDAb W) the design bearing resistance becomes 9650 kN

                        812

                        ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                        qs frac14 cu frac14 040 105 frac14 42 kN=m2

                        For a single pile

                        Qf frac14 Abqb thorn Asqs

                        frac14

                        4 062 1305

                        thorn eth 06 15 42THORN

                        frac14 369thorn 1187 frac14 1556 kN

                        Bearing capacity 69

                        Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                        eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                        (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                        (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                        qbkfrac14 9cuk frac14 9 220

                        150frac14 1320 kN=m2

                        qskfrac14cuk frac14 040 105

                        150frac14 28 kN=m2

                        Rbkfrac14

                        4 0602 1320 frac14 373 kN

                        Rskfrac14 060 15 28 frac14 791 kN

                        Rcdfrac14 373

                        160thorn 791

                        130frac14 233thorn 608 frac14 841 kN

                        Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                        Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                        (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                        q frac14 21 000

                        1762frac14 68 kN=m2

                        Immediate settlement

                        H

                        Bfrac14 15

                        176frac14 085

                        D

                        Bfrac14 13

                        176frac14 074

                        L

                        Bfrac14 1

                        Hence from Figure 515

                        130 frac14 078 and 131 frac14 041

                        70 Bearing capacity

                        Thus using Equation 528

                        si frac14 078 041 68 176

                        65frac14 6mm

                        Consolidation settlement

                        Layer z (m) Area (m2) (kNm2) mvH (mm)

                        1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                        434 (sod)

                        Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                        sc frac14 056 434 frac14 24mm

                        The total settlement is (6thorn 24) frac14 30mm

                        813

                        At base level N frac14 26 Then using Equation 830

                        qb frac14 40NDb

                        Bfrac14 40 26 2

                        025frac14 8320 kN=m2

                        ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                        Figure Q812

                        Bearing capacity 71

                        Over the length embedded in sand

                        N frac14 21 ie18thorn 24

                        2

                        Using Equation 831

                        qs frac14 2N frac14 2 21 frac14 42 kN=m2

                        For a single pile

                        Qf frac14 Abqb thorn Asqs

                        frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                        For the pile group assuming a group efficiency of 12

                        XQf frac14 12 9 604 frac14 6523 kN

                        Then the load factor is

                        F frac14 6523

                        2000thorn 1000frac14 21

                        (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                        Characteristic base resistance per unit area qbk frac14 8320

                        150frac14 5547 kNm2

                        Characteristic shaft resistance per unit area qsk frac14 42

                        150frac14 28 kNm2

                        Characteristic base and shaft resistances for a single pile

                        Rbk frac14 0252 5547 frac14 347 kN

                        Rsk frac14 4 025 2 28 frac14 56 kN

                        For a driven pile the partial factors are b frac14 s frac14 130

                        Design bearing resistance Rcd frac14 347

                        130thorn 56

                        130frac14 310 kN

                        For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                        Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                        (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                        From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                        72 Bearing capacity

                        N frac14 24thorn 26thorn 34

                        3frac14 28

                        Ic frac14 171

                        2814frac14 0016 ethEquation 818THORN

                        s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                        814

                        Using Equation 841

                        Tf frac14 DLcu thorn

                        4ethD2 d2THORNcuNc

                        frac14 eth 02 5 06 110THORN thorn

                        4eth022 012THORN110 9

                        frac14 207thorn 23 frac14 230 kN

                        Figure Q813

                        Bearing capacity 73

                        Chapter 9

                        Stability of slopes

                        91

                        Referring to Figure Q91

                        W frac14 417 19 frac14 792 kN=m

                        Q frac14 20 28 frac14 56 kN=m

                        Arc lengthAB frac14

                        180 73 90 frac14 115m

                        Arc length BC frac14

                        180 28 90 frac14 44m

                        The factor of safety is given by

                        F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                        90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                        Depth of tension crack z0 frac14 2cu

                        frac14 2 20

                        19frac14 21m

                        Arc length BD frac14

                        180 13

                        1

                        2 90 frac14 21m

                        F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                        The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                        Design resisting moment frac14 rXethcudLaTHORN frac14 90

                        14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                        Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                        The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                        92

                        u frac14 0

                        Depth factor D frac14 11

                        9frac14 122

                        Using Equation 92 with F frac14 10

                        Ns frac14 cu

                        FHfrac14 30

                        10 19 9frac14 0175

                        Hence from Figure 93

                        frac14 50

                        For F frac14 12

                        Ns frac14 30

                        12 19 9frac14 0146

                        frac14 27

                        93

                        Refer to Figure Q93

                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                        1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                        74 m

                        214 1deg

                        213 1deg

                        39 m

                        WB

                        D

                        C

                        28 m

                        21 m

                        A

                        Q

                        Soil (1)Soil (2)

                        73deg

                        Figure Q91

                        Stability of slopes 75

                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                        9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                        599 256 328 1372

                        Figure Q93

                        76 Stability of slopes

                        XW cos frac14 b

                        Xh cos frac14 21 2 599 frac14 2516 kN=mX

                        W sin frac14 bX

                        h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                        Arc length La frac14

                        180 57

                        1

                        2 326 frac14 327m

                        The factor of safety is given by

                        F frac14 c0La thorn tan0ethW cos ulTHORN

                        W sin

                        frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                        frac14 091

                        According to the limit state method

                        0d frac14 tan1tan 32

                        125

                        frac14 265

                        c0 frac14 8

                        160frac14 5 kN=m2

                        Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                        Design disturbing moment frac14 1075 kN=m

                        The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                        94

                        F frac14 1

                        W sin

                        Xfc0bthorn ethW ubTHORN tan0g sec

                        1thorn ethtan tan0=FTHORN

                        c0 frac14 8 kN=m2

                        0 frac14 32

                        c0b frac14 8 2 frac14 16 kN=m

                        W frac14 bh frac14 21 2 h frac14 42h kN=m

                        Try F frac14 100

                        tan0

                        Ffrac14 0625

                        Stability of slopes 77

                        Values of u are as obtained in Figure Q93

                        SliceNo

                        h(m)

                        W frac14 bh(kNm)

                        W sin(kNm)

                        ub(kNm)

                        c0bthorn (W ub) tan0(kNm)

                        sec

                        1thorn (tan tan0)FProduct(kNm)

                        1 05 21 6 2 8 24 1078 262 13 55 31

                        23 33 30 1042 31

                        3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                        224 92 72 0931 67

                        6 50 210 11 40 100 85 0907 777 55 231 14

                        12 58 112 90 0889 80

                        8 60 252 1812

                        80 114 102 0874 899 63 265 22 99 116 109 0861 94

                        10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                        2154 88 116 0853 99

                        14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                        1074 1091

                        F frac14 1091

                        1074frac14 102 (assumed value 100)

                        Thus

                        F frac14 101

                        95

                        F frac14 1

                        W sin

                        XfWeth1 ruTHORN tan0g sec

                        1thorn ethtan tan0THORN=F

                        0 frac14 33

                        ru frac14 020

                        W frac14 bh frac14 20 5 h frac14 100h kN=m

                        eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                        Try F frac14 110

                        tan 0

                        Ffrac14 tan 33

                        110frac14 0590

                        78 Stability of slopes

                        Referring to Figure Q95

                        SliceNo

                        h(m)

                        W frac14 bh(kNm)

                        W sin(kNm)

                        W(1 ru) tan0(kNm)

                        sec

                        1thorn ( tan tan0)FProduct(kNm)

                        1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                        2120 234 0892 209

                        4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                        1185 1271

                        Figure Q95

                        Stability of slopes 79

                        F frac14 1271

                        1185frac14 107

                        The trial value was 110 therefore take F to be 108

                        96

                        (a) Water table at surface the factor of safety is given by Equation 912

                        F frac14 0

                        sat

                        tan0

                        tan

                        ptie 15 frac14 92

                        19

                        tan 36

                        tan

                        tan frac14 0234

                        frac14 13

                        Water table well below surface the factor of safety is given by Equation 911

                        F frac14 tan0

                        tan

                        frac14 tan 36

                        tan 13

                        frac14 31

                        (b) 0d frac14 tan1tan 36

                        125

                        frac14 30

                        Depth of potential failure surface frac14 z

                        Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                        frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                        frac14 504z kN

                        Design disturbing moment per unit area Sd frac14 sat sin cos

                        frac14 19 z sin 13 cos 13

                        frac14 416z kN

                        Rd gtSd therefore the limit state for overall stability is satisfied

                        80 Stability of slopes

                        • Book Cover
                        • Title
                        • Contents
                        • Basic characteristics of soils
                        • Seepage
                        • Effective stress
                        • Shear strength
                        • Stresses and displacements
                        • Lateral earth pressure
                        • Consolidation theory
                        • Bearing capacity
                        • Stability of slopes

                          23

                          The flow net is drawn in Figure Q23 from which Nffrac14 35 and Ndfrac14 9 The overall lossin total head is 300m Then

                          q frac14 kh Nf

                          Ndfrac14 5 105 300 35

                          9frac14 58 105 m3=s per m

                          The pore water pressure is determined at the points of intersection of the equipoten-tials with the base of the structure The total head (h) at each point is obtained fromthe flow net The elevation head (z) at each point on the base of the structure is250m The calculations are tabulated below and the distribution of pressure (u) isplotted to scale in the figure

                          Point h (m) h z (m) u frac14 w(h z)(kNm2)

                          1 233 483 472 200 450 443 167 417 414 133 383 375 100 350 346 067 317 31

                          eg for Point 1

                          h1 frac14 7

                          9 300 frac14 233m

                          h1 z1 frac14 233 eth250THORN frac14 483m

                          Figure Q23

                          Seepage 7

                          u1 frac14 98 483 frac14 47 kN=m2

                          The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

                          24

                          The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

                          q frac14 kh Nf

                          Ndfrac14 40 107 550 10

                          11frac14 20 106 m3=s per m

                          25

                          The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

                          q frac14 kh Nf

                          Ndfrac14 20 106 500 42

                          9frac14 47 106 m3=s per m

                          Figure Q24

                          8 Seepage

                          26

                          The scale transformation factor in the x direction is given by Equation 221 ie

                          xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                          ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

                          Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

                          k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                          pfrac14

                          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

                          p 107 frac14 30 107 m=s

                          Thus the quantity of seepage is given by

                          q frac14 k0h Nf

                          Ndfrac14 30 107 1400 325

                          12frac14 11 106 m3=s per m

                          Figure Q25

                          Seepage 9

                          27

                          The scale transformation factor in the x direction is

                          xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                          ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

                          Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

                          k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                          pfrac14

                          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

                          p 106 frac14 45 106 m=s

                          The focus of the basic parabola is at point A The parabola passes through point Gsuch that

                          GC frac14 03HC frac14 03 30 frac14 90m

                          Thus the coordinates of G are

                          x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

                          480 frac14 x0 2002

                          4x0

                          Figure Q26

                          10 Seepage

                          Hence

                          x0 frac14 20m

                          Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

                          x frac14 20 z2

                          80

                          x 20 0 50 100 200 300z 0 400 748 980 1327 1600

                          The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

                          No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

                          q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

                          28

                          The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

                          Figure Q27

                          Seepage 11

                          q frac14 kh Nf

                          Ndfrac14 45 105 28 33

                          7

                          frac14 59 105 m3=s per m

                          29

                          The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

                          kx frac14 H1k1 thornH2k2

                          H1 thornH2frac14 106

                          10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

                          kz frac14 H1 thornH2

                          H1

                          k1thornH2

                          k2

                          frac14 10

                          5

                          eth2 106THORN thorn5

                          eth16 106THORNfrac14 36 106 m=s

                          Then the scale transformation factor is given by

                          xt frac14 xffiffiffiffiffikz

                          pffiffiffiffiffikx

                          p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

                          Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

                          Figure Q28

                          12 Seepage

                          k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                          qfrac14

                          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

                          p 106 frac14 57 106 m=s

                          Then the quantity of seepage is given by

                          q frac14 k0h Nf

                          Ndfrac14 57 106 350 56

                          11

                          frac14 10 105 m3=s per m

                          Figure Q29

                          Seepage 13

                          Chapter 3

                          Effective stress

                          31

                          Buoyant unit weight

                          0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                          Effective vertical stress

                          0v frac14 5 102 frac14 51 kN=m2 or

                          Total vertical stress

                          v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

                          Pore water pressure

                          u frac14 7 98 frac14 686 kN=m2

                          Effective vertical stress

                          0v frac14 v u frac14 1196 686 frac14 51 kN=m2

                          32

                          Buoyant unit weight

                          0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                          Effective vertical stress

                          0v frac14 5 102 frac14 51 kN=m2 or

                          Total vertical stress

                          v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

                          Pore water pressure

                          u frac14 205 98 frac14 2009 kN=m2

                          Effective vertical stress

                          0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

                          33

                          At top of the clay

                          v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

                          u frac14 2 98 frac14 196 kN=m2

                          0v frac14 v u frac14 710 196 frac14 514 kN=m2

                          Alternatively

                          0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                          0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

                          At bottom of the clay

                          v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

                          u frac14 12 98 frac14 1176 kN=m2

                          0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

                          NB The alternative method of calculation is not applicable because of the artesiancondition

                          Figure Q3132

                          Effective stress 15

                          34

                          0 frac14 20 98 frac14 102 kN=m3

                          At 8m depth

                          0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                          35

                          0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                          0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                          Figure Q33

                          Figure Q34

                          16 Effective stress

                          (a) Immediately after WT rise

                          At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                          0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                          At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                          0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                          (b) Several years after WT rise

                          At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                          At 8m depth

                          0v frac14 940 kN=m2 (as above)

                          At 12m depth

                          0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                          Figure Q35

                          Effective stress 17

                          36

                          Total weight

                          ab frac14 210 kN

                          Effective weight

                          ac frac14 112 kN

                          Resultant boundary water force

                          be frac14 119 kN

                          Seepage force

                          ce frac14 34 kN

                          Resultant body force

                          ae frac14 99 kN eth73 to horizontalTHORN

                          (Refer to Figure Q36)

                          Figure Q36

                          18 Effective stress

                          37

                          Situation (1)(a)

                          frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                          u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                          0 frac14 u frac14 694 392 frac14 302 kN=m2

                          (b)

                          i frac14 2

                          4frac14 05

                          j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                          Situation (2)(a)

                          frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                          u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                          0 frac14 u frac14 498 392 frac14 106 kN=m2

                          (b)

                          i frac14 2

                          4frac14 05

                          j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                          38

                          The flow net is drawn in Figure Q24

                          Loss in total head between adjacent equipotentials

                          h frac14 550

                          Ndfrac14 550

                          11frac14 050m

                          Exit hydraulic gradient

                          ie frac14 h

                          sfrac14 050

                          070frac14 071

                          Effective stress 19

                          The critical hydraulic gradient is given by Equation 39

                          ic frac14 0

                          wfrac14 102

                          98frac14 104

                          Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                          F frac14 iciefrac14 104

                          071frac14 15

                          Total head at C

                          hC frac14 nd

                          Ndh frac14 24

                          11 550 frac14 120m

                          Elevation head at C

                          zC frac14 250m

                          Pore water pressure at C

                          uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                          Therefore effective vertical stress at C

                          0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                          For point D

                          hD frac14 73

                          11 550 frac14 365m

                          zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                          0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                          39

                          The flow net is drawn in Figure Q25

                          For a soil prism 150 300m adjacent to the piling

                          hm frac14 26

                          9 500 frac14 145m

                          20 Effective stress

                          Factor of safety against lsquoheavingrsquo (Equation 310)

                          F frac14 ic

                          imfrac14 0d

                          whmfrac14 97 300

                          98 145frac14 20

                          With a filter

                          F frac14 0d thorn wwhm

                          3 frac14 eth97 300THORN thorn w98 145

                          w frac14 135 kN=m2

                          Depth of filterfrac14 13521frac14 065m (if above water level)

                          Effective stress 21

                          Chapter 4

                          Shear strength

                          41

                          frac14 295 kN=m2

                          u frac14 120 kN=m2

                          0 frac14 u frac14 295 120 frac14 175 kN=m2

                          f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                          42

                          03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                          100 452 552200 908 1108400 1810 2210800 3624 4424

                          The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                          Figure Q42

                          43

                          3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                          200 222 422400 218 618600 220 820

                          The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                          44

                          The modified shear strength parameters are

                          0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                          a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                          The coordinates of the stress point representing failure conditions in the test are

                          1

                          2eth1 2THORN frac14 1

                          2 170 frac14 85 kN=m2

                          1

                          2eth1 thorn 3THORN frac14 1

                          2eth270thorn 100THORN frac14 185 kN=m2

                          The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                          uf frac14 36 kN=m2

                          Figure Q43

                          Figure Q44

                          Shear strength 23

                          45

                          3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                          150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                          The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                          The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                          46

                          03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                          200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                          The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                          Figure Q45

                          24 Shear strength

                          47

                          The torque required to produce shear failure is given by

                          T frac14 dh cud

                          2thorn 2

                          Z d=2

                          0

                          2r drcur

                          frac14 cud2h

                          2thorn 4cu

                          Z d=2

                          0

                          r2dr

                          frac14 cud2h

                          2thorn d

                          3

                          6

                          Then

                          35 frac14 cu52 10

                          2thorn 53

                          6

                          103

                          cu frac14 76 kN=m3

                          400

                          0 400 800 1200 1600

                          τ (k

                          Nm

                          2 )

                          σprime (kNm2)

                          34deg

                          315deg29deg

                          (a)

                          (b)

                          0 400

                          400

                          800 1200 1600

                          Failure envelope

                          300 500

                          σprime (kNm2)

                          τ (k

                          Nm

                          2 )

                          20 (kNm2)

                          31deg

                          Figure Q46

                          Shear strength 25

                          48

                          The relevant stress values are calculated as follows

                          3 frac14 600 kN=m2

                          1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                          2(1 3) 0 40 79 107 139 159

                          1

                          2(01 thorn 03) 400 411 402 389 351 326

                          1

                          2(1 thorn 3) 600 640 679 707 739 759

                          The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                          Af frac14 433 200

                          319frac14 073

                          49

                          B frac14 u33

                          frac14 144

                          350 200frac14 096

                          a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                          0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                          10 283 65 023

                          Figure Q48

                          26 Shear strength

                          The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                          Figure Q49

                          Shear strength 27

                          Chapter 5

                          Stresses and displacements

                          51

                          Vertical stress is given by

                          z frac14 Qz2Ip frac14 5000

                          52Ip

                          Values of Ip are obtained from Table 51

                          r (m) rz Ip z (kNm2)

                          0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                          10 20 0009 2

                          The variation of z with radial distance (r) is plotted in Figure Q51

                          Figure Q51

                          52

                          Below the centre load (Figure Q52)

                          r

                          zfrac14 0 for the 7500-kN load

                          Ip frac14 0478

                          r

                          zfrac14 5

                          4frac14 125 for the 10 000- and 9000-kN loads

                          Ip frac14 0045

                          Then

                          z frac14X Q

                          z2Ip

                          frac14 7500 0478

                          42thorn 10 000 0045

                          42thorn 9000 0045

                          42

                          frac14 224thorn 28thorn 25 frac14 277 kN=m2

                          53

                          The vertical stress under a corner of a rectangular area is given by

                          z frac14 qIr

                          where values of Ir are obtained from Figure 510 In this case

                          z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                          z

                          Figure Q52

                          Stresses and displacements 29

                          z (m) m n Ir z (kNm2)

                          0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                          10 010 0005 5

                          z is plotted against z in Figure Q53

                          54

                          (a)

                          m frac14 125

                          12frac14 104

                          n frac14 18

                          12frac14 150

                          From Figure 510 Irfrac14 0196

                          z frac14 2 175 0196 frac14 68 kN=m2

                          Figure Q53

                          30 Stresses and displacements

                          (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                          z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                          55

                          Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                          Px frac14 2Q

                          1

                          m2 thorn 1frac14 2 150

                          125frac14 76 kN=m

                          Equation 517 is used to obtain the pressure distribution

                          px frac14 4Q

                          h

                          m2n

                          ethm2 thorn n2THORN2 frac14150

                          m2n

                          ethm2 thorn n2THORN2 ethkN=m2THORN

                          Figure Q54

                          Stresses and displacements 31

                          n m2n

                          (m2 thorn n2)2

                          px(kNm2)

                          0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                          The pressure distribution is plotted in Figure Q55

                          56

                          H

                          Bfrac14 10

                          2frac14 5

                          L

                          Bfrac14 4

                          2frac14 2

                          D

                          Bfrac14 1

                          2frac14 05

                          Hence from Figure 515

                          131 frac14 082

                          130 frac14 094

                          Figure Q55

                          32 Stresses and displacements

                          The immediate settlement is given by Equation 528

                          si frac14 130131qB

                          Eu

                          frac14 094 082 200 2

                          45frac14 7mm

                          Stresses and displacements 33

                          Chapter 6

                          Lateral earth pressure

                          61

                          For 0 frac14 37 the active pressure coefficient is given by

                          Ka frac14 1 sin 37

                          1thorn sin 37frac14 025

                          The total active thrust (Equation 66a with c0 frac14 0) is

                          Pa frac14 1

                          2KaH

                          2 frac14 1

                          2 025 17 62 frac14 765 kN=m

                          If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                          K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                          and the thrust on the wall is

                          P0 frac14 1

                          2K0H

                          2 frac14 1

                          2 040 17 62 frac14 122 kN=m

                          62

                          The active pressure coefficients for the three soil types are as follows

                          Ka1 frac141 sin 35

                          1thorn sin 35frac14 0271

                          Ka2 frac141 sin 27

                          1thorn sin 27frac14 0375

                          ffiffiffiffiffiffiffiKa2

                          p frac14 0613

                          Ka3 frac141 sin 42

                          1thorn sin 42frac14 0198

                          Distribution of active pressure (plotted in Figure Q62)

                          Depth (m) Soil Active pressure (kNm2)

                          3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                          12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                          At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                          Total thrust frac14 571 kNm

                          Point of application is (4893571) m from the top of the wall ie 857m

                          Force (kN) Arm (m) Moment (kN m)

                          (1)1

                          2 0271 16 32 frac14 195 20 390

                          (2) 0271 16 3 2 frac14 260 40 1040

                          (3)1

                          2 0271 92 22 frac14 50 433 217

                          (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                          (5)1

                          2 0375 102 32 frac14 172 70 1204

                          (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                          (7)1

                          2 0198 112 42 frac14 177 1067 1889

                          (8)1

                          2 98 92 frac14 3969 90 35721

                          5713 48934

                          Figure Q62

                          Lateral earth pressure 35

                          63

                          (a) For u frac14 0 Ka frac14 Kp frac14 1

                          Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                          frac14 245

                          At the lower end of the piling

                          pa frac14 Kaqthorn Kasatz Kaccu

                          frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                          frac14 115 kN=m2

                          pp frac14 Kpsatzthorn Kpccu

                          frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                          frac14 202 kN=m2

                          (b) For 0 frac14 26 and frac14 1

                          20

                          Ka frac14 035

                          Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                          pfrac14 145 ethEquation 619THORN

                          Kp frac14 37

                          Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                          pfrac14 47 ethEquation 624THORN

                          At the lower end of the piling

                          pa frac14 Kaqthorn Ka0z Kacc

                          0

                          frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                          frac14 187 kN=m2

                          pp frac14 Kp0zthorn Kpcc

                          0

                          frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                          frac14 198 kN=m2

                          36 Lateral earth pressure

                          64

                          (a) For 0 frac14 38 Ka frac14 024

                          0 frac14 20 98 frac14 102 kN=m3

                          The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                          Force (kN) Arm (m) Moment (kN m)

                          (1) 024 10 66 frac14 159 33 525

                          (2)1

                          2 024 17 392 frac14 310 400 1240

                          (3) 024 17 39 27 frac14 430 135 580

                          (4)1

                          2 024 102 272 frac14 89 090 80

                          (5)1

                          2 98 272 frac14 357 090 321

                          Hfrac14 1345 MH frac14 2746

                          (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                          (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                          XM frac14MV MH frac14 7790 kNm

                          Lever arm of base resultant

                          M

                          Vfrac14 779

                          488frac14 160

                          Eccentricity of base resultant

                          e frac14 200 160 frac14 040m

                          39 m

                          27 m

                          40 m

                          04 m

                          04 m

                          26 m

                          (7)

                          (9)

                          (1)(2)

                          (3)

                          (4)

                          (5)

                          (8)(6)

                          (10)

                          WT

                          10 kNm2

                          Hydrostatic

                          Figure Q64

                          Lateral earth pressure 37

                          Base pressures (Equation 627)

                          p frac14 VB

                          1 6e

                          B

                          frac14 488

                          4eth1 060THORN

                          frac14 195 kN=m2 and 49 kN=m2

                          Factor of safety against sliding (Equation 628)

                          F frac14 V tan

                          Hfrac14 488 tan 25

                          1345frac14 17

                          (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                          Hfrac14 1633 kN

                          V frac14 4879 kN

                          MH frac14 3453 kNm

                          MV frac14 10536 kNm

                          The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                          65

                          For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                          Kp

                          Ffrac14 385

                          2

                          0 frac14 20 98 frac14 102 kN=m3

                          The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                          Force (kN) Arm (m) Moment (kN m)

                          (1)1

                          2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                          (2) 026 17 45 d frac14 199d d2 995d2

                          (3)1

                          2 026 102 d2 frac14 133d2 d3 044d3

                          (4)1

                          2 385

                          2 17 152 frac14 368 dthorn 05 368d 184

                          (5)385

                          2 17 15 d frac14 491d d2 2455d2

                          (6)1

                          2 385

                          2 102 d2 frac14 982d2 d3 327d3

                          38 Lateral earth pressure

                          XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                          d3 thorn 516d2 283d 1724 frac14 0

                          d frac14 179m

                          Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                          Over additional 20 embedded depth

                          pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                          Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                          66

                          The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                          Ka frac14 sin 69=sin 105

                          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                          pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                          26664

                          37775

                          2

                          frac14 050

                          The total active thrust (acting at 25 above the normal) is given by Equation 616

                          Pa frac14 1

                          2 050 19 7502 frac14 267 kN=m

                          Figure Q65

                          Lateral earth pressure 39

                          Horizontal component

                          Ph frac14 267 cos 40 frac14 205 kN=m

                          Vertical component

                          Pv frac14 267 sin 40 frac14 172 kN=m

                          Consider moments about the toe of the wall (Figure Q66) (per m)

                          Force (kN) Arm (m) Moment (kN m)

                          (1)1

                          2 175 650 235 frac14 1337 258 345

                          (2) 050 650 235 frac14 764 175 134

                          (3)1

                          2 070 650 235 frac14 535 127 68

                          (4) 100 400 235 frac14 940 200 188

                          (5) 1

                          2 080 050 235 frac14 47 027 1

                          Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                          Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                          Lever arm of base resultant

                          M

                          Vfrac14 795

                          525frac14 151m

                          Eccentricity of base resultant

                          e frac14 200 151 frac14 049m

                          Figure Q66

                          40 Lateral earth pressure

                          Base pressures (Equation 627)

                          p frac14 525

                          41 6 049

                          4

                          frac14 228 kN=m2 and 35 kN=m2

                          The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                          The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                          The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                          67

                          For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                          Force (kN) Arm (m) Moment (kNm)

                          (1)1

                          2 027 17 52 frac14 574 183 1050

                          (2) 027 17 5 3 frac14 689 500 3445

                          (3)1

                          2 027 102 32 frac14 124 550 682

                          (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                          (5)1

                          2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                          (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                          (7) 1

                          2 267

                          2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                          (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                          2 d frac14 163d d2thorn 650 82d2 1060d

                          Tie rod force per m frac14 T 0 0

                          XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                          d3 thorn 77d2 269d 1438 frac14 0

                          d frac14 467m

                          Depth of penetration frac14 12d frac14 560m

                          Lateral earth pressure 41

                          Algebraic sum of forces for d frac14 467m isX

                          F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                          T frac14 905 kN=m

                          Force in each tie rod frac14 25T frac14 226 kN

                          68

                          (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                          0 frac14 21 98 frac14 112 kN=m3

                          The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                          uC frac14 150

                          165 15 98 frac14 134 kN=m2

                          The average seepage pressure is

                          j frac14 15

                          165 98 frac14 09 kN=m3

                          Hence

                          0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                          0 j frac14 112 09 frac14 103 kN=m3

                          Figure Q67

                          42 Lateral earth pressure

                          Consider moments about the anchor point A (per m)

                          Force (kN) Arm (m) Moment (kN m)

                          (1) 10 026 150 frac14 390 60 2340

                          (2)1

                          2 026 18 452 frac14 474 15 711

                          (3) 026 18 45 105 frac14 2211 825 18240

                          (4)1

                          2 026 121 1052 frac14 1734 100 17340

                          (5)1

                          2 134 15 frac14 101 40 404

                          (6) 134 30 frac14 402 60 2412

                          (7)1

                          2 134 60 frac14 402 95 3819

                          571 4527(8) Ppm

                          115 115PPm

                          XM frac14 0

                          Ppm frac144527

                          115frac14 394 kN=m

                          Available passive resistance

                          Pp frac14 1

                          2 385 103 62 frac14 714 kN=m

                          Factor of safety

                          Fp frac14 Pp

                          Ppm

                          frac14 714

                          394frac14 18

                          Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                          Figure Q68

                          Lateral earth pressure 43

                          (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                          Consider moments (per m) about the tie point A

                          Force (kN) Arm (m)

                          (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                          (2)1

                          2 033 18 452 frac14 601 15

                          (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                          (4)1

                          2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                          (5)1

                          2 134 15 frac14 101 40

                          (6) 134 30 frac14 402 60

                          (7)1

                          2 134 d frac14 67d d3thorn 75

                          (8) 1

                          2 30 103 d2 frac141545d2 2d3thorn 75

                          Moment (kN m)

                          (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                          XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                          d3 thorn 827d2 466d 1518 frac14 0

                          By trial

                          d frac14 544m

                          The minimum depth of embedment required is 544m

                          69

                          For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                          0 frac14 20 98 frac14 102 kN=m3

                          The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                          44 Lateral earth pressure

                          uC frac14 147

                          173 26 98 frac14 216 kN=m2

                          and the average seepage pressure around the wall is

                          j frac14 26

                          173 98 frac14 15 kN=m3

                          Consider moments about the prop (A) (per m)

                          Force (kN) Arm (m) Moment (kN m)

                          (1)1

                          2 03 17 272 frac14 186 020 37

                          (2) 03 17 27 53 frac14 730 335 2445

                          (3)1

                          2 03 (102thorn 15) 532 frac14 493 423 2085

                          (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                          (5)1

                          2 216 26 frac14 281 243 684

                          (6) 216 27 frac14 583 465 2712

                          (7)1

                          2 216 60 frac14 648 800 5184

                          3055(8)

                          1

                          2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                          Factor of safety

                          Fr frac14 6885

                          3055frac14 225

                          Figure Q69

                          Lateral earth pressure 45

                          610

                          For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                          p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                          Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                          Using the recommendations of Twine and Roscoe

                          p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                          Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                          611

                          frac14 18 kN=m3 0 frac14 34

                          H frac14 350m nH frac14 335m mH frac14 185m

                          Consider a trial value of F frac14 20 Refer to Figure 635

                          0m frac14 tan1tan 34

                          20

                          frac14 186

                          Then

                          frac14 45 thorn 0m2frac14 543

                          W frac14 1

                          2 18 3502 cot 543 frac14 792 kN=m

                          Figure Q610

                          46 Lateral earth pressure

                          P frac14 1

                          2 s 3352 frac14 561s kN=m

                          U frac14 1

                          2 98 1852 cosec 543 frac14 206 kN=m

                          Equations 630 and 631 then become

                          561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                          792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                          ie

                          561s 0616N 405 frac14 0

                          792 0857N thorn 563 frac14 0

                          N frac14 848

                          0857frac14 989 kN=m

                          Then

                          561s 609 405 frac14 0

                          s frac14 649

                          561frac14 116 kN=m3

                          The calculations for trial values of F of 20 15 and 10 are summarized below

                          F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                          20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                          s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                          Figure Q611

                          Lateral earth pressure 47

                          612

                          For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                          45 thorn 0

                          2frac14 63

                          For the retained material between the surface and a depth of 36m

                          Pa frac14 1

                          2 030 18 362 frac14 350 kN=m

                          Weight of reinforced fill between the surface and a depth of 36m is

                          Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                          eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                          Eccentricity of Rv

                          e frac14 263 250 frac14 013m

                          The average vertical stress at a depth of 36m is

                          z frac14 Rv

                          L 2efrac14 324

                          474frac14 68 kN=m2

                          (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                          Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                          Tensile stress in the element frac14 138 103

                          65 3frac14 71N=mm2

                          Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                          Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                          Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                          The weight of ABC is

                          W frac14 1

                          2 18 52 265 frac14 124 kN=m

                          From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                          48 Lateral earth pressure

                          (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                          Tp frac14 032 68 120 065 frac14 170 kN

                          Tr frac14 213 420

                          418frac14 214 kN

                          Again the tensile failure and slipping limit states are satisfied for this element

                          Figure Q612

                          Lateral earth pressure 49

                          Chapter 7

                          Consolidation theory

                          71

                          Total change in thickness

                          H frac14 782 602 frac14 180mm

                          Average thickness frac14 1530thorn 180

                          2frac14 1620mm

                          Length of drainage path d frac14 1620

                          2frac14 810mm

                          Root time plot (Figure Q71a)

                          ffiffiffiffiffiffit90p frac14 33

                          t90 frac14 109min

                          cv frac14 0848d2

                          t90frac14 0848 8102

                          109 1440 365

                          106frac14 27m2=year

                          r0 frac14 782 764

                          782 602frac14 018

                          180frac14 0100

                          rp frac14 10eth764 645THORN9eth782 602THORN frac14

                          10 119

                          9 180frac14 0735

                          rs frac14 1 eth0100thorn 0735THORN frac14 0165

                          Log time plot (Figure Q71b)

                          t50 frac14 26min

                          cv frac14 0196d2

                          t50frac14 0196 8102

                          26 1440 365

                          106frac14 26m2=year

                          r0 frac14 782 763

                          782 602frac14 019

                          180frac14 0106

                          rp frac14 763 623

                          782 602frac14 140

                          180frac14 0778

                          rs frac14 1 eth0106thorn 0778THORN frac14 0116

                          Figure Q71(a)

                          Figure Q71(b)

                          Final void ratio

                          e1 frac14 w1Gs frac14 0232 272 frac14 0631

                          e

                          Hfrac14 1thorn e0

                          H0frac14 1thorn e1 thorne

                          H0

                          ie

                          e

                          180frac14 1631thorne

                          1710

                          e frac14 2936

                          1530frac14 0192

                          Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                          mv frac14 1

                          1thorn e0 e0 e101 00

                          frac14 1

                          1823 0192

                          0107frac14 098m2=MN

                          k frac14 cvmvw frac14 265 098 98

                          60 1440 365 103frac14 81 1010 m=s

                          72

                          Using Equation 77 (one-dimensional method)

                          sc frac14 e0 e11thorn e0 H

                          Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                          Figure Q72

                          52 Consolidation theory

                          Settlement

                          Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                          1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                          318

                          Notes 5 92y 460thorn 84

                          Heave

                          Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                          1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                          38

                          73

                          U frac14 f ethTvTHORN frac14 f cvt

                          d2

                          Hence if cv is constant

                          t1

                          t2frac14 d

                          21

                          d22

                          where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                          d1 frac14 95mm and d2 frac14 2500mm

                          for U frac14 050 t2 frac14 t1 d22

                          d21

                          frac14 20

                          60 24 365 25002

                          952frac14 263 years

                          for U lt 060 Tv frac14

                          4U2 (Equation 724(a))

                          t030 frac14 t050 0302

                          0502

                          frac14 263 036 frac14 095 years

                          Consolidation theory 53

                          74

                          The layer is open

                          d frac14 8

                          2frac14 4m

                          Tv frac14 cvtd2frac14 24 3

                          42frac14 0450

                          ui frac14 frac14 84 kN=m2

                          The excess pore water pressure is given by Equation 721

                          ue frac14Xmfrac141mfrac140

                          2ui

                          Msin

                          Mz

                          d

                          expethM2TvTHORN

                          In this case z frac14 d

                          sinMz

                          d

                          frac14 sinM

                          where

                          M frac14

                          23

                          25

                          2

                          M sin M M2Tv exp (M2Tv)

                          2thorn1 1110 0329

                          3

                          21 9993 457 105

                          ue frac14 2 84 2

                          1 0329 ethother terms negligibleTHORN

                          frac14 352 kN=m2

                          75

                          The layer is open

                          d frac14 6

                          2frac14 3m

                          Tv frac14 cvtd2frac14 10 3

                          32frac14 0333

                          The layer thickness will be divided into six equal parts ie m frac14 6

                          54 Consolidation theory

                          For an open layer

                          Tv frac14 4n

                          m2

                          n frac14 0333 62

                          4frac14 300

                          The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                          ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                          i j

                          0 1 2 3 4 5 6 7 8 9 10 11 12

                          0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                          The initial and 3-year isochrones are plotted in Figure Q75

                          Area under initial isochrone frac14 180 units

                          Area under 3-year isochrone frac14 63 units

                          The average degree of consolidation is given by Equation 725Thus

                          U frac14 1 63

                          180frac14 065

                          Figure Q75

                          Consolidation theory 55

                          76

                          At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                          0 frac14 2w frac14 2 98 frac14 196 kN=m2

                          The final consolidation settlement (one-dimensional method) is

                          sc frac14 mv0H frac14 083 196 8 frac14 130mm

                          Corrected time t frac14 2 1

                          2

                          40

                          52

                          frac14 1615 years

                          Tv frac14 cvtd2frac14 44 1615

                          42frac14 0444

                          From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                          77

                          The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                          Figure Q77

                          56 Consolidation theory

                          Point m n Ir (kNm2) sc (mm)

                          13020frac14 15 20

                          20frac14 10 0194 (4) 113 124

                          260

                          20frac14 30

                          20

                          20frac14 10 0204 (2) 59 65

                          360

                          20frac14 30

                          40

                          20frac14 20 0238 (1) 35 38

                          430

                          20frac14 15

                          40

                          20frac14 20 0224 (2) 65 72

                          Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                          78

                          Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                          (a) Immediate settlement

                          H

                          Bfrac14 30

                          35frac14 086

                          D

                          Bfrac14 2

                          35frac14 006

                          Figure Q78

                          Consolidation theory 57

                          From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                          si frac14 130131qB

                          Eufrac14 10 032 105 35

                          40frac14 30mm

                          (b) Consolidation settlement

                          Layer z (m) Dz Ic (kNm2) syod (mm)

                          1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                          3150

                          Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                          Now

                          H

                          Bfrac14 30

                          35frac14 086 and A frac14 065

                          from Figure 712 13 frac14 079

                          sc frac14 13sod frac14 079 315 frac14 250mm

                          Total settlement

                          s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                          79

                          Without sand drains

                          Uv frac14 025

                          Tv frac14 0049 ethfrom Figure 718THORN

                          t frac14 Tvd2

                          cvfrac14 0049 82

                          cvWith sand drains

                          R frac14 0564S frac14 0564 3 frac14 169m

                          n frac14 Rrfrac14 169

                          015frac14 113

                          Tr frac14 cht

                          4R2frac14 ch

                          4 1692 0049 82

                          cvethand ch frac14 cvTHORN

                          frac14 0275

                          Ur frac14 073 (from Figure 730)

                          58 Consolidation theory

                          Using Equation 740

                          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                          U frac14 080

                          710

                          Without sand drains

                          Uv frac14 090

                          Tv frac14 0848

                          t frac14 Tvd2

                          cvfrac14 0848 102

                          96frac14 88 years

                          With sand drains

                          R frac14 0564S frac14 0564 4 frac14 226m

                          n frac14 Rrfrac14 226

                          015frac14 15

                          Tr

                          Tvfrac14 chcv

                          d2

                          4R2ethsame tTHORN

                          Tr

                          Tvfrac14 140

                          96 102

                          4 2262frac14 714 eth1THORN

                          Using Equation 740

                          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                          An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                          Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                          040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                          Thus

                          Uv frac14 0295 and Ur frac14 086

                          t frac14 88 00683

                          0848frac14 07 years

                          Consolidation theory 59

                          Chapter 8

                          Bearing capacity

                          81

                          (a) The ultimate bearing capacity is given by Equation 83

                          qf frac14 cNc thorn DNq thorn 1

                          2BN

                          For u frac14 0

                          Nc frac14 514 Nq frac14 1 N frac14 0

                          qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                          The net ultimate bearing capacity is

                          qnf frac14 qf D frac14 540 kN=m2

                          The net foundation pressure is

                          qn frac14 q D frac14 425

                          2 eth21 1THORN frac14 192 kN=m2

                          The factor of safety (Equation 86) is

                          F frac14 qnfqnfrac14 540

                          192frac14 28

                          (b) For 0 frac14 28

                          Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                          qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                          2 112 2 13

                          frac14 260thorn 168thorn 146 frac14 574 kN=m2

                          qnf frac14 574 112 frac14 563 kN=m2

                          F frac14 563

                          192frac14 29

                          (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                          82

                          For 0 frac14 38

                          Nq frac14 49 N frac14 67

                          qnf frac14 DethNq 1THORN thorn 1

                          2BN ethfrom Equation 83THORN

                          frac14 eth18 075 48THORN thorn 1

                          2 18 15 67

                          frac14 648thorn 905 frac14 1553 kN=m2

                          qn frac14 500

                          15 eth18 075THORN frac14 320 kN=m2

                          F frac14 qnfqnfrac14 1553

                          320frac14 48

                          0d frac14 tan1tan 38

                          125

                          frac14 32 therefore Nq frac14 23 and N frac14 25

                          Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                          2 18 15 25

                          frac14 15eth310thorn 337THORNfrac14 970 kN=m

                          Design load (action) Vd frac14 500 kN=m

                          The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                          83

                          D

                          Bfrac14 350

                          225frac14 155

                          From Figure 85 for a square foundation

                          Nc frac14 81

                          Bearing capacity 61

                          For a rectangular foundation (L frac14 450m B frac14 225m)

                          Nc frac14 084thorn 016B

                          L

                          81 frac14 745

                          Using Equation 810

                          qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                          For F frac14 3

                          qn frac14 1006

                          3frac14 335 kN=m2

                          q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                          Design load frac14 405 450 225 frac14 4100 kN

                          Design undrained strength cud frac14 135

                          14frac14 96 kN=m2

                          Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                          frac14 7241 kN

                          Design load Vd frac14 4100 kN

                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                          84

                          For 0 frac14 40

                          Nq frac14 64 N frac14 95

                          qnf frac14 DethNq 1THORN thorn 04BN

                          (a) Water table 5m below ground level

                          qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                          qn frac14 400 17 frac14 383 kN=m2

                          F frac14 2686

                          383frac14 70

                          (b) Water table 1m below ground level (ie at foundation level)

                          0 frac14 20 98 frac14 102 kN=m3

                          62 Bearing capacity

                          qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                          F frac14 2040

                          383frac14 53

                          (c) Water table at ground level with upward hydraulic gradient 02

                          eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                          qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                          F frac14 1296

                          392frac14 33

                          85

                          The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                          Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                          Design value of 0 frac14 tan1tan 39

                          125

                          frac14 33

                          For 0 frac14 33 Nq frac14 26 and N frac14 29

                          Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                          86

                          (a) Undrained shear for u frac14 0

                          Nc frac14 514 Nq frac14 1 N frac14 0

                          qnf frac14 12cuNc

                          frac14 12 100 514 frac14 617 kN=m2

                          qn frac14 qnfFfrac14 617

                          3frac14 206 kN=m2

                          q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                          Bearing capacity 63

                          Drained shear for 0 frac14 32

                          Nq frac14 23 N frac14 25

                          0 frac14 21 98 frac14 112 kN=m3

                          qnf frac14 0DethNq 1THORN thorn 040BN

                          frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                          frac14 694 kN=m2

                          q frac14 694

                          3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                          Design load frac14 42 227 frac14 3632 kN

                          (b) Design undrained strength cud frac14 100

                          14frac14 71 kNm2

                          Design bearing resistance Rd frac14 12cudNe area

                          frac14 12 71 514 42

                          frac14 7007 kN

                          For drained shear 0d frac14 tan1tan 32

                          125

                          frac14 26

                          Nq frac14 12 N frac14 10

                          Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                          (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                          Layer z (m) m n Ir 0 (kNm2) sod (mm)

                          1 2 100 0175 0700qn 0182qn

                          2 6 033 0044 0176qn 0046qn

                          3 10 020 0017 0068qn 0018qn

                          0246qn

                          Diameter of equivalent circle B frac14 45m

                          H

                          Bfrac14 12

                          45frac14 27 and A frac14 042

                          13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                          64 Bearing capacity

                          For sc frac14 30mm

                          qn frac14 30

                          0147frac14 204 kN=m2

                          q frac14 204thorn 21 frac14 225 kN=m2

                          Design load frac14 42 225 frac14 3600 kN

                          The design load is 3600 kN settlement being the limiting criterion

                          87

                          D

                          Bfrac14 8

                          4frac14 20

                          From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                          F frac14 cuNc

                          Dfrac14 40 71

                          20 8frac14 18

                          88

                          Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                          Design value of 0 frac14 tan1tan 38

                          125

                          frac14 32

                          Figure Q86

                          Bearing capacity 65

                          For 0 frac14 32 Nq frac14 23 and N frac14 25

                          Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                          The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                          Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                          For B frac14 250m qn frac14 3750

                          2502 17 frac14 583 kN=m2

                          From Figure 510 m frac14 n frac14 126

                          6frac14 021

                          Ir frac14 0019

                          Stress increment frac14 4 0019 583 frac14 44 kN=m2

                          Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                          The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                          89

                          Depth (m) N 0v (kNm2) CN N1

                          070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                          Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                          (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                          Cw frac14 05thorn 0530

                          47

                          frac14 082

                          66 Bearing capacity

                          Thus

                          qa frac14 150 082 frac14 120 kN=m2

                          (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                          Thus

                          qa frac14 90 15 frac14 135 kN=m2

                          (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                          Ic frac14 171

                          1014frac14 0068

                          From Equation 819(a) with s frac14 25mm

                          q frac14 25

                          3507 0068frac14 150 kN=m2

                          810

                          Peak value of strain influence factor occurs at a depth of 27m and is given by

                          Izp frac14 05thorn 01130

                          16 27

                          05

                          frac14 067

                          Refer to Figure Q810

                          E frac14 25qc

                          Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                          Ez (mm3MN)

                          1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                          0203

                          C1 frac14 1 0500qnfrac14 1 05 12 16

                          130frac14 093

                          C2 frac14 1 ethsayTHORN

                          s frac14 C1C2qnX Iz

                          Ez frac14 093 1 130 0203 frac14 25mm

                          Bearing capacity 67

                          811

                          At pile base level

                          cu frac14 220 kN=m2

                          qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                          Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                          00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                          qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                          Then

                          Qf frac14 Abqb thorn Asqs

                          frac14

                          4 32 1980

                          thorn eth 105 139 86THORN

                          frac14 13 996thorn 3941 frac14 17 937 kN

                          0 01 02 03 04 05 06 07

                          0 2 4 6 8 10 12 14

                          1

                          2

                          3

                          4

                          5

                          6

                          7

                          8

                          (1)

                          (2)

                          (3)

                          (4)

                          (5)

                          qc

                          qc

                          Iz

                          Iz

                          (MNm2)

                          z (m)

                          Figure Q810

                          68 Bearing capacity

                          Allowable load

                          ethaTHORN Qf

                          2frac14 17 937

                          2frac14 8968 kN

                          ethbTHORN Abqb

                          3thorn Asqs frac14 13 996

                          3thorn 3941 frac14 8606 kN

                          ie allowable load frac14 8600 kN

                          Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                          According to the limit state method

                          Characteristic undrained strength at base level cuk frac14 220

                          150kN=m2

                          Characteristic base resistance qbk frac14 9cuk frac14 9 220

                          150frac14 1320 kN=m2

                          Characteristic shaft resistance qsk frac14 00150

                          frac14 86

                          150frac14 57 kN=m2

                          Characteristic base and shaft resistances

                          Rbk frac14

                          4 32 1320 frac14 9330 kN

                          Rsk frac14 105 139 86

                          150frac14 2629 kN

                          For a bored pile the partial factors are b frac14 160 and s frac14 130

                          Design bearing resistance Rcd frac14 9330

                          160thorn 2629

                          130

                          frac14 5831thorn 2022

                          frac14 7850 kN

                          Adding ethDAb W) the design bearing resistance becomes 9650 kN

                          812

                          ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                          qs frac14 cu frac14 040 105 frac14 42 kN=m2

                          For a single pile

                          Qf frac14 Abqb thorn Asqs

                          frac14

                          4 062 1305

                          thorn eth 06 15 42THORN

                          frac14 369thorn 1187 frac14 1556 kN

                          Bearing capacity 69

                          Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                          eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                          (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                          (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                          qbkfrac14 9cuk frac14 9 220

                          150frac14 1320 kN=m2

                          qskfrac14cuk frac14 040 105

                          150frac14 28 kN=m2

                          Rbkfrac14

                          4 0602 1320 frac14 373 kN

                          Rskfrac14 060 15 28 frac14 791 kN

                          Rcdfrac14 373

                          160thorn 791

                          130frac14 233thorn 608 frac14 841 kN

                          Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                          Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                          (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                          q frac14 21 000

                          1762frac14 68 kN=m2

                          Immediate settlement

                          H

                          Bfrac14 15

                          176frac14 085

                          D

                          Bfrac14 13

                          176frac14 074

                          L

                          Bfrac14 1

                          Hence from Figure 515

                          130 frac14 078 and 131 frac14 041

                          70 Bearing capacity

                          Thus using Equation 528

                          si frac14 078 041 68 176

                          65frac14 6mm

                          Consolidation settlement

                          Layer z (m) Area (m2) (kNm2) mvH (mm)

                          1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                          434 (sod)

                          Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                          sc frac14 056 434 frac14 24mm

                          The total settlement is (6thorn 24) frac14 30mm

                          813

                          At base level N frac14 26 Then using Equation 830

                          qb frac14 40NDb

                          Bfrac14 40 26 2

                          025frac14 8320 kN=m2

                          ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                          Figure Q812

                          Bearing capacity 71

                          Over the length embedded in sand

                          N frac14 21 ie18thorn 24

                          2

                          Using Equation 831

                          qs frac14 2N frac14 2 21 frac14 42 kN=m2

                          For a single pile

                          Qf frac14 Abqb thorn Asqs

                          frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                          For the pile group assuming a group efficiency of 12

                          XQf frac14 12 9 604 frac14 6523 kN

                          Then the load factor is

                          F frac14 6523

                          2000thorn 1000frac14 21

                          (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                          Characteristic base resistance per unit area qbk frac14 8320

                          150frac14 5547 kNm2

                          Characteristic shaft resistance per unit area qsk frac14 42

                          150frac14 28 kNm2

                          Characteristic base and shaft resistances for a single pile

                          Rbk frac14 0252 5547 frac14 347 kN

                          Rsk frac14 4 025 2 28 frac14 56 kN

                          For a driven pile the partial factors are b frac14 s frac14 130

                          Design bearing resistance Rcd frac14 347

                          130thorn 56

                          130frac14 310 kN

                          For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                          Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                          (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                          From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                          72 Bearing capacity

                          N frac14 24thorn 26thorn 34

                          3frac14 28

                          Ic frac14 171

                          2814frac14 0016 ethEquation 818THORN

                          s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                          814

                          Using Equation 841

                          Tf frac14 DLcu thorn

                          4ethD2 d2THORNcuNc

                          frac14 eth 02 5 06 110THORN thorn

                          4eth022 012THORN110 9

                          frac14 207thorn 23 frac14 230 kN

                          Figure Q813

                          Bearing capacity 73

                          Chapter 9

                          Stability of slopes

                          91

                          Referring to Figure Q91

                          W frac14 417 19 frac14 792 kN=m

                          Q frac14 20 28 frac14 56 kN=m

                          Arc lengthAB frac14

                          180 73 90 frac14 115m

                          Arc length BC frac14

                          180 28 90 frac14 44m

                          The factor of safety is given by

                          F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                          90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                          Depth of tension crack z0 frac14 2cu

                          frac14 2 20

                          19frac14 21m

                          Arc length BD frac14

                          180 13

                          1

                          2 90 frac14 21m

                          F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                          The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                          Design resisting moment frac14 rXethcudLaTHORN frac14 90

                          14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                          Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                          The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                          92

                          u frac14 0

                          Depth factor D frac14 11

                          9frac14 122

                          Using Equation 92 with F frac14 10

                          Ns frac14 cu

                          FHfrac14 30

                          10 19 9frac14 0175

                          Hence from Figure 93

                          frac14 50

                          For F frac14 12

                          Ns frac14 30

                          12 19 9frac14 0146

                          frac14 27

                          93

                          Refer to Figure Q93

                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                          1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                          74 m

                          214 1deg

                          213 1deg

                          39 m

                          WB

                          D

                          C

                          28 m

                          21 m

                          A

                          Q

                          Soil (1)Soil (2)

                          73deg

                          Figure Q91

                          Stability of slopes 75

                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                          9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                          599 256 328 1372

                          Figure Q93

                          76 Stability of slopes

                          XW cos frac14 b

                          Xh cos frac14 21 2 599 frac14 2516 kN=mX

                          W sin frac14 bX

                          h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                          Arc length La frac14

                          180 57

                          1

                          2 326 frac14 327m

                          The factor of safety is given by

                          F frac14 c0La thorn tan0ethW cos ulTHORN

                          W sin

                          frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                          frac14 091

                          According to the limit state method

                          0d frac14 tan1tan 32

                          125

                          frac14 265

                          c0 frac14 8

                          160frac14 5 kN=m2

                          Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                          Design disturbing moment frac14 1075 kN=m

                          The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                          94

                          F frac14 1

                          W sin

                          Xfc0bthorn ethW ubTHORN tan0g sec

                          1thorn ethtan tan0=FTHORN

                          c0 frac14 8 kN=m2

                          0 frac14 32

                          c0b frac14 8 2 frac14 16 kN=m

                          W frac14 bh frac14 21 2 h frac14 42h kN=m

                          Try F frac14 100

                          tan0

                          Ffrac14 0625

                          Stability of slopes 77

                          Values of u are as obtained in Figure Q93

                          SliceNo

                          h(m)

                          W frac14 bh(kNm)

                          W sin(kNm)

                          ub(kNm)

                          c0bthorn (W ub) tan0(kNm)

                          sec

                          1thorn (tan tan0)FProduct(kNm)

                          1 05 21 6 2 8 24 1078 262 13 55 31

                          23 33 30 1042 31

                          3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                          224 92 72 0931 67

                          6 50 210 11 40 100 85 0907 777 55 231 14

                          12 58 112 90 0889 80

                          8 60 252 1812

                          80 114 102 0874 899 63 265 22 99 116 109 0861 94

                          10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                          2154 88 116 0853 99

                          14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                          1074 1091

                          F frac14 1091

                          1074frac14 102 (assumed value 100)

                          Thus

                          F frac14 101

                          95

                          F frac14 1

                          W sin

                          XfWeth1 ruTHORN tan0g sec

                          1thorn ethtan tan0THORN=F

                          0 frac14 33

                          ru frac14 020

                          W frac14 bh frac14 20 5 h frac14 100h kN=m

                          eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                          Try F frac14 110

                          tan 0

                          Ffrac14 tan 33

                          110frac14 0590

                          78 Stability of slopes

                          Referring to Figure Q95

                          SliceNo

                          h(m)

                          W frac14 bh(kNm)

                          W sin(kNm)

                          W(1 ru) tan0(kNm)

                          sec

                          1thorn ( tan tan0)FProduct(kNm)

                          1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                          2120 234 0892 209

                          4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                          1185 1271

                          Figure Q95

                          Stability of slopes 79

                          F frac14 1271

                          1185frac14 107

                          The trial value was 110 therefore take F to be 108

                          96

                          (a) Water table at surface the factor of safety is given by Equation 912

                          F frac14 0

                          sat

                          tan0

                          tan

                          ptie 15 frac14 92

                          19

                          tan 36

                          tan

                          tan frac14 0234

                          frac14 13

                          Water table well below surface the factor of safety is given by Equation 911

                          F frac14 tan0

                          tan

                          frac14 tan 36

                          tan 13

                          frac14 31

                          (b) 0d frac14 tan1tan 36

                          125

                          frac14 30

                          Depth of potential failure surface frac14 z

                          Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                          frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                          frac14 504z kN

                          Design disturbing moment per unit area Sd frac14 sat sin cos

                          frac14 19 z sin 13 cos 13

                          frac14 416z kN

                          Rd gtSd therefore the limit state for overall stability is satisfied

                          80 Stability of slopes

                          • Book Cover
                          • Title
                          • Contents
                          • Basic characteristics of soils
                          • Seepage
                          • Effective stress
                          • Shear strength
                          • Stresses and displacements
                          • Lateral earth pressure
                          • Consolidation theory
                          • Bearing capacity
                          • Stability of slopes

                            u1 frac14 98 483 frac14 47 kN=m2

                            The uplift force on the base of the structure is equal to the area of the pressure diagramand is 316 kN per unit length

                            24

                            The flow net is drawn in Figure Q24 from which Nffrac14 100 and Ndfrac14 11 The overallloss in total head is 550m Then

                            q frac14 kh Nf

                            Ndfrac14 40 107 550 10

                            11frac14 20 106 m3=s per m

                            25

                            The flow net is drawn in Figure Q25 from which Nffrac14 42 and Ndfrac14 9 The overall lossin total head is 500m Then

                            q frac14 kh Nf

                            Ndfrac14 20 106 500 42

                            9frac14 47 106 m3=s per m

                            Figure Q24

                            8 Seepage

                            26

                            The scale transformation factor in the x direction is given by Equation 221 ie

                            xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                            ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

                            Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

                            k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                            pfrac14

                            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

                            p 107 frac14 30 107 m=s

                            Thus the quantity of seepage is given by

                            q frac14 k0h Nf

                            Ndfrac14 30 107 1400 325

                            12frac14 11 106 m3=s per m

                            Figure Q25

                            Seepage 9

                            27

                            The scale transformation factor in the x direction is

                            xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                            ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

                            Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

                            k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                            pfrac14

                            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

                            p 106 frac14 45 106 m=s

                            The focus of the basic parabola is at point A The parabola passes through point Gsuch that

                            GC frac14 03HC frac14 03 30 frac14 90m

                            Thus the coordinates of G are

                            x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

                            480 frac14 x0 2002

                            4x0

                            Figure Q26

                            10 Seepage

                            Hence

                            x0 frac14 20m

                            Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

                            x frac14 20 z2

                            80

                            x 20 0 50 100 200 300z 0 400 748 980 1327 1600

                            The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

                            No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

                            q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

                            28

                            The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

                            Figure Q27

                            Seepage 11

                            q frac14 kh Nf

                            Ndfrac14 45 105 28 33

                            7

                            frac14 59 105 m3=s per m

                            29

                            The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

                            kx frac14 H1k1 thornH2k2

                            H1 thornH2frac14 106

                            10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

                            kz frac14 H1 thornH2

                            H1

                            k1thornH2

                            k2

                            frac14 10

                            5

                            eth2 106THORN thorn5

                            eth16 106THORNfrac14 36 106 m=s

                            Then the scale transformation factor is given by

                            xt frac14 xffiffiffiffiffikz

                            pffiffiffiffiffikx

                            p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

                            Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

                            Figure Q28

                            12 Seepage

                            k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                            qfrac14

                            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

                            p 106 frac14 57 106 m=s

                            Then the quantity of seepage is given by

                            q frac14 k0h Nf

                            Ndfrac14 57 106 350 56

                            11

                            frac14 10 105 m3=s per m

                            Figure Q29

                            Seepage 13

                            Chapter 3

                            Effective stress

                            31

                            Buoyant unit weight

                            0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                            Effective vertical stress

                            0v frac14 5 102 frac14 51 kN=m2 or

                            Total vertical stress

                            v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

                            Pore water pressure

                            u frac14 7 98 frac14 686 kN=m2

                            Effective vertical stress

                            0v frac14 v u frac14 1196 686 frac14 51 kN=m2

                            32

                            Buoyant unit weight

                            0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                            Effective vertical stress

                            0v frac14 5 102 frac14 51 kN=m2 or

                            Total vertical stress

                            v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

                            Pore water pressure

                            u frac14 205 98 frac14 2009 kN=m2

                            Effective vertical stress

                            0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

                            33

                            At top of the clay

                            v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

                            u frac14 2 98 frac14 196 kN=m2

                            0v frac14 v u frac14 710 196 frac14 514 kN=m2

                            Alternatively

                            0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                            0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

                            At bottom of the clay

                            v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

                            u frac14 12 98 frac14 1176 kN=m2

                            0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

                            NB The alternative method of calculation is not applicable because of the artesiancondition

                            Figure Q3132

                            Effective stress 15

                            34

                            0 frac14 20 98 frac14 102 kN=m3

                            At 8m depth

                            0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                            35

                            0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                            0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                            Figure Q33

                            Figure Q34

                            16 Effective stress

                            (a) Immediately after WT rise

                            At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                            0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                            At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                            0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                            (b) Several years after WT rise

                            At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                            At 8m depth

                            0v frac14 940 kN=m2 (as above)

                            At 12m depth

                            0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                            Figure Q35

                            Effective stress 17

                            36

                            Total weight

                            ab frac14 210 kN

                            Effective weight

                            ac frac14 112 kN

                            Resultant boundary water force

                            be frac14 119 kN

                            Seepage force

                            ce frac14 34 kN

                            Resultant body force

                            ae frac14 99 kN eth73 to horizontalTHORN

                            (Refer to Figure Q36)

                            Figure Q36

                            18 Effective stress

                            37

                            Situation (1)(a)

                            frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                            u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                            0 frac14 u frac14 694 392 frac14 302 kN=m2

                            (b)

                            i frac14 2

                            4frac14 05

                            j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                            Situation (2)(a)

                            frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                            u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                            0 frac14 u frac14 498 392 frac14 106 kN=m2

                            (b)

                            i frac14 2

                            4frac14 05

                            j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                            38

                            The flow net is drawn in Figure Q24

                            Loss in total head between adjacent equipotentials

                            h frac14 550

                            Ndfrac14 550

                            11frac14 050m

                            Exit hydraulic gradient

                            ie frac14 h

                            sfrac14 050

                            070frac14 071

                            Effective stress 19

                            The critical hydraulic gradient is given by Equation 39

                            ic frac14 0

                            wfrac14 102

                            98frac14 104

                            Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                            F frac14 iciefrac14 104

                            071frac14 15

                            Total head at C

                            hC frac14 nd

                            Ndh frac14 24

                            11 550 frac14 120m

                            Elevation head at C

                            zC frac14 250m

                            Pore water pressure at C

                            uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                            Therefore effective vertical stress at C

                            0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                            For point D

                            hD frac14 73

                            11 550 frac14 365m

                            zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                            0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                            39

                            The flow net is drawn in Figure Q25

                            For a soil prism 150 300m adjacent to the piling

                            hm frac14 26

                            9 500 frac14 145m

                            20 Effective stress

                            Factor of safety against lsquoheavingrsquo (Equation 310)

                            F frac14 ic

                            imfrac14 0d

                            whmfrac14 97 300

                            98 145frac14 20

                            With a filter

                            F frac14 0d thorn wwhm

                            3 frac14 eth97 300THORN thorn w98 145

                            w frac14 135 kN=m2

                            Depth of filterfrac14 13521frac14 065m (if above water level)

                            Effective stress 21

                            Chapter 4

                            Shear strength

                            41

                            frac14 295 kN=m2

                            u frac14 120 kN=m2

                            0 frac14 u frac14 295 120 frac14 175 kN=m2

                            f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                            42

                            03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                            100 452 552200 908 1108400 1810 2210800 3624 4424

                            The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                            Figure Q42

                            43

                            3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                            200 222 422400 218 618600 220 820

                            The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                            44

                            The modified shear strength parameters are

                            0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                            a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                            The coordinates of the stress point representing failure conditions in the test are

                            1

                            2eth1 2THORN frac14 1

                            2 170 frac14 85 kN=m2

                            1

                            2eth1 thorn 3THORN frac14 1

                            2eth270thorn 100THORN frac14 185 kN=m2

                            The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                            uf frac14 36 kN=m2

                            Figure Q43

                            Figure Q44

                            Shear strength 23

                            45

                            3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                            150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                            The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                            The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                            46

                            03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                            200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                            The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                            Figure Q45

                            24 Shear strength

                            47

                            The torque required to produce shear failure is given by

                            T frac14 dh cud

                            2thorn 2

                            Z d=2

                            0

                            2r drcur

                            frac14 cud2h

                            2thorn 4cu

                            Z d=2

                            0

                            r2dr

                            frac14 cud2h

                            2thorn d

                            3

                            6

                            Then

                            35 frac14 cu52 10

                            2thorn 53

                            6

                            103

                            cu frac14 76 kN=m3

                            400

                            0 400 800 1200 1600

                            τ (k

                            Nm

                            2 )

                            σprime (kNm2)

                            34deg

                            315deg29deg

                            (a)

                            (b)

                            0 400

                            400

                            800 1200 1600

                            Failure envelope

                            300 500

                            σprime (kNm2)

                            τ (k

                            Nm

                            2 )

                            20 (kNm2)

                            31deg

                            Figure Q46

                            Shear strength 25

                            48

                            The relevant stress values are calculated as follows

                            3 frac14 600 kN=m2

                            1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                            2(1 3) 0 40 79 107 139 159

                            1

                            2(01 thorn 03) 400 411 402 389 351 326

                            1

                            2(1 thorn 3) 600 640 679 707 739 759

                            The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                            Af frac14 433 200

                            319frac14 073

                            49

                            B frac14 u33

                            frac14 144

                            350 200frac14 096

                            a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                            0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                            10 283 65 023

                            Figure Q48

                            26 Shear strength

                            The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                            Figure Q49

                            Shear strength 27

                            Chapter 5

                            Stresses and displacements

                            51

                            Vertical stress is given by

                            z frac14 Qz2Ip frac14 5000

                            52Ip

                            Values of Ip are obtained from Table 51

                            r (m) rz Ip z (kNm2)

                            0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                            10 20 0009 2

                            The variation of z with radial distance (r) is plotted in Figure Q51

                            Figure Q51

                            52

                            Below the centre load (Figure Q52)

                            r

                            zfrac14 0 for the 7500-kN load

                            Ip frac14 0478

                            r

                            zfrac14 5

                            4frac14 125 for the 10 000- and 9000-kN loads

                            Ip frac14 0045

                            Then

                            z frac14X Q

                            z2Ip

                            frac14 7500 0478

                            42thorn 10 000 0045

                            42thorn 9000 0045

                            42

                            frac14 224thorn 28thorn 25 frac14 277 kN=m2

                            53

                            The vertical stress under a corner of a rectangular area is given by

                            z frac14 qIr

                            where values of Ir are obtained from Figure 510 In this case

                            z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                            z

                            Figure Q52

                            Stresses and displacements 29

                            z (m) m n Ir z (kNm2)

                            0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                            10 010 0005 5

                            z is plotted against z in Figure Q53

                            54

                            (a)

                            m frac14 125

                            12frac14 104

                            n frac14 18

                            12frac14 150

                            From Figure 510 Irfrac14 0196

                            z frac14 2 175 0196 frac14 68 kN=m2

                            Figure Q53

                            30 Stresses and displacements

                            (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                            z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                            55

                            Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                            Px frac14 2Q

                            1

                            m2 thorn 1frac14 2 150

                            125frac14 76 kN=m

                            Equation 517 is used to obtain the pressure distribution

                            px frac14 4Q

                            h

                            m2n

                            ethm2 thorn n2THORN2 frac14150

                            m2n

                            ethm2 thorn n2THORN2 ethkN=m2THORN

                            Figure Q54

                            Stresses and displacements 31

                            n m2n

                            (m2 thorn n2)2

                            px(kNm2)

                            0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                            The pressure distribution is plotted in Figure Q55

                            56

                            H

                            Bfrac14 10

                            2frac14 5

                            L

                            Bfrac14 4

                            2frac14 2

                            D

                            Bfrac14 1

                            2frac14 05

                            Hence from Figure 515

                            131 frac14 082

                            130 frac14 094

                            Figure Q55

                            32 Stresses and displacements

                            The immediate settlement is given by Equation 528

                            si frac14 130131qB

                            Eu

                            frac14 094 082 200 2

                            45frac14 7mm

                            Stresses and displacements 33

                            Chapter 6

                            Lateral earth pressure

                            61

                            For 0 frac14 37 the active pressure coefficient is given by

                            Ka frac14 1 sin 37

                            1thorn sin 37frac14 025

                            The total active thrust (Equation 66a with c0 frac14 0) is

                            Pa frac14 1

                            2KaH

                            2 frac14 1

                            2 025 17 62 frac14 765 kN=m

                            If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                            K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                            and the thrust on the wall is

                            P0 frac14 1

                            2K0H

                            2 frac14 1

                            2 040 17 62 frac14 122 kN=m

                            62

                            The active pressure coefficients for the three soil types are as follows

                            Ka1 frac141 sin 35

                            1thorn sin 35frac14 0271

                            Ka2 frac141 sin 27

                            1thorn sin 27frac14 0375

                            ffiffiffiffiffiffiffiKa2

                            p frac14 0613

                            Ka3 frac141 sin 42

                            1thorn sin 42frac14 0198

                            Distribution of active pressure (plotted in Figure Q62)

                            Depth (m) Soil Active pressure (kNm2)

                            3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                            12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                            At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                            Total thrust frac14 571 kNm

                            Point of application is (4893571) m from the top of the wall ie 857m

                            Force (kN) Arm (m) Moment (kN m)

                            (1)1

                            2 0271 16 32 frac14 195 20 390

                            (2) 0271 16 3 2 frac14 260 40 1040

                            (3)1

                            2 0271 92 22 frac14 50 433 217

                            (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                            (5)1

                            2 0375 102 32 frac14 172 70 1204

                            (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                            (7)1

                            2 0198 112 42 frac14 177 1067 1889

                            (8)1

                            2 98 92 frac14 3969 90 35721

                            5713 48934

                            Figure Q62

                            Lateral earth pressure 35

                            63

                            (a) For u frac14 0 Ka frac14 Kp frac14 1

                            Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                            frac14 245

                            At the lower end of the piling

                            pa frac14 Kaqthorn Kasatz Kaccu

                            frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                            frac14 115 kN=m2

                            pp frac14 Kpsatzthorn Kpccu

                            frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                            frac14 202 kN=m2

                            (b) For 0 frac14 26 and frac14 1

                            20

                            Ka frac14 035

                            Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                            pfrac14 145 ethEquation 619THORN

                            Kp frac14 37

                            Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                            pfrac14 47 ethEquation 624THORN

                            At the lower end of the piling

                            pa frac14 Kaqthorn Ka0z Kacc

                            0

                            frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                            frac14 187 kN=m2

                            pp frac14 Kp0zthorn Kpcc

                            0

                            frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                            frac14 198 kN=m2

                            36 Lateral earth pressure

                            64

                            (a) For 0 frac14 38 Ka frac14 024

                            0 frac14 20 98 frac14 102 kN=m3

                            The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                            Force (kN) Arm (m) Moment (kN m)

                            (1) 024 10 66 frac14 159 33 525

                            (2)1

                            2 024 17 392 frac14 310 400 1240

                            (3) 024 17 39 27 frac14 430 135 580

                            (4)1

                            2 024 102 272 frac14 89 090 80

                            (5)1

                            2 98 272 frac14 357 090 321

                            Hfrac14 1345 MH frac14 2746

                            (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                            (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                            XM frac14MV MH frac14 7790 kNm

                            Lever arm of base resultant

                            M

                            Vfrac14 779

                            488frac14 160

                            Eccentricity of base resultant

                            e frac14 200 160 frac14 040m

                            39 m

                            27 m

                            40 m

                            04 m

                            04 m

                            26 m

                            (7)

                            (9)

                            (1)(2)

                            (3)

                            (4)

                            (5)

                            (8)(6)

                            (10)

                            WT

                            10 kNm2

                            Hydrostatic

                            Figure Q64

                            Lateral earth pressure 37

                            Base pressures (Equation 627)

                            p frac14 VB

                            1 6e

                            B

                            frac14 488

                            4eth1 060THORN

                            frac14 195 kN=m2 and 49 kN=m2

                            Factor of safety against sliding (Equation 628)

                            F frac14 V tan

                            Hfrac14 488 tan 25

                            1345frac14 17

                            (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                            Hfrac14 1633 kN

                            V frac14 4879 kN

                            MH frac14 3453 kNm

                            MV frac14 10536 kNm

                            The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                            65

                            For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                            Kp

                            Ffrac14 385

                            2

                            0 frac14 20 98 frac14 102 kN=m3

                            The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                            Force (kN) Arm (m) Moment (kN m)

                            (1)1

                            2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                            (2) 026 17 45 d frac14 199d d2 995d2

                            (3)1

                            2 026 102 d2 frac14 133d2 d3 044d3

                            (4)1

                            2 385

                            2 17 152 frac14 368 dthorn 05 368d 184

                            (5)385

                            2 17 15 d frac14 491d d2 2455d2

                            (6)1

                            2 385

                            2 102 d2 frac14 982d2 d3 327d3

                            38 Lateral earth pressure

                            XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                            d3 thorn 516d2 283d 1724 frac14 0

                            d frac14 179m

                            Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                            Over additional 20 embedded depth

                            pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                            Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                            66

                            The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                            Ka frac14 sin 69=sin 105

                            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                            pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                            26664

                            37775

                            2

                            frac14 050

                            The total active thrust (acting at 25 above the normal) is given by Equation 616

                            Pa frac14 1

                            2 050 19 7502 frac14 267 kN=m

                            Figure Q65

                            Lateral earth pressure 39

                            Horizontal component

                            Ph frac14 267 cos 40 frac14 205 kN=m

                            Vertical component

                            Pv frac14 267 sin 40 frac14 172 kN=m

                            Consider moments about the toe of the wall (Figure Q66) (per m)

                            Force (kN) Arm (m) Moment (kN m)

                            (1)1

                            2 175 650 235 frac14 1337 258 345

                            (2) 050 650 235 frac14 764 175 134

                            (3)1

                            2 070 650 235 frac14 535 127 68

                            (4) 100 400 235 frac14 940 200 188

                            (5) 1

                            2 080 050 235 frac14 47 027 1

                            Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                            Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                            Lever arm of base resultant

                            M

                            Vfrac14 795

                            525frac14 151m

                            Eccentricity of base resultant

                            e frac14 200 151 frac14 049m

                            Figure Q66

                            40 Lateral earth pressure

                            Base pressures (Equation 627)

                            p frac14 525

                            41 6 049

                            4

                            frac14 228 kN=m2 and 35 kN=m2

                            The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                            The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                            The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                            67

                            For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                            Force (kN) Arm (m) Moment (kNm)

                            (1)1

                            2 027 17 52 frac14 574 183 1050

                            (2) 027 17 5 3 frac14 689 500 3445

                            (3)1

                            2 027 102 32 frac14 124 550 682

                            (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                            (5)1

                            2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                            (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                            (7) 1

                            2 267

                            2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                            (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                            2 d frac14 163d d2thorn 650 82d2 1060d

                            Tie rod force per m frac14 T 0 0

                            XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                            d3 thorn 77d2 269d 1438 frac14 0

                            d frac14 467m

                            Depth of penetration frac14 12d frac14 560m

                            Lateral earth pressure 41

                            Algebraic sum of forces for d frac14 467m isX

                            F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                            T frac14 905 kN=m

                            Force in each tie rod frac14 25T frac14 226 kN

                            68

                            (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                            0 frac14 21 98 frac14 112 kN=m3

                            The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                            uC frac14 150

                            165 15 98 frac14 134 kN=m2

                            The average seepage pressure is

                            j frac14 15

                            165 98 frac14 09 kN=m3

                            Hence

                            0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                            0 j frac14 112 09 frac14 103 kN=m3

                            Figure Q67

                            42 Lateral earth pressure

                            Consider moments about the anchor point A (per m)

                            Force (kN) Arm (m) Moment (kN m)

                            (1) 10 026 150 frac14 390 60 2340

                            (2)1

                            2 026 18 452 frac14 474 15 711

                            (3) 026 18 45 105 frac14 2211 825 18240

                            (4)1

                            2 026 121 1052 frac14 1734 100 17340

                            (5)1

                            2 134 15 frac14 101 40 404

                            (6) 134 30 frac14 402 60 2412

                            (7)1

                            2 134 60 frac14 402 95 3819

                            571 4527(8) Ppm

                            115 115PPm

                            XM frac14 0

                            Ppm frac144527

                            115frac14 394 kN=m

                            Available passive resistance

                            Pp frac14 1

                            2 385 103 62 frac14 714 kN=m

                            Factor of safety

                            Fp frac14 Pp

                            Ppm

                            frac14 714

                            394frac14 18

                            Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                            Figure Q68

                            Lateral earth pressure 43

                            (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                            Consider moments (per m) about the tie point A

                            Force (kN) Arm (m)

                            (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                            (2)1

                            2 033 18 452 frac14 601 15

                            (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                            (4)1

                            2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                            (5)1

                            2 134 15 frac14 101 40

                            (6) 134 30 frac14 402 60

                            (7)1

                            2 134 d frac14 67d d3thorn 75

                            (8) 1

                            2 30 103 d2 frac141545d2 2d3thorn 75

                            Moment (kN m)

                            (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                            XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                            d3 thorn 827d2 466d 1518 frac14 0

                            By trial

                            d frac14 544m

                            The minimum depth of embedment required is 544m

                            69

                            For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                            0 frac14 20 98 frac14 102 kN=m3

                            The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                            44 Lateral earth pressure

                            uC frac14 147

                            173 26 98 frac14 216 kN=m2

                            and the average seepage pressure around the wall is

                            j frac14 26

                            173 98 frac14 15 kN=m3

                            Consider moments about the prop (A) (per m)

                            Force (kN) Arm (m) Moment (kN m)

                            (1)1

                            2 03 17 272 frac14 186 020 37

                            (2) 03 17 27 53 frac14 730 335 2445

                            (3)1

                            2 03 (102thorn 15) 532 frac14 493 423 2085

                            (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                            (5)1

                            2 216 26 frac14 281 243 684

                            (6) 216 27 frac14 583 465 2712

                            (7)1

                            2 216 60 frac14 648 800 5184

                            3055(8)

                            1

                            2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                            Factor of safety

                            Fr frac14 6885

                            3055frac14 225

                            Figure Q69

                            Lateral earth pressure 45

                            610

                            For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                            p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                            Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                            Using the recommendations of Twine and Roscoe

                            p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                            Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                            611

                            frac14 18 kN=m3 0 frac14 34

                            H frac14 350m nH frac14 335m mH frac14 185m

                            Consider a trial value of F frac14 20 Refer to Figure 635

                            0m frac14 tan1tan 34

                            20

                            frac14 186

                            Then

                            frac14 45 thorn 0m2frac14 543

                            W frac14 1

                            2 18 3502 cot 543 frac14 792 kN=m

                            Figure Q610

                            46 Lateral earth pressure

                            P frac14 1

                            2 s 3352 frac14 561s kN=m

                            U frac14 1

                            2 98 1852 cosec 543 frac14 206 kN=m

                            Equations 630 and 631 then become

                            561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                            792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                            ie

                            561s 0616N 405 frac14 0

                            792 0857N thorn 563 frac14 0

                            N frac14 848

                            0857frac14 989 kN=m

                            Then

                            561s 609 405 frac14 0

                            s frac14 649

                            561frac14 116 kN=m3

                            The calculations for trial values of F of 20 15 and 10 are summarized below

                            F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                            20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                            s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                            Figure Q611

                            Lateral earth pressure 47

                            612

                            For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                            45 thorn 0

                            2frac14 63

                            For the retained material between the surface and a depth of 36m

                            Pa frac14 1

                            2 030 18 362 frac14 350 kN=m

                            Weight of reinforced fill between the surface and a depth of 36m is

                            Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                            eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                            Eccentricity of Rv

                            e frac14 263 250 frac14 013m

                            The average vertical stress at a depth of 36m is

                            z frac14 Rv

                            L 2efrac14 324

                            474frac14 68 kN=m2

                            (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                            Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                            Tensile stress in the element frac14 138 103

                            65 3frac14 71N=mm2

                            Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                            Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                            Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                            The weight of ABC is

                            W frac14 1

                            2 18 52 265 frac14 124 kN=m

                            From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                            48 Lateral earth pressure

                            (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                            Tp frac14 032 68 120 065 frac14 170 kN

                            Tr frac14 213 420

                            418frac14 214 kN

                            Again the tensile failure and slipping limit states are satisfied for this element

                            Figure Q612

                            Lateral earth pressure 49

                            Chapter 7

                            Consolidation theory

                            71

                            Total change in thickness

                            H frac14 782 602 frac14 180mm

                            Average thickness frac14 1530thorn 180

                            2frac14 1620mm

                            Length of drainage path d frac14 1620

                            2frac14 810mm

                            Root time plot (Figure Q71a)

                            ffiffiffiffiffiffit90p frac14 33

                            t90 frac14 109min

                            cv frac14 0848d2

                            t90frac14 0848 8102

                            109 1440 365

                            106frac14 27m2=year

                            r0 frac14 782 764

                            782 602frac14 018

                            180frac14 0100

                            rp frac14 10eth764 645THORN9eth782 602THORN frac14

                            10 119

                            9 180frac14 0735

                            rs frac14 1 eth0100thorn 0735THORN frac14 0165

                            Log time plot (Figure Q71b)

                            t50 frac14 26min

                            cv frac14 0196d2

                            t50frac14 0196 8102

                            26 1440 365

                            106frac14 26m2=year

                            r0 frac14 782 763

                            782 602frac14 019

                            180frac14 0106

                            rp frac14 763 623

                            782 602frac14 140

                            180frac14 0778

                            rs frac14 1 eth0106thorn 0778THORN frac14 0116

                            Figure Q71(a)

                            Figure Q71(b)

                            Final void ratio

                            e1 frac14 w1Gs frac14 0232 272 frac14 0631

                            e

                            Hfrac14 1thorn e0

                            H0frac14 1thorn e1 thorne

                            H0

                            ie

                            e

                            180frac14 1631thorne

                            1710

                            e frac14 2936

                            1530frac14 0192

                            Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                            mv frac14 1

                            1thorn e0 e0 e101 00

                            frac14 1

                            1823 0192

                            0107frac14 098m2=MN

                            k frac14 cvmvw frac14 265 098 98

                            60 1440 365 103frac14 81 1010 m=s

                            72

                            Using Equation 77 (one-dimensional method)

                            sc frac14 e0 e11thorn e0 H

                            Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                            Figure Q72

                            52 Consolidation theory

                            Settlement

                            Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                            1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                            318

                            Notes 5 92y 460thorn 84

                            Heave

                            Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                            1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                            38

                            73

                            U frac14 f ethTvTHORN frac14 f cvt

                            d2

                            Hence if cv is constant

                            t1

                            t2frac14 d

                            21

                            d22

                            where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                            d1 frac14 95mm and d2 frac14 2500mm

                            for U frac14 050 t2 frac14 t1 d22

                            d21

                            frac14 20

                            60 24 365 25002

                            952frac14 263 years

                            for U lt 060 Tv frac14

                            4U2 (Equation 724(a))

                            t030 frac14 t050 0302

                            0502

                            frac14 263 036 frac14 095 years

                            Consolidation theory 53

                            74

                            The layer is open

                            d frac14 8

                            2frac14 4m

                            Tv frac14 cvtd2frac14 24 3

                            42frac14 0450

                            ui frac14 frac14 84 kN=m2

                            The excess pore water pressure is given by Equation 721

                            ue frac14Xmfrac141mfrac140

                            2ui

                            Msin

                            Mz

                            d

                            expethM2TvTHORN

                            In this case z frac14 d

                            sinMz

                            d

                            frac14 sinM

                            where

                            M frac14

                            23

                            25

                            2

                            M sin M M2Tv exp (M2Tv)

                            2thorn1 1110 0329

                            3

                            21 9993 457 105

                            ue frac14 2 84 2

                            1 0329 ethother terms negligibleTHORN

                            frac14 352 kN=m2

                            75

                            The layer is open

                            d frac14 6

                            2frac14 3m

                            Tv frac14 cvtd2frac14 10 3

                            32frac14 0333

                            The layer thickness will be divided into six equal parts ie m frac14 6

                            54 Consolidation theory

                            For an open layer

                            Tv frac14 4n

                            m2

                            n frac14 0333 62

                            4frac14 300

                            The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                            ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                            i j

                            0 1 2 3 4 5 6 7 8 9 10 11 12

                            0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                            The initial and 3-year isochrones are plotted in Figure Q75

                            Area under initial isochrone frac14 180 units

                            Area under 3-year isochrone frac14 63 units

                            The average degree of consolidation is given by Equation 725Thus

                            U frac14 1 63

                            180frac14 065

                            Figure Q75

                            Consolidation theory 55

                            76

                            At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                            0 frac14 2w frac14 2 98 frac14 196 kN=m2

                            The final consolidation settlement (one-dimensional method) is

                            sc frac14 mv0H frac14 083 196 8 frac14 130mm

                            Corrected time t frac14 2 1

                            2

                            40

                            52

                            frac14 1615 years

                            Tv frac14 cvtd2frac14 44 1615

                            42frac14 0444

                            From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                            77

                            The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                            Figure Q77

                            56 Consolidation theory

                            Point m n Ir (kNm2) sc (mm)

                            13020frac14 15 20

                            20frac14 10 0194 (4) 113 124

                            260

                            20frac14 30

                            20

                            20frac14 10 0204 (2) 59 65

                            360

                            20frac14 30

                            40

                            20frac14 20 0238 (1) 35 38

                            430

                            20frac14 15

                            40

                            20frac14 20 0224 (2) 65 72

                            Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                            78

                            Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                            (a) Immediate settlement

                            H

                            Bfrac14 30

                            35frac14 086

                            D

                            Bfrac14 2

                            35frac14 006

                            Figure Q78

                            Consolidation theory 57

                            From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                            si frac14 130131qB

                            Eufrac14 10 032 105 35

                            40frac14 30mm

                            (b) Consolidation settlement

                            Layer z (m) Dz Ic (kNm2) syod (mm)

                            1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                            3150

                            Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                            Now

                            H

                            Bfrac14 30

                            35frac14 086 and A frac14 065

                            from Figure 712 13 frac14 079

                            sc frac14 13sod frac14 079 315 frac14 250mm

                            Total settlement

                            s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                            79

                            Without sand drains

                            Uv frac14 025

                            Tv frac14 0049 ethfrom Figure 718THORN

                            t frac14 Tvd2

                            cvfrac14 0049 82

                            cvWith sand drains

                            R frac14 0564S frac14 0564 3 frac14 169m

                            n frac14 Rrfrac14 169

                            015frac14 113

                            Tr frac14 cht

                            4R2frac14 ch

                            4 1692 0049 82

                            cvethand ch frac14 cvTHORN

                            frac14 0275

                            Ur frac14 073 (from Figure 730)

                            58 Consolidation theory

                            Using Equation 740

                            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                            U frac14 080

                            710

                            Without sand drains

                            Uv frac14 090

                            Tv frac14 0848

                            t frac14 Tvd2

                            cvfrac14 0848 102

                            96frac14 88 years

                            With sand drains

                            R frac14 0564S frac14 0564 4 frac14 226m

                            n frac14 Rrfrac14 226

                            015frac14 15

                            Tr

                            Tvfrac14 chcv

                            d2

                            4R2ethsame tTHORN

                            Tr

                            Tvfrac14 140

                            96 102

                            4 2262frac14 714 eth1THORN

                            Using Equation 740

                            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                            An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                            Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                            040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                            Thus

                            Uv frac14 0295 and Ur frac14 086

                            t frac14 88 00683

                            0848frac14 07 years

                            Consolidation theory 59

                            Chapter 8

                            Bearing capacity

                            81

                            (a) The ultimate bearing capacity is given by Equation 83

                            qf frac14 cNc thorn DNq thorn 1

                            2BN

                            For u frac14 0

                            Nc frac14 514 Nq frac14 1 N frac14 0

                            qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                            The net ultimate bearing capacity is

                            qnf frac14 qf D frac14 540 kN=m2

                            The net foundation pressure is

                            qn frac14 q D frac14 425

                            2 eth21 1THORN frac14 192 kN=m2

                            The factor of safety (Equation 86) is

                            F frac14 qnfqnfrac14 540

                            192frac14 28

                            (b) For 0 frac14 28

                            Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                            qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                            2 112 2 13

                            frac14 260thorn 168thorn 146 frac14 574 kN=m2

                            qnf frac14 574 112 frac14 563 kN=m2

                            F frac14 563

                            192frac14 29

                            (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                            82

                            For 0 frac14 38

                            Nq frac14 49 N frac14 67

                            qnf frac14 DethNq 1THORN thorn 1

                            2BN ethfrom Equation 83THORN

                            frac14 eth18 075 48THORN thorn 1

                            2 18 15 67

                            frac14 648thorn 905 frac14 1553 kN=m2

                            qn frac14 500

                            15 eth18 075THORN frac14 320 kN=m2

                            F frac14 qnfqnfrac14 1553

                            320frac14 48

                            0d frac14 tan1tan 38

                            125

                            frac14 32 therefore Nq frac14 23 and N frac14 25

                            Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                            2 18 15 25

                            frac14 15eth310thorn 337THORNfrac14 970 kN=m

                            Design load (action) Vd frac14 500 kN=m

                            The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                            83

                            D

                            Bfrac14 350

                            225frac14 155

                            From Figure 85 for a square foundation

                            Nc frac14 81

                            Bearing capacity 61

                            For a rectangular foundation (L frac14 450m B frac14 225m)

                            Nc frac14 084thorn 016B

                            L

                            81 frac14 745

                            Using Equation 810

                            qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                            For F frac14 3

                            qn frac14 1006

                            3frac14 335 kN=m2

                            q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                            Design load frac14 405 450 225 frac14 4100 kN

                            Design undrained strength cud frac14 135

                            14frac14 96 kN=m2

                            Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                            frac14 7241 kN

                            Design load Vd frac14 4100 kN

                            Rd gt Vd therefore the bearing resistance limit state is satisfied

                            84

                            For 0 frac14 40

                            Nq frac14 64 N frac14 95

                            qnf frac14 DethNq 1THORN thorn 04BN

                            (a) Water table 5m below ground level

                            qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                            qn frac14 400 17 frac14 383 kN=m2

                            F frac14 2686

                            383frac14 70

                            (b) Water table 1m below ground level (ie at foundation level)

                            0 frac14 20 98 frac14 102 kN=m3

                            62 Bearing capacity

                            qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                            F frac14 2040

                            383frac14 53

                            (c) Water table at ground level with upward hydraulic gradient 02

                            eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                            qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                            F frac14 1296

                            392frac14 33

                            85

                            The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                            Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                            Design value of 0 frac14 tan1tan 39

                            125

                            frac14 33

                            For 0 frac14 33 Nq frac14 26 and N frac14 29

                            Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                            Rd gt Vd therefore the bearing resistance limit state is satisfied

                            86

                            (a) Undrained shear for u frac14 0

                            Nc frac14 514 Nq frac14 1 N frac14 0

                            qnf frac14 12cuNc

                            frac14 12 100 514 frac14 617 kN=m2

                            qn frac14 qnfFfrac14 617

                            3frac14 206 kN=m2

                            q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                            Bearing capacity 63

                            Drained shear for 0 frac14 32

                            Nq frac14 23 N frac14 25

                            0 frac14 21 98 frac14 112 kN=m3

                            qnf frac14 0DethNq 1THORN thorn 040BN

                            frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                            frac14 694 kN=m2

                            q frac14 694

                            3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                            Design load frac14 42 227 frac14 3632 kN

                            (b) Design undrained strength cud frac14 100

                            14frac14 71 kNm2

                            Design bearing resistance Rd frac14 12cudNe area

                            frac14 12 71 514 42

                            frac14 7007 kN

                            For drained shear 0d frac14 tan1tan 32

                            125

                            frac14 26

                            Nq frac14 12 N frac14 10

                            Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                            (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                            Layer z (m) m n Ir 0 (kNm2) sod (mm)

                            1 2 100 0175 0700qn 0182qn

                            2 6 033 0044 0176qn 0046qn

                            3 10 020 0017 0068qn 0018qn

                            0246qn

                            Diameter of equivalent circle B frac14 45m

                            H

                            Bfrac14 12

                            45frac14 27 and A frac14 042

                            13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                            64 Bearing capacity

                            For sc frac14 30mm

                            qn frac14 30

                            0147frac14 204 kN=m2

                            q frac14 204thorn 21 frac14 225 kN=m2

                            Design load frac14 42 225 frac14 3600 kN

                            The design load is 3600 kN settlement being the limiting criterion

                            87

                            D

                            Bfrac14 8

                            4frac14 20

                            From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                            F frac14 cuNc

                            Dfrac14 40 71

                            20 8frac14 18

                            88

                            Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                            Design value of 0 frac14 tan1tan 38

                            125

                            frac14 32

                            Figure Q86

                            Bearing capacity 65

                            For 0 frac14 32 Nq frac14 23 and N frac14 25

                            Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                            The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                            Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                            For B frac14 250m qn frac14 3750

                            2502 17 frac14 583 kN=m2

                            From Figure 510 m frac14 n frac14 126

                            6frac14 021

                            Ir frac14 0019

                            Stress increment frac14 4 0019 583 frac14 44 kN=m2

                            Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                            The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                            89

                            Depth (m) N 0v (kNm2) CN N1

                            070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                            Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                            (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                            Cw frac14 05thorn 0530

                            47

                            frac14 082

                            66 Bearing capacity

                            Thus

                            qa frac14 150 082 frac14 120 kN=m2

                            (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                            Thus

                            qa frac14 90 15 frac14 135 kN=m2

                            (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                            Ic frac14 171

                            1014frac14 0068

                            From Equation 819(a) with s frac14 25mm

                            q frac14 25

                            3507 0068frac14 150 kN=m2

                            810

                            Peak value of strain influence factor occurs at a depth of 27m and is given by

                            Izp frac14 05thorn 01130

                            16 27

                            05

                            frac14 067

                            Refer to Figure Q810

                            E frac14 25qc

                            Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                            Ez (mm3MN)

                            1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                            0203

                            C1 frac14 1 0500qnfrac14 1 05 12 16

                            130frac14 093

                            C2 frac14 1 ethsayTHORN

                            s frac14 C1C2qnX Iz

                            Ez frac14 093 1 130 0203 frac14 25mm

                            Bearing capacity 67

                            811

                            At pile base level

                            cu frac14 220 kN=m2

                            qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                            Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                            00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                            qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                            Then

                            Qf frac14 Abqb thorn Asqs

                            frac14

                            4 32 1980

                            thorn eth 105 139 86THORN

                            frac14 13 996thorn 3941 frac14 17 937 kN

                            0 01 02 03 04 05 06 07

                            0 2 4 6 8 10 12 14

                            1

                            2

                            3

                            4

                            5

                            6

                            7

                            8

                            (1)

                            (2)

                            (3)

                            (4)

                            (5)

                            qc

                            qc

                            Iz

                            Iz

                            (MNm2)

                            z (m)

                            Figure Q810

                            68 Bearing capacity

                            Allowable load

                            ethaTHORN Qf

                            2frac14 17 937

                            2frac14 8968 kN

                            ethbTHORN Abqb

                            3thorn Asqs frac14 13 996

                            3thorn 3941 frac14 8606 kN

                            ie allowable load frac14 8600 kN

                            Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                            According to the limit state method

                            Characteristic undrained strength at base level cuk frac14 220

                            150kN=m2

                            Characteristic base resistance qbk frac14 9cuk frac14 9 220

                            150frac14 1320 kN=m2

                            Characteristic shaft resistance qsk frac14 00150

                            frac14 86

                            150frac14 57 kN=m2

                            Characteristic base and shaft resistances

                            Rbk frac14

                            4 32 1320 frac14 9330 kN

                            Rsk frac14 105 139 86

                            150frac14 2629 kN

                            For a bored pile the partial factors are b frac14 160 and s frac14 130

                            Design bearing resistance Rcd frac14 9330

                            160thorn 2629

                            130

                            frac14 5831thorn 2022

                            frac14 7850 kN

                            Adding ethDAb W) the design bearing resistance becomes 9650 kN

                            812

                            ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                            qs frac14 cu frac14 040 105 frac14 42 kN=m2

                            For a single pile

                            Qf frac14 Abqb thorn Asqs

                            frac14

                            4 062 1305

                            thorn eth 06 15 42THORN

                            frac14 369thorn 1187 frac14 1556 kN

                            Bearing capacity 69

                            Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                            eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                            (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                            (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                            qbkfrac14 9cuk frac14 9 220

                            150frac14 1320 kN=m2

                            qskfrac14cuk frac14 040 105

                            150frac14 28 kN=m2

                            Rbkfrac14

                            4 0602 1320 frac14 373 kN

                            Rskfrac14 060 15 28 frac14 791 kN

                            Rcdfrac14 373

                            160thorn 791

                            130frac14 233thorn 608 frac14 841 kN

                            Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                            Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                            (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                            q frac14 21 000

                            1762frac14 68 kN=m2

                            Immediate settlement

                            H

                            Bfrac14 15

                            176frac14 085

                            D

                            Bfrac14 13

                            176frac14 074

                            L

                            Bfrac14 1

                            Hence from Figure 515

                            130 frac14 078 and 131 frac14 041

                            70 Bearing capacity

                            Thus using Equation 528

                            si frac14 078 041 68 176

                            65frac14 6mm

                            Consolidation settlement

                            Layer z (m) Area (m2) (kNm2) mvH (mm)

                            1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                            434 (sod)

                            Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                            sc frac14 056 434 frac14 24mm

                            The total settlement is (6thorn 24) frac14 30mm

                            813

                            At base level N frac14 26 Then using Equation 830

                            qb frac14 40NDb

                            Bfrac14 40 26 2

                            025frac14 8320 kN=m2

                            ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                            Figure Q812

                            Bearing capacity 71

                            Over the length embedded in sand

                            N frac14 21 ie18thorn 24

                            2

                            Using Equation 831

                            qs frac14 2N frac14 2 21 frac14 42 kN=m2

                            For a single pile

                            Qf frac14 Abqb thorn Asqs

                            frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                            For the pile group assuming a group efficiency of 12

                            XQf frac14 12 9 604 frac14 6523 kN

                            Then the load factor is

                            F frac14 6523

                            2000thorn 1000frac14 21

                            (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                            Characteristic base resistance per unit area qbk frac14 8320

                            150frac14 5547 kNm2

                            Characteristic shaft resistance per unit area qsk frac14 42

                            150frac14 28 kNm2

                            Characteristic base and shaft resistances for a single pile

                            Rbk frac14 0252 5547 frac14 347 kN

                            Rsk frac14 4 025 2 28 frac14 56 kN

                            For a driven pile the partial factors are b frac14 s frac14 130

                            Design bearing resistance Rcd frac14 347

                            130thorn 56

                            130frac14 310 kN

                            For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                            Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                            (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                            From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                            72 Bearing capacity

                            N frac14 24thorn 26thorn 34

                            3frac14 28

                            Ic frac14 171

                            2814frac14 0016 ethEquation 818THORN

                            s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                            814

                            Using Equation 841

                            Tf frac14 DLcu thorn

                            4ethD2 d2THORNcuNc

                            frac14 eth 02 5 06 110THORN thorn

                            4eth022 012THORN110 9

                            frac14 207thorn 23 frac14 230 kN

                            Figure Q813

                            Bearing capacity 73

                            Chapter 9

                            Stability of slopes

                            91

                            Referring to Figure Q91

                            W frac14 417 19 frac14 792 kN=m

                            Q frac14 20 28 frac14 56 kN=m

                            Arc lengthAB frac14

                            180 73 90 frac14 115m

                            Arc length BC frac14

                            180 28 90 frac14 44m

                            The factor of safety is given by

                            F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                            90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                            Depth of tension crack z0 frac14 2cu

                            frac14 2 20

                            19frac14 21m

                            Arc length BD frac14

                            180 13

                            1

                            2 90 frac14 21m

                            F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                            The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                            Design resisting moment frac14 rXethcudLaTHORN frac14 90

                            14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                            Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                            The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                            92

                            u frac14 0

                            Depth factor D frac14 11

                            9frac14 122

                            Using Equation 92 with F frac14 10

                            Ns frac14 cu

                            FHfrac14 30

                            10 19 9frac14 0175

                            Hence from Figure 93

                            frac14 50

                            For F frac14 12

                            Ns frac14 30

                            12 19 9frac14 0146

                            frac14 27

                            93

                            Refer to Figure Q93

                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                            1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                            74 m

                            214 1deg

                            213 1deg

                            39 m

                            WB

                            D

                            C

                            28 m

                            21 m

                            A

                            Q

                            Soil (1)Soil (2)

                            73deg

                            Figure Q91

                            Stability of slopes 75

                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                            9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                            599 256 328 1372

                            Figure Q93

                            76 Stability of slopes

                            XW cos frac14 b

                            Xh cos frac14 21 2 599 frac14 2516 kN=mX

                            W sin frac14 bX

                            h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                            Arc length La frac14

                            180 57

                            1

                            2 326 frac14 327m

                            The factor of safety is given by

                            F frac14 c0La thorn tan0ethW cos ulTHORN

                            W sin

                            frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                            frac14 091

                            According to the limit state method

                            0d frac14 tan1tan 32

                            125

                            frac14 265

                            c0 frac14 8

                            160frac14 5 kN=m2

                            Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                            Design disturbing moment frac14 1075 kN=m

                            The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                            94

                            F frac14 1

                            W sin

                            Xfc0bthorn ethW ubTHORN tan0g sec

                            1thorn ethtan tan0=FTHORN

                            c0 frac14 8 kN=m2

                            0 frac14 32

                            c0b frac14 8 2 frac14 16 kN=m

                            W frac14 bh frac14 21 2 h frac14 42h kN=m

                            Try F frac14 100

                            tan0

                            Ffrac14 0625

                            Stability of slopes 77

                            Values of u are as obtained in Figure Q93

                            SliceNo

                            h(m)

                            W frac14 bh(kNm)

                            W sin(kNm)

                            ub(kNm)

                            c0bthorn (W ub) tan0(kNm)

                            sec

                            1thorn (tan tan0)FProduct(kNm)

                            1 05 21 6 2 8 24 1078 262 13 55 31

                            23 33 30 1042 31

                            3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                            224 92 72 0931 67

                            6 50 210 11 40 100 85 0907 777 55 231 14

                            12 58 112 90 0889 80

                            8 60 252 1812

                            80 114 102 0874 899 63 265 22 99 116 109 0861 94

                            10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                            2154 88 116 0853 99

                            14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                            1074 1091

                            F frac14 1091

                            1074frac14 102 (assumed value 100)

                            Thus

                            F frac14 101

                            95

                            F frac14 1

                            W sin

                            XfWeth1 ruTHORN tan0g sec

                            1thorn ethtan tan0THORN=F

                            0 frac14 33

                            ru frac14 020

                            W frac14 bh frac14 20 5 h frac14 100h kN=m

                            eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                            Try F frac14 110

                            tan 0

                            Ffrac14 tan 33

                            110frac14 0590

                            78 Stability of slopes

                            Referring to Figure Q95

                            SliceNo

                            h(m)

                            W frac14 bh(kNm)

                            W sin(kNm)

                            W(1 ru) tan0(kNm)

                            sec

                            1thorn ( tan tan0)FProduct(kNm)

                            1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                            2120 234 0892 209

                            4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                            1185 1271

                            Figure Q95

                            Stability of slopes 79

                            F frac14 1271

                            1185frac14 107

                            The trial value was 110 therefore take F to be 108

                            96

                            (a) Water table at surface the factor of safety is given by Equation 912

                            F frac14 0

                            sat

                            tan0

                            tan

                            ptie 15 frac14 92

                            19

                            tan 36

                            tan

                            tan frac14 0234

                            frac14 13

                            Water table well below surface the factor of safety is given by Equation 911

                            F frac14 tan0

                            tan

                            frac14 tan 36

                            tan 13

                            frac14 31

                            (b) 0d frac14 tan1tan 36

                            125

                            frac14 30

                            Depth of potential failure surface frac14 z

                            Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                            frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                            frac14 504z kN

                            Design disturbing moment per unit area Sd frac14 sat sin cos

                            frac14 19 z sin 13 cos 13

                            frac14 416z kN

                            Rd gtSd therefore the limit state for overall stability is satisfied

                            80 Stability of slopes

                            • Book Cover
                            • Title
                            • Contents
                            • Basic characteristics of soils
                            • Seepage
                            • Effective stress
                            • Shear strength
                            • Stresses and displacements
                            • Lateral earth pressure
                            • Consolidation theory
                            • Bearing capacity
                            • Stability of slopes

                              26

                              The scale transformation factor in the x direction is given by Equation 221 ie

                              xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                              ffiffiffiffiffiffiffi18pffiffiffiffiffiffiffi50p frac14 060x

                              Thus in the transformed section the horizontal dimension 3300m becomes(3300 060) ie 1980m and the slope 15 becomes 13 All dimensions in thevertical direction are unchanged The transformed section is shown in Figure Q26and the flow net is drawn as for the isotropic case From the flow net Nffrac14 325 andNdfrac14 12 The overall loss in total head is 1400m The equivalent isotropic perme-ability applying to the transformed section is given by Equation 223 ie

                              k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                              pfrac14

                              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth50 18THORN

                              p 107 frac14 30 107 m=s

                              Thus the quantity of seepage is given by

                              q frac14 k0h Nf

                              Ndfrac14 30 107 1400 325

                              12frac14 11 106 m3=s per m

                              Figure Q25

                              Seepage 9

                              27

                              The scale transformation factor in the x direction is

                              xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                              ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

                              Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

                              k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                              pfrac14

                              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

                              p 106 frac14 45 106 m=s

                              The focus of the basic parabola is at point A The parabola passes through point Gsuch that

                              GC frac14 03HC frac14 03 30 frac14 90m

                              Thus the coordinates of G are

                              x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

                              480 frac14 x0 2002

                              4x0

                              Figure Q26

                              10 Seepage

                              Hence

                              x0 frac14 20m

                              Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

                              x frac14 20 z2

                              80

                              x 20 0 50 100 200 300z 0 400 748 980 1327 1600

                              The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

                              No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

                              q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

                              28

                              The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

                              Figure Q27

                              Seepage 11

                              q frac14 kh Nf

                              Ndfrac14 45 105 28 33

                              7

                              frac14 59 105 m3=s per m

                              29

                              The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

                              kx frac14 H1k1 thornH2k2

                              H1 thornH2frac14 106

                              10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

                              kz frac14 H1 thornH2

                              H1

                              k1thornH2

                              k2

                              frac14 10

                              5

                              eth2 106THORN thorn5

                              eth16 106THORNfrac14 36 106 m=s

                              Then the scale transformation factor is given by

                              xt frac14 xffiffiffiffiffikz

                              pffiffiffiffiffikx

                              p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

                              Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

                              Figure Q28

                              12 Seepage

                              k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                              qfrac14

                              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

                              p 106 frac14 57 106 m=s

                              Then the quantity of seepage is given by

                              q frac14 k0h Nf

                              Ndfrac14 57 106 350 56

                              11

                              frac14 10 105 m3=s per m

                              Figure Q29

                              Seepage 13

                              Chapter 3

                              Effective stress

                              31

                              Buoyant unit weight

                              0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                              Effective vertical stress

                              0v frac14 5 102 frac14 51 kN=m2 or

                              Total vertical stress

                              v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

                              Pore water pressure

                              u frac14 7 98 frac14 686 kN=m2

                              Effective vertical stress

                              0v frac14 v u frac14 1196 686 frac14 51 kN=m2

                              32

                              Buoyant unit weight

                              0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                              Effective vertical stress

                              0v frac14 5 102 frac14 51 kN=m2 or

                              Total vertical stress

                              v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

                              Pore water pressure

                              u frac14 205 98 frac14 2009 kN=m2

                              Effective vertical stress

                              0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

                              33

                              At top of the clay

                              v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

                              u frac14 2 98 frac14 196 kN=m2

                              0v frac14 v u frac14 710 196 frac14 514 kN=m2

                              Alternatively

                              0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                              0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

                              At bottom of the clay

                              v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

                              u frac14 12 98 frac14 1176 kN=m2

                              0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

                              NB The alternative method of calculation is not applicable because of the artesiancondition

                              Figure Q3132

                              Effective stress 15

                              34

                              0 frac14 20 98 frac14 102 kN=m3

                              At 8m depth

                              0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                              35

                              0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                              0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                              Figure Q33

                              Figure Q34

                              16 Effective stress

                              (a) Immediately after WT rise

                              At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                              0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                              At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                              0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                              (b) Several years after WT rise

                              At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                              At 8m depth

                              0v frac14 940 kN=m2 (as above)

                              At 12m depth

                              0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                              Figure Q35

                              Effective stress 17

                              36

                              Total weight

                              ab frac14 210 kN

                              Effective weight

                              ac frac14 112 kN

                              Resultant boundary water force

                              be frac14 119 kN

                              Seepage force

                              ce frac14 34 kN

                              Resultant body force

                              ae frac14 99 kN eth73 to horizontalTHORN

                              (Refer to Figure Q36)

                              Figure Q36

                              18 Effective stress

                              37

                              Situation (1)(a)

                              frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                              u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                              0 frac14 u frac14 694 392 frac14 302 kN=m2

                              (b)

                              i frac14 2

                              4frac14 05

                              j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                              Situation (2)(a)

                              frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                              u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                              0 frac14 u frac14 498 392 frac14 106 kN=m2

                              (b)

                              i frac14 2

                              4frac14 05

                              j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                              38

                              The flow net is drawn in Figure Q24

                              Loss in total head between adjacent equipotentials

                              h frac14 550

                              Ndfrac14 550

                              11frac14 050m

                              Exit hydraulic gradient

                              ie frac14 h

                              sfrac14 050

                              070frac14 071

                              Effective stress 19

                              The critical hydraulic gradient is given by Equation 39

                              ic frac14 0

                              wfrac14 102

                              98frac14 104

                              Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                              F frac14 iciefrac14 104

                              071frac14 15

                              Total head at C

                              hC frac14 nd

                              Ndh frac14 24

                              11 550 frac14 120m

                              Elevation head at C

                              zC frac14 250m

                              Pore water pressure at C

                              uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                              Therefore effective vertical stress at C

                              0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                              For point D

                              hD frac14 73

                              11 550 frac14 365m

                              zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                              0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                              39

                              The flow net is drawn in Figure Q25

                              For a soil prism 150 300m adjacent to the piling

                              hm frac14 26

                              9 500 frac14 145m

                              20 Effective stress

                              Factor of safety against lsquoheavingrsquo (Equation 310)

                              F frac14 ic

                              imfrac14 0d

                              whmfrac14 97 300

                              98 145frac14 20

                              With a filter

                              F frac14 0d thorn wwhm

                              3 frac14 eth97 300THORN thorn w98 145

                              w frac14 135 kN=m2

                              Depth of filterfrac14 13521frac14 065m (if above water level)

                              Effective stress 21

                              Chapter 4

                              Shear strength

                              41

                              frac14 295 kN=m2

                              u frac14 120 kN=m2

                              0 frac14 u frac14 295 120 frac14 175 kN=m2

                              f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                              42

                              03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                              100 452 552200 908 1108400 1810 2210800 3624 4424

                              The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                              Figure Q42

                              43

                              3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                              200 222 422400 218 618600 220 820

                              The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                              44

                              The modified shear strength parameters are

                              0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                              a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                              The coordinates of the stress point representing failure conditions in the test are

                              1

                              2eth1 2THORN frac14 1

                              2 170 frac14 85 kN=m2

                              1

                              2eth1 thorn 3THORN frac14 1

                              2eth270thorn 100THORN frac14 185 kN=m2

                              The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                              uf frac14 36 kN=m2

                              Figure Q43

                              Figure Q44

                              Shear strength 23

                              45

                              3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                              150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                              The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                              The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                              46

                              03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                              200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                              The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                              Figure Q45

                              24 Shear strength

                              47

                              The torque required to produce shear failure is given by

                              T frac14 dh cud

                              2thorn 2

                              Z d=2

                              0

                              2r drcur

                              frac14 cud2h

                              2thorn 4cu

                              Z d=2

                              0

                              r2dr

                              frac14 cud2h

                              2thorn d

                              3

                              6

                              Then

                              35 frac14 cu52 10

                              2thorn 53

                              6

                              103

                              cu frac14 76 kN=m3

                              400

                              0 400 800 1200 1600

                              τ (k

                              Nm

                              2 )

                              σprime (kNm2)

                              34deg

                              315deg29deg

                              (a)

                              (b)

                              0 400

                              400

                              800 1200 1600

                              Failure envelope

                              300 500

                              σprime (kNm2)

                              τ (k

                              Nm

                              2 )

                              20 (kNm2)

                              31deg

                              Figure Q46

                              Shear strength 25

                              48

                              The relevant stress values are calculated as follows

                              3 frac14 600 kN=m2

                              1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                              2(1 3) 0 40 79 107 139 159

                              1

                              2(01 thorn 03) 400 411 402 389 351 326

                              1

                              2(1 thorn 3) 600 640 679 707 739 759

                              The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                              Af frac14 433 200

                              319frac14 073

                              49

                              B frac14 u33

                              frac14 144

                              350 200frac14 096

                              a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                              0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                              10 283 65 023

                              Figure Q48

                              26 Shear strength

                              The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                              Figure Q49

                              Shear strength 27

                              Chapter 5

                              Stresses and displacements

                              51

                              Vertical stress is given by

                              z frac14 Qz2Ip frac14 5000

                              52Ip

                              Values of Ip are obtained from Table 51

                              r (m) rz Ip z (kNm2)

                              0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                              10 20 0009 2

                              The variation of z with radial distance (r) is plotted in Figure Q51

                              Figure Q51

                              52

                              Below the centre load (Figure Q52)

                              r

                              zfrac14 0 for the 7500-kN load

                              Ip frac14 0478

                              r

                              zfrac14 5

                              4frac14 125 for the 10 000- and 9000-kN loads

                              Ip frac14 0045

                              Then

                              z frac14X Q

                              z2Ip

                              frac14 7500 0478

                              42thorn 10 000 0045

                              42thorn 9000 0045

                              42

                              frac14 224thorn 28thorn 25 frac14 277 kN=m2

                              53

                              The vertical stress under a corner of a rectangular area is given by

                              z frac14 qIr

                              where values of Ir are obtained from Figure 510 In this case

                              z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                              z

                              Figure Q52

                              Stresses and displacements 29

                              z (m) m n Ir z (kNm2)

                              0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                              10 010 0005 5

                              z is plotted against z in Figure Q53

                              54

                              (a)

                              m frac14 125

                              12frac14 104

                              n frac14 18

                              12frac14 150

                              From Figure 510 Irfrac14 0196

                              z frac14 2 175 0196 frac14 68 kN=m2

                              Figure Q53

                              30 Stresses and displacements

                              (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                              z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                              55

                              Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                              Px frac14 2Q

                              1

                              m2 thorn 1frac14 2 150

                              125frac14 76 kN=m

                              Equation 517 is used to obtain the pressure distribution

                              px frac14 4Q

                              h

                              m2n

                              ethm2 thorn n2THORN2 frac14150

                              m2n

                              ethm2 thorn n2THORN2 ethkN=m2THORN

                              Figure Q54

                              Stresses and displacements 31

                              n m2n

                              (m2 thorn n2)2

                              px(kNm2)

                              0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                              The pressure distribution is plotted in Figure Q55

                              56

                              H

                              Bfrac14 10

                              2frac14 5

                              L

                              Bfrac14 4

                              2frac14 2

                              D

                              Bfrac14 1

                              2frac14 05

                              Hence from Figure 515

                              131 frac14 082

                              130 frac14 094

                              Figure Q55

                              32 Stresses and displacements

                              The immediate settlement is given by Equation 528

                              si frac14 130131qB

                              Eu

                              frac14 094 082 200 2

                              45frac14 7mm

                              Stresses and displacements 33

                              Chapter 6

                              Lateral earth pressure

                              61

                              For 0 frac14 37 the active pressure coefficient is given by

                              Ka frac14 1 sin 37

                              1thorn sin 37frac14 025

                              The total active thrust (Equation 66a with c0 frac14 0) is

                              Pa frac14 1

                              2KaH

                              2 frac14 1

                              2 025 17 62 frac14 765 kN=m

                              If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                              K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                              and the thrust on the wall is

                              P0 frac14 1

                              2K0H

                              2 frac14 1

                              2 040 17 62 frac14 122 kN=m

                              62

                              The active pressure coefficients for the three soil types are as follows

                              Ka1 frac141 sin 35

                              1thorn sin 35frac14 0271

                              Ka2 frac141 sin 27

                              1thorn sin 27frac14 0375

                              ffiffiffiffiffiffiffiKa2

                              p frac14 0613

                              Ka3 frac141 sin 42

                              1thorn sin 42frac14 0198

                              Distribution of active pressure (plotted in Figure Q62)

                              Depth (m) Soil Active pressure (kNm2)

                              3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                              12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                              At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                              Total thrust frac14 571 kNm

                              Point of application is (4893571) m from the top of the wall ie 857m

                              Force (kN) Arm (m) Moment (kN m)

                              (1)1

                              2 0271 16 32 frac14 195 20 390

                              (2) 0271 16 3 2 frac14 260 40 1040

                              (3)1

                              2 0271 92 22 frac14 50 433 217

                              (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                              (5)1

                              2 0375 102 32 frac14 172 70 1204

                              (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                              (7)1

                              2 0198 112 42 frac14 177 1067 1889

                              (8)1

                              2 98 92 frac14 3969 90 35721

                              5713 48934

                              Figure Q62

                              Lateral earth pressure 35

                              63

                              (a) For u frac14 0 Ka frac14 Kp frac14 1

                              Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                              frac14 245

                              At the lower end of the piling

                              pa frac14 Kaqthorn Kasatz Kaccu

                              frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                              frac14 115 kN=m2

                              pp frac14 Kpsatzthorn Kpccu

                              frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                              frac14 202 kN=m2

                              (b) For 0 frac14 26 and frac14 1

                              20

                              Ka frac14 035

                              Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                              pfrac14 145 ethEquation 619THORN

                              Kp frac14 37

                              Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                              pfrac14 47 ethEquation 624THORN

                              At the lower end of the piling

                              pa frac14 Kaqthorn Ka0z Kacc

                              0

                              frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                              frac14 187 kN=m2

                              pp frac14 Kp0zthorn Kpcc

                              0

                              frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                              frac14 198 kN=m2

                              36 Lateral earth pressure

                              64

                              (a) For 0 frac14 38 Ka frac14 024

                              0 frac14 20 98 frac14 102 kN=m3

                              The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                              Force (kN) Arm (m) Moment (kN m)

                              (1) 024 10 66 frac14 159 33 525

                              (2)1

                              2 024 17 392 frac14 310 400 1240

                              (3) 024 17 39 27 frac14 430 135 580

                              (4)1

                              2 024 102 272 frac14 89 090 80

                              (5)1

                              2 98 272 frac14 357 090 321

                              Hfrac14 1345 MH frac14 2746

                              (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                              (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                              XM frac14MV MH frac14 7790 kNm

                              Lever arm of base resultant

                              M

                              Vfrac14 779

                              488frac14 160

                              Eccentricity of base resultant

                              e frac14 200 160 frac14 040m

                              39 m

                              27 m

                              40 m

                              04 m

                              04 m

                              26 m

                              (7)

                              (9)

                              (1)(2)

                              (3)

                              (4)

                              (5)

                              (8)(6)

                              (10)

                              WT

                              10 kNm2

                              Hydrostatic

                              Figure Q64

                              Lateral earth pressure 37

                              Base pressures (Equation 627)

                              p frac14 VB

                              1 6e

                              B

                              frac14 488

                              4eth1 060THORN

                              frac14 195 kN=m2 and 49 kN=m2

                              Factor of safety against sliding (Equation 628)

                              F frac14 V tan

                              Hfrac14 488 tan 25

                              1345frac14 17

                              (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                              Hfrac14 1633 kN

                              V frac14 4879 kN

                              MH frac14 3453 kNm

                              MV frac14 10536 kNm

                              The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                              65

                              For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                              Kp

                              Ffrac14 385

                              2

                              0 frac14 20 98 frac14 102 kN=m3

                              The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                              Force (kN) Arm (m) Moment (kN m)

                              (1)1

                              2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                              (2) 026 17 45 d frac14 199d d2 995d2

                              (3)1

                              2 026 102 d2 frac14 133d2 d3 044d3

                              (4)1

                              2 385

                              2 17 152 frac14 368 dthorn 05 368d 184

                              (5)385

                              2 17 15 d frac14 491d d2 2455d2

                              (6)1

                              2 385

                              2 102 d2 frac14 982d2 d3 327d3

                              38 Lateral earth pressure

                              XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                              d3 thorn 516d2 283d 1724 frac14 0

                              d frac14 179m

                              Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                              Over additional 20 embedded depth

                              pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                              Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                              66

                              The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                              Ka frac14 sin 69=sin 105

                              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                              pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                              26664

                              37775

                              2

                              frac14 050

                              The total active thrust (acting at 25 above the normal) is given by Equation 616

                              Pa frac14 1

                              2 050 19 7502 frac14 267 kN=m

                              Figure Q65

                              Lateral earth pressure 39

                              Horizontal component

                              Ph frac14 267 cos 40 frac14 205 kN=m

                              Vertical component

                              Pv frac14 267 sin 40 frac14 172 kN=m

                              Consider moments about the toe of the wall (Figure Q66) (per m)

                              Force (kN) Arm (m) Moment (kN m)

                              (1)1

                              2 175 650 235 frac14 1337 258 345

                              (2) 050 650 235 frac14 764 175 134

                              (3)1

                              2 070 650 235 frac14 535 127 68

                              (4) 100 400 235 frac14 940 200 188

                              (5) 1

                              2 080 050 235 frac14 47 027 1

                              Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                              Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                              Lever arm of base resultant

                              M

                              Vfrac14 795

                              525frac14 151m

                              Eccentricity of base resultant

                              e frac14 200 151 frac14 049m

                              Figure Q66

                              40 Lateral earth pressure

                              Base pressures (Equation 627)

                              p frac14 525

                              41 6 049

                              4

                              frac14 228 kN=m2 and 35 kN=m2

                              The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                              The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                              The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                              67

                              For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                              Force (kN) Arm (m) Moment (kNm)

                              (1)1

                              2 027 17 52 frac14 574 183 1050

                              (2) 027 17 5 3 frac14 689 500 3445

                              (3)1

                              2 027 102 32 frac14 124 550 682

                              (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                              (5)1

                              2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                              (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                              (7) 1

                              2 267

                              2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                              (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                              2 d frac14 163d d2thorn 650 82d2 1060d

                              Tie rod force per m frac14 T 0 0

                              XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                              d3 thorn 77d2 269d 1438 frac14 0

                              d frac14 467m

                              Depth of penetration frac14 12d frac14 560m

                              Lateral earth pressure 41

                              Algebraic sum of forces for d frac14 467m isX

                              F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                              T frac14 905 kN=m

                              Force in each tie rod frac14 25T frac14 226 kN

                              68

                              (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                              0 frac14 21 98 frac14 112 kN=m3

                              The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                              uC frac14 150

                              165 15 98 frac14 134 kN=m2

                              The average seepage pressure is

                              j frac14 15

                              165 98 frac14 09 kN=m3

                              Hence

                              0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                              0 j frac14 112 09 frac14 103 kN=m3

                              Figure Q67

                              42 Lateral earth pressure

                              Consider moments about the anchor point A (per m)

                              Force (kN) Arm (m) Moment (kN m)

                              (1) 10 026 150 frac14 390 60 2340

                              (2)1

                              2 026 18 452 frac14 474 15 711

                              (3) 026 18 45 105 frac14 2211 825 18240

                              (4)1

                              2 026 121 1052 frac14 1734 100 17340

                              (5)1

                              2 134 15 frac14 101 40 404

                              (6) 134 30 frac14 402 60 2412

                              (7)1

                              2 134 60 frac14 402 95 3819

                              571 4527(8) Ppm

                              115 115PPm

                              XM frac14 0

                              Ppm frac144527

                              115frac14 394 kN=m

                              Available passive resistance

                              Pp frac14 1

                              2 385 103 62 frac14 714 kN=m

                              Factor of safety

                              Fp frac14 Pp

                              Ppm

                              frac14 714

                              394frac14 18

                              Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                              Figure Q68

                              Lateral earth pressure 43

                              (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                              Consider moments (per m) about the tie point A

                              Force (kN) Arm (m)

                              (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                              (2)1

                              2 033 18 452 frac14 601 15

                              (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                              (4)1

                              2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                              (5)1

                              2 134 15 frac14 101 40

                              (6) 134 30 frac14 402 60

                              (7)1

                              2 134 d frac14 67d d3thorn 75

                              (8) 1

                              2 30 103 d2 frac141545d2 2d3thorn 75

                              Moment (kN m)

                              (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                              XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                              d3 thorn 827d2 466d 1518 frac14 0

                              By trial

                              d frac14 544m

                              The minimum depth of embedment required is 544m

                              69

                              For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                              0 frac14 20 98 frac14 102 kN=m3

                              The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                              44 Lateral earth pressure

                              uC frac14 147

                              173 26 98 frac14 216 kN=m2

                              and the average seepage pressure around the wall is

                              j frac14 26

                              173 98 frac14 15 kN=m3

                              Consider moments about the prop (A) (per m)

                              Force (kN) Arm (m) Moment (kN m)

                              (1)1

                              2 03 17 272 frac14 186 020 37

                              (2) 03 17 27 53 frac14 730 335 2445

                              (3)1

                              2 03 (102thorn 15) 532 frac14 493 423 2085

                              (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                              (5)1

                              2 216 26 frac14 281 243 684

                              (6) 216 27 frac14 583 465 2712

                              (7)1

                              2 216 60 frac14 648 800 5184

                              3055(8)

                              1

                              2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                              Factor of safety

                              Fr frac14 6885

                              3055frac14 225

                              Figure Q69

                              Lateral earth pressure 45

                              610

                              For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                              p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                              Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                              Using the recommendations of Twine and Roscoe

                              p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                              Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                              611

                              frac14 18 kN=m3 0 frac14 34

                              H frac14 350m nH frac14 335m mH frac14 185m

                              Consider a trial value of F frac14 20 Refer to Figure 635

                              0m frac14 tan1tan 34

                              20

                              frac14 186

                              Then

                              frac14 45 thorn 0m2frac14 543

                              W frac14 1

                              2 18 3502 cot 543 frac14 792 kN=m

                              Figure Q610

                              46 Lateral earth pressure

                              P frac14 1

                              2 s 3352 frac14 561s kN=m

                              U frac14 1

                              2 98 1852 cosec 543 frac14 206 kN=m

                              Equations 630 and 631 then become

                              561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                              792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                              ie

                              561s 0616N 405 frac14 0

                              792 0857N thorn 563 frac14 0

                              N frac14 848

                              0857frac14 989 kN=m

                              Then

                              561s 609 405 frac14 0

                              s frac14 649

                              561frac14 116 kN=m3

                              The calculations for trial values of F of 20 15 and 10 are summarized below

                              F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                              20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                              s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                              Figure Q611

                              Lateral earth pressure 47

                              612

                              For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                              45 thorn 0

                              2frac14 63

                              For the retained material between the surface and a depth of 36m

                              Pa frac14 1

                              2 030 18 362 frac14 350 kN=m

                              Weight of reinforced fill between the surface and a depth of 36m is

                              Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                              eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                              Eccentricity of Rv

                              e frac14 263 250 frac14 013m

                              The average vertical stress at a depth of 36m is

                              z frac14 Rv

                              L 2efrac14 324

                              474frac14 68 kN=m2

                              (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                              Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                              Tensile stress in the element frac14 138 103

                              65 3frac14 71N=mm2

                              Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                              Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                              Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                              The weight of ABC is

                              W frac14 1

                              2 18 52 265 frac14 124 kN=m

                              From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                              48 Lateral earth pressure

                              (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                              Tp frac14 032 68 120 065 frac14 170 kN

                              Tr frac14 213 420

                              418frac14 214 kN

                              Again the tensile failure and slipping limit states are satisfied for this element

                              Figure Q612

                              Lateral earth pressure 49

                              Chapter 7

                              Consolidation theory

                              71

                              Total change in thickness

                              H frac14 782 602 frac14 180mm

                              Average thickness frac14 1530thorn 180

                              2frac14 1620mm

                              Length of drainage path d frac14 1620

                              2frac14 810mm

                              Root time plot (Figure Q71a)

                              ffiffiffiffiffiffit90p frac14 33

                              t90 frac14 109min

                              cv frac14 0848d2

                              t90frac14 0848 8102

                              109 1440 365

                              106frac14 27m2=year

                              r0 frac14 782 764

                              782 602frac14 018

                              180frac14 0100

                              rp frac14 10eth764 645THORN9eth782 602THORN frac14

                              10 119

                              9 180frac14 0735

                              rs frac14 1 eth0100thorn 0735THORN frac14 0165

                              Log time plot (Figure Q71b)

                              t50 frac14 26min

                              cv frac14 0196d2

                              t50frac14 0196 8102

                              26 1440 365

                              106frac14 26m2=year

                              r0 frac14 782 763

                              782 602frac14 019

                              180frac14 0106

                              rp frac14 763 623

                              782 602frac14 140

                              180frac14 0778

                              rs frac14 1 eth0106thorn 0778THORN frac14 0116

                              Figure Q71(a)

                              Figure Q71(b)

                              Final void ratio

                              e1 frac14 w1Gs frac14 0232 272 frac14 0631

                              e

                              Hfrac14 1thorn e0

                              H0frac14 1thorn e1 thorne

                              H0

                              ie

                              e

                              180frac14 1631thorne

                              1710

                              e frac14 2936

                              1530frac14 0192

                              Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                              mv frac14 1

                              1thorn e0 e0 e101 00

                              frac14 1

                              1823 0192

                              0107frac14 098m2=MN

                              k frac14 cvmvw frac14 265 098 98

                              60 1440 365 103frac14 81 1010 m=s

                              72

                              Using Equation 77 (one-dimensional method)

                              sc frac14 e0 e11thorn e0 H

                              Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                              Figure Q72

                              52 Consolidation theory

                              Settlement

                              Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                              1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                              318

                              Notes 5 92y 460thorn 84

                              Heave

                              Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                              1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                              38

                              73

                              U frac14 f ethTvTHORN frac14 f cvt

                              d2

                              Hence if cv is constant

                              t1

                              t2frac14 d

                              21

                              d22

                              where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                              d1 frac14 95mm and d2 frac14 2500mm

                              for U frac14 050 t2 frac14 t1 d22

                              d21

                              frac14 20

                              60 24 365 25002

                              952frac14 263 years

                              for U lt 060 Tv frac14

                              4U2 (Equation 724(a))

                              t030 frac14 t050 0302

                              0502

                              frac14 263 036 frac14 095 years

                              Consolidation theory 53

                              74

                              The layer is open

                              d frac14 8

                              2frac14 4m

                              Tv frac14 cvtd2frac14 24 3

                              42frac14 0450

                              ui frac14 frac14 84 kN=m2

                              The excess pore water pressure is given by Equation 721

                              ue frac14Xmfrac141mfrac140

                              2ui

                              Msin

                              Mz

                              d

                              expethM2TvTHORN

                              In this case z frac14 d

                              sinMz

                              d

                              frac14 sinM

                              where

                              M frac14

                              23

                              25

                              2

                              M sin M M2Tv exp (M2Tv)

                              2thorn1 1110 0329

                              3

                              21 9993 457 105

                              ue frac14 2 84 2

                              1 0329 ethother terms negligibleTHORN

                              frac14 352 kN=m2

                              75

                              The layer is open

                              d frac14 6

                              2frac14 3m

                              Tv frac14 cvtd2frac14 10 3

                              32frac14 0333

                              The layer thickness will be divided into six equal parts ie m frac14 6

                              54 Consolidation theory

                              For an open layer

                              Tv frac14 4n

                              m2

                              n frac14 0333 62

                              4frac14 300

                              The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                              ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                              i j

                              0 1 2 3 4 5 6 7 8 9 10 11 12

                              0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                              The initial and 3-year isochrones are plotted in Figure Q75

                              Area under initial isochrone frac14 180 units

                              Area under 3-year isochrone frac14 63 units

                              The average degree of consolidation is given by Equation 725Thus

                              U frac14 1 63

                              180frac14 065

                              Figure Q75

                              Consolidation theory 55

                              76

                              At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                              0 frac14 2w frac14 2 98 frac14 196 kN=m2

                              The final consolidation settlement (one-dimensional method) is

                              sc frac14 mv0H frac14 083 196 8 frac14 130mm

                              Corrected time t frac14 2 1

                              2

                              40

                              52

                              frac14 1615 years

                              Tv frac14 cvtd2frac14 44 1615

                              42frac14 0444

                              From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                              77

                              The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                              Figure Q77

                              56 Consolidation theory

                              Point m n Ir (kNm2) sc (mm)

                              13020frac14 15 20

                              20frac14 10 0194 (4) 113 124

                              260

                              20frac14 30

                              20

                              20frac14 10 0204 (2) 59 65

                              360

                              20frac14 30

                              40

                              20frac14 20 0238 (1) 35 38

                              430

                              20frac14 15

                              40

                              20frac14 20 0224 (2) 65 72

                              Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                              78

                              Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                              (a) Immediate settlement

                              H

                              Bfrac14 30

                              35frac14 086

                              D

                              Bfrac14 2

                              35frac14 006

                              Figure Q78

                              Consolidation theory 57

                              From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                              si frac14 130131qB

                              Eufrac14 10 032 105 35

                              40frac14 30mm

                              (b) Consolidation settlement

                              Layer z (m) Dz Ic (kNm2) syod (mm)

                              1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                              3150

                              Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                              Now

                              H

                              Bfrac14 30

                              35frac14 086 and A frac14 065

                              from Figure 712 13 frac14 079

                              sc frac14 13sod frac14 079 315 frac14 250mm

                              Total settlement

                              s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                              79

                              Without sand drains

                              Uv frac14 025

                              Tv frac14 0049 ethfrom Figure 718THORN

                              t frac14 Tvd2

                              cvfrac14 0049 82

                              cvWith sand drains

                              R frac14 0564S frac14 0564 3 frac14 169m

                              n frac14 Rrfrac14 169

                              015frac14 113

                              Tr frac14 cht

                              4R2frac14 ch

                              4 1692 0049 82

                              cvethand ch frac14 cvTHORN

                              frac14 0275

                              Ur frac14 073 (from Figure 730)

                              58 Consolidation theory

                              Using Equation 740

                              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                              U frac14 080

                              710

                              Without sand drains

                              Uv frac14 090

                              Tv frac14 0848

                              t frac14 Tvd2

                              cvfrac14 0848 102

                              96frac14 88 years

                              With sand drains

                              R frac14 0564S frac14 0564 4 frac14 226m

                              n frac14 Rrfrac14 226

                              015frac14 15

                              Tr

                              Tvfrac14 chcv

                              d2

                              4R2ethsame tTHORN

                              Tr

                              Tvfrac14 140

                              96 102

                              4 2262frac14 714 eth1THORN

                              Using Equation 740

                              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                              An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                              Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                              040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                              Thus

                              Uv frac14 0295 and Ur frac14 086

                              t frac14 88 00683

                              0848frac14 07 years

                              Consolidation theory 59

                              Chapter 8

                              Bearing capacity

                              81

                              (a) The ultimate bearing capacity is given by Equation 83

                              qf frac14 cNc thorn DNq thorn 1

                              2BN

                              For u frac14 0

                              Nc frac14 514 Nq frac14 1 N frac14 0

                              qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                              The net ultimate bearing capacity is

                              qnf frac14 qf D frac14 540 kN=m2

                              The net foundation pressure is

                              qn frac14 q D frac14 425

                              2 eth21 1THORN frac14 192 kN=m2

                              The factor of safety (Equation 86) is

                              F frac14 qnfqnfrac14 540

                              192frac14 28

                              (b) For 0 frac14 28

                              Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                              qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                              2 112 2 13

                              frac14 260thorn 168thorn 146 frac14 574 kN=m2

                              qnf frac14 574 112 frac14 563 kN=m2

                              F frac14 563

                              192frac14 29

                              (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                              82

                              For 0 frac14 38

                              Nq frac14 49 N frac14 67

                              qnf frac14 DethNq 1THORN thorn 1

                              2BN ethfrom Equation 83THORN

                              frac14 eth18 075 48THORN thorn 1

                              2 18 15 67

                              frac14 648thorn 905 frac14 1553 kN=m2

                              qn frac14 500

                              15 eth18 075THORN frac14 320 kN=m2

                              F frac14 qnfqnfrac14 1553

                              320frac14 48

                              0d frac14 tan1tan 38

                              125

                              frac14 32 therefore Nq frac14 23 and N frac14 25

                              Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                              2 18 15 25

                              frac14 15eth310thorn 337THORNfrac14 970 kN=m

                              Design load (action) Vd frac14 500 kN=m

                              The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                              83

                              D

                              Bfrac14 350

                              225frac14 155

                              From Figure 85 for a square foundation

                              Nc frac14 81

                              Bearing capacity 61

                              For a rectangular foundation (L frac14 450m B frac14 225m)

                              Nc frac14 084thorn 016B

                              L

                              81 frac14 745

                              Using Equation 810

                              qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                              For F frac14 3

                              qn frac14 1006

                              3frac14 335 kN=m2

                              q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                              Design load frac14 405 450 225 frac14 4100 kN

                              Design undrained strength cud frac14 135

                              14frac14 96 kN=m2

                              Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                              frac14 7241 kN

                              Design load Vd frac14 4100 kN

                              Rd gt Vd therefore the bearing resistance limit state is satisfied

                              84

                              For 0 frac14 40

                              Nq frac14 64 N frac14 95

                              qnf frac14 DethNq 1THORN thorn 04BN

                              (a) Water table 5m below ground level

                              qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                              qn frac14 400 17 frac14 383 kN=m2

                              F frac14 2686

                              383frac14 70

                              (b) Water table 1m below ground level (ie at foundation level)

                              0 frac14 20 98 frac14 102 kN=m3

                              62 Bearing capacity

                              qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                              F frac14 2040

                              383frac14 53

                              (c) Water table at ground level with upward hydraulic gradient 02

                              eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                              qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                              F frac14 1296

                              392frac14 33

                              85

                              The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                              Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                              Design value of 0 frac14 tan1tan 39

                              125

                              frac14 33

                              For 0 frac14 33 Nq frac14 26 and N frac14 29

                              Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                              Rd gt Vd therefore the bearing resistance limit state is satisfied

                              86

                              (a) Undrained shear for u frac14 0

                              Nc frac14 514 Nq frac14 1 N frac14 0

                              qnf frac14 12cuNc

                              frac14 12 100 514 frac14 617 kN=m2

                              qn frac14 qnfFfrac14 617

                              3frac14 206 kN=m2

                              q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                              Bearing capacity 63

                              Drained shear for 0 frac14 32

                              Nq frac14 23 N frac14 25

                              0 frac14 21 98 frac14 112 kN=m3

                              qnf frac14 0DethNq 1THORN thorn 040BN

                              frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                              frac14 694 kN=m2

                              q frac14 694

                              3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                              Design load frac14 42 227 frac14 3632 kN

                              (b) Design undrained strength cud frac14 100

                              14frac14 71 kNm2

                              Design bearing resistance Rd frac14 12cudNe area

                              frac14 12 71 514 42

                              frac14 7007 kN

                              For drained shear 0d frac14 tan1tan 32

                              125

                              frac14 26

                              Nq frac14 12 N frac14 10

                              Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                              (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                              Layer z (m) m n Ir 0 (kNm2) sod (mm)

                              1 2 100 0175 0700qn 0182qn

                              2 6 033 0044 0176qn 0046qn

                              3 10 020 0017 0068qn 0018qn

                              0246qn

                              Diameter of equivalent circle B frac14 45m

                              H

                              Bfrac14 12

                              45frac14 27 and A frac14 042

                              13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                              64 Bearing capacity

                              For sc frac14 30mm

                              qn frac14 30

                              0147frac14 204 kN=m2

                              q frac14 204thorn 21 frac14 225 kN=m2

                              Design load frac14 42 225 frac14 3600 kN

                              The design load is 3600 kN settlement being the limiting criterion

                              87

                              D

                              Bfrac14 8

                              4frac14 20

                              From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                              F frac14 cuNc

                              Dfrac14 40 71

                              20 8frac14 18

                              88

                              Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                              Design value of 0 frac14 tan1tan 38

                              125

                              frac14 32

                              Figure Q86

                              Bearing capacity 65

                              For 0 frac14 32 Nq frac14 23 and N frac14 25

                              Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                              The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                              Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                              For B frac14 250m qn frac14 3750

                              2502 17 frac14 583 kN=m2

                              From Figure 510 m frac14 n frac14 126

                              6frac14 021

                              Ir frac14 0019

                              Stress increment frac14 4 0019 583 frac14 44 kN=m2

                              Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                              The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                              89

                              Depth (m) N 0v (kNm2) CN N1

                              070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                              Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                              (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                              Cw frac14 05thorn 0530

                              47

                              frac14 082

                              66 Bearing capacity

                              Thus

                              qa frac14 150 082 frac14 120 kN=m2

                              (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                              Thus

                              qa frac14 90 15 frac14 135 kN=m2

                              (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                              Ic frac14 171

                              1014frac14 0068

                              From Equation 819(a) with s frac14 25mm

                              q frac14 25

                              3507 0068frac14 150 kN=m2

                              810

                              Peak value of strain influence factor occurs at a depth of 27m and is given by

                              Izp frac14 05thorn 01130

                              16 27

                              05

                              frac14 067

                              Refer to Figure Q810

                              E frac14 25qc

                              Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                              Ez (mm3MN)

                              1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                              0203

                              C1 frac14 1 0500qnfrac14 1 05 12 16

                              130frac14 093

                              C2 frac14 1 ethsayTHORN

                              s frac14 C1C2qnX Iz

                              Ez frac14 093 1 130 0203 frac14 25mm

                              Bearing capacity 67

                              811

                              At pile base level

                              cu frac14 220 kN=m2

                              qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                              Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                              00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                              qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                              Then

                              Qf frac14 Abqb thorn Asqs

                              frac14

                              4 32 1980

                              thorn eth 105 139 86THORN

                              frac14 13 996thorn 3941 frac14 17 937 kN

                              0 01 02 03 04 05 06 07

                              0 2 4 6 8 10 12 14

                              1

                              2

                              3

                              4

                              5

                              6

                              7

                              8

                              (1)

                              (2)

                              (3)

                              (4)

                              (5)

                              qc

                              qc

                              Iz

                              Iz

                              (MNm2)

                              z (m)

                              Figure Q810

                              68 Bearing capacity

                              Allowable load

                              ethaTHORN Qf

                              2frac14 17 937

                              2frac14 8968 kN

                              ethbTHORN Abqb

                              3thorn Asqs frac14 13 996

                              3thorn 3941 frac14 8606 kN

                              ie allowable load frac14 8600 kN

                              Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                              According to the limit state method

                              Characteristic undrained strength at base level cuk frac14 220

                              150kN=m2

                              Characteristic base resistance qbk frac14 9cuk frac14 9 220

                              150frac14 1320 kN=m2

                              Characteristic shaft resistance qsk frac14 00150

                              frac14 86

                              150frac14 57 kN=m2

                              Characteristic base and shaft resistances

                              Rbk frac14

                              4 32 1320 frac14 9330 kN

                              Rsk frac14 105 139 86

                              150frac14 2629 kN

                              For a bored pile the partial factors are b frac14 160 and s frac14 130

                              Design bearing resistance Rcd frac14 9330

                              160thorn 2629

                              130

                              frac14 5831thorn 2022

                              frac14 7850 kN

                              Adding ethDAb W) the design bearing resistance becomes 9650 kN

                              812

                              ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                              qs frac14 cu frac14 040 105 frac14 42 kN=m2

                              For a single pile

                              Qf frac14 Abqb thorn Asqs

                              frac14

                              4 062 1305

                              thorn eth 06 15 42THORN

                              frac14 369thorn 1187 frac14 1556 kN

                              Bearing capacity 69

                              Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                              eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                              (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                              (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                              qbkfrac14 9cuk frac14 9 220

                              150frac14 1320 kN=m2

                              qskfrac14cuk frac14 040 105

                              150frac14 28 kN=m2

                              Rbkfrac14

                              4 0602 1320 frac14 373 kN

                              Rskfrac14 060 15 28 frac14 791 kN

                              Rcdfrac14 373

                              160thorn 791

                              130frac14 233thorn 608 frac14 841 kN

                              Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                              Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                              (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                              q frac14 21 000

                              1762frac14 68 kN=m2

                              Immediate settlement

                              H

                              Bfrac14 15

                              176frac14 085

                              D

                              Bfrac14 13

                              176frac14 074

                              L

                              Bfrac14 1

                              Hence from Figure 515

                              130 frac14 078 and 131 frac14 041

                              70 Bearing capacity

                              Thus using Equation 528

                              si frac14 078 041 68 176

                              65frac14 6mm

                              Consolidation settlement

                              Layer z (m) Area (m2) (kNm2) mvH (mm)

                              1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                              434 (sod)

                              Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                              sc frac14 056 434 frac14 24mm

                              The total settlement is (6thorn 24) frac14 30mm

                              813

                              At base level N frac14 26 Then using Equation 830

                              qb frac14 40NDb

                              Bfrac14 40 26 2

                              025frac14 8320 kN=m2

                              ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                              Figure Q812

                              Bearing capacity 71

                              Over the length embedded in sand

                              N frac14 21 ie18thorn 24

                              2

                              Using Equation 831

                              qs frac14 2N frac14 2 21 frac14 42 kN=m2

                              For a single pile

                              Qf frac14 Abqb thorn Asqs

                              frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                              For the pile group assuming a group efficiency of 12

                              XQf frac14 12 9 604 frac14 6523 kN

                              Then the load factor is

                              F frac14 6523

                              2000thorn 1000frac14 21

                              (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                              Characteristic base resistance per unit area qbk frac14 8320

                              150frac14 5547 kNm2

                              Characteristic shaft resistance per unit area qsk frac14 42

                              150frac14 28 kNm2

                              Characteristic base and shaft resistances for a single pile

                              Rbk frac14 0252 5547 frac14 347 kN

                              Rsk frac14 4 025 2 28 frac14 56 kN

                              For a driven pile the partial factors are b frac14 s frac14 130

                              Design bearing resistance Rcd frac14 347

                              130thorn 56

                              130frac14 310 kN

                              For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                              Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                              (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                              From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                              72 Bearing capacity

                              N frac14 24thorn 26thorn 34

                              3frac14 28

                              Ic frac14 171

                              2814frac14 0016 ethEquation 818THORN

                              s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                              814

                              Using Equation 841

                              Tf frac14 DLcu thorn

                              4ethD2 d2THORNcuNc

                              frac14 eth 02 5 06 110THORN thorn

                              4eth022 012THORN110 9

                              frac14 207thorn 23 frac14 230 kN

                              Figure Q813

                              Bearing capacity 73

                              Chapter 9

                              Stability of slopes

                              91

                              Referring to Figure Q91

                              W frac14 417 19 frac14 792 kN=m

                              Q frac14 20 28 frac14 56 kN=m

                              Arc lengthAB frac14

                              180 73 90 frac14 115m

                              Arc length BC frac14

                              180 28 90 frac14 44m

                              The factor of safety is given by

                              F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                              90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                              Depth of tension crack z0 frac14 2cu

                              frac14 2 20

                              19frac14 21m

                              Arc length BD frac14

                              180 13

                              1

                              2 90 frac14 21m

                              F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                              The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                              Design resisting moment frac14 rXethcudLaTHORN frac14 90

                              14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                              Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                              The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                              92

                              u frac14 0

                              Depth factor D frac14 11

                              9frac14 122

                              Using Equation 92 with F frac14 10

                              Ns frac14 cu

                              FHfrac14 30

                              10 19 9frac14 0175

                              Hence from Figure 93

                              frac14 50

                              For F frac14 12

                              Ns frac14 30

                              12 19 9frac14 0146

                              frac14 27

                              93

                              Refer to Figure Q93

                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                              1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                              74 m

                              214 1deg

                              213 1deg

                              39 m

                              WB

                              D

                              C

                              28 m

                              21 m

                              A

                              Q

                              Soil (1)Soil (2)

                              73deg

                              Figure Q91

                              Stability of slopes 75

                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                              9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                              599 256 328 1372

                              Figure Q93

                              76 Stability of slopes

                              XW cos frac14 b

                              Xh cos frac14 21 2 599 frac14 2516 kN=mX

                              W sin frac14 bX

                              h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                              Arc length La frac14

                              180 57

                              1

                              2 326 frac14 327m

                              The factor of safety is given by

                              F frac14 c0La thorn tan0ethW cos ulTHORN

                              W sin

                              frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                              frac14 091

                              According to the limit state method

                              0d frac14 tan1tan 32

                              125

                              frac14 265

                              c0 frac14 8

                              160frac14 5 kN=m2

                              Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                              Design disturbing moment frac14 1075 kN=m

                              The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                              94

                              F frac14 1

                              W sin

                              Xfc0bthorn ethW ubTHORN tan0g sec

                              1thorn ethtan tan0=FTHORN

                              c0 frac14 8 kN=m2

                              0 frac14 32

                              c0b frac14 8 2 frac14 16 kN=m

                              W frac14 bh frac14 21 2 h frac14 42h kN=m

                              Try F frac14 100

                              tan0

                              Ffrac14 0625

                              Stability of slopes 77

                              Values of u are as obtained in Figure Q93

                              SliceNo

                              h(m)

                              W frac14 bh(kNm)

                              W sin(kNm)

                              ub(kNm)

                              c0bthorn (W ub) tan0(kNm)

                              sec

                              1thorn (tan tan0)FProduct(kNm)

                              1 05 21 6 2 8 24 1078 262 13 55 31

                              23 33 30 1042 31

                              3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                              224 92 72 0931 67

                              6 50 210 11 40 100 85 0907 777 55 231 14

                              12 58 112 90 0889 80

                              8 60 252 1812

                              80 114 102 0874 899 63 265 22 99 116 109 0861 94

                              10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                              2154 88 116 0853 99

                              14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                              1074 1091

                              F frac14 1091

                              1074frac14 102 (assumed value 100)

                              Thus

                              F frac14 101

                              95

                              F frac14 1

                              W sin

                              XfWeth1 ruTHORN tan0g sec

                              1thorn ethtan tan0THORN=F

                              0 frac14 33

                              ru frac14 020

                              W frac14 bh frac14 20 5 h frac14 100h kN=m

                              eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                              Try F frac14 110

                              tan 0

                              Ffrac14 tan 33

                              110frac14 0590

                              78 Stability of slopes

                              Referring to Figure Q95

                              SliceNo

                              h(m)

                              W frac14 bh(kNm)

                              W sin(kNm)

                              W(1 ru) tan0(kNm)

                              sec

                              1thorn ( tan tan0)FProduct(kNm)

                              1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                              2120 234 0892 209

                              4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                              1185 1271

                              Figure Q95

                              Stability of slopes 79

                              F frac14 1271

                              1185frac14 107

                              The trial value was 110 therefore take F to be 108

                              96

                              (a) Water table at surface the factor of safety is given by Equation 912

                              F frac14 0

                              sat

                              tan0

                              tan

                              ptie 15 frac14 92

                              19

                              tan 36

                              tan

                              tan frac14 0234

                              frac14 13

                              Water table well below surface the factor of safety is given by Equation 911

                              F frac14 tan0

                              tan

                              frac14 tan 36

                              tan 13

                              frac14 31

                              (b) 0d frac14 tan1tan 36

                              125

                              frac14 30

                              Depth of potential failure surface frac14 z

                              Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                              frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                              frac14 504z kN

                              Design disturbing moment per unit area Sd frac14 sat sin cos

                              frac14 19 z sin 13 cos 13

                              frac14 416z kN

                              Rd gtSd therefore the limit state for overall stability is satisfied

                              80 Stability of slopes

                              • Book Cover
                              • Title
                              • Contents
                              • Basic characteristics of soils
                              • Seepage
                              • Effective stress
                              • Shear strength
                              • Stresses and displacements
                              • Lateral earth pressure
                              • Consolidation theory
                              • Bearing capacity
                              • Stability of slopes

                                27

                                The scale transformation factor in the x direction is

                                xt frac14 xffiffiffiffiffikzpffiffiffiffiffikxp frac14 x

                                ffiffiffiffiffiffiffi27pffiffiffiffiffiffiffi75p frac14 060x

                                Thus all dimensions in the x direction are multipled by 060 All dimensions in thez direction are unchanged The transformed section is shown in Figure Q27 Theequivalent isotropic permeability is

                                k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                                pfrac14

                                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth75 27THORN

                                p 106 frac14 45 106 m=s

                                The focus of the basic parabola is at point A The parabola passes through point Gsuch that

                                GC frac14 03HC frac14 03 30 frac14 90m

                                Thus the coordinates of G are

                                x frac14 480 and z frac14 thorn200Substituting these coordinates in Equation 234

                                480 frac14 x0 2002

                                4x0

                                Figure Q26

                                10 Seepage

                                Hence

                                x0 frac14 20m

                                Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

                                x frac14 20 z2

                                80

                                x 20 0 50 100 200 300z 0 400 748 980 1327 1600

                                The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

                                No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

                                q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

                                28

                                The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

                                Figure Q27

                                Seepage 11

                                q frac14 kh Nf

                                Ndfrac14 45 105 28 33

                                7

                                frac14 59 105 m3=s per m

                                29

                                The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

                                kx frac14 H1k1 thornH2k2

                                H1 thornH2frac14 106

                                10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

                                kz frac14 H1 thornH2

                                H1

                                k1thornH2

                                k2

                                frac14 10

                                5

                                eth2 106THORN thorn5

                                eth16 106THORNfrac14 36 106 m=s

                                Then the scale transformation factor is given by

                                xt frac14 xffiffiffiffiffikz

                                pffiffiffiffiffikx

                                p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

                                Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

                                Figure Q28

                                12 Seepage

                                k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                                qfrac14

                                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

                                p 106 frac14 57 106 m=s

                                Then the quantity of seepage is given by

                                q frac14 k0h Nf

                                Ndfrac14 57 106 350 56

                                11

                                frac14 10 105 m3=s per m

                                Figure Q29

                                Seepage 13

                                Chapter 3

                                Effective stress

                                31

                                Buoyant unit weight

                                0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                                Effective vertical stress

                                0v frac14 5 102 frac14 51 kN=m2 or

                                Total vertical stress

                                v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

                                Pore water pressure

                                u frac14 7 98 frac14 686 kN=m2

                                Effective vertical stress

                                0v frac14 v u frac14 1196 686 frac14 51 kN=m2

                                32

                                Buoyant unit weight

                                0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                                Effective vertical stress

                                0v frac14 5 102 frac14 51 kN=m2 or

                                Total vertical stress

                                v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

                                Pore water pressure

                                u frac14 205 98 frac14 2009 kN=m2

                                Effective vertical stress

                                0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

                                33

                                At top of the clay

                                v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

                                u frac14 2 98 frac14 196 kN=m2

                                0v frac14 v u frac14 710 196 frac14 514 kN=m2

                                Alternatively

                                0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                                0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

                                At bottom of the clay

                                v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

                                u frac14 12 98 frac14 1176 kN=m2

                                0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

                                NB The alternative method of calculation is not applicable because of the artesiancondition

                                Figure Q3132

                                Effective stress 15

                                34

                                0 frac14 20 98 frac14 102 kN=m3

                                At 8m depth

                                0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                                35

                                0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                                0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                                Figure Q33

                                Figure Q34

                                16 Effective stress

                                (a) Immediately after WT rise

                                At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                                0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                                At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                                0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                                (b) Several years after WT rise

                                At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                                At 8m depth

                                0v frac14 940 kN=m2 (as above)

                                At 12m depth

                                0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                                Figure Q35

                                Effective stress 17

                                36

                                Total weight

                                ab frac14 210 kN

                                Effective weight

                                ac frac14 112 kN

                                Resultant boundary water force

                                be frac14 119 kN

                                Seepage force

                                ce frac14 34 kN

                                Resultant body force

                                ae frac14 99 kN eth73 to horizontalTHORN

                                (Refer to Figure Q36)

                                Figure Q36

                                18 Effective stress

                                37

                                Situation (1)(a)

                                frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                                u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                0 frac14 u frac14 694 392 frac14 302 kN=m2

                                (b)

                                i frac14 2

                                4frac14 05

                                j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                                Situation (2)(a)

                                frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                                u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                0 frac14 u frac14 498 392 frac14 106 kN=m2

                                (b)

                                i frac14 2

                                4frac14 05

                                j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                                38

                                The flow net is drawn in Figure Q24

                                Loss in total head between adjacent equipotentials

                                h frac14 550

                                Ndfrac14 550

                                11frac14 050m

                                Exit hydraulic gradient

                                ie frac14 h

                                sfrac14 050

                                070frac14 071

                                Effective stress 19

                                The critical hydraulic gradient is given by Equation 39

                                ic frac14 0

                                wfrac14 102

                                98frac14 104

                                Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                                F frac14 iciefrac14 104

                                071frac14 15

                                Total head at C

                                hC frac14 nd

                                Ndh frac14 24

                                11 550 frac14 120m

                                Elevation head at C

                                zC frac14 250m

                                Pore water pressure at C

                                uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                                Therefore effective vertical stress at C

                                0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                                For point D

                                hD frac14 73

                                11 550 frac14 365m

                                zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                                0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                                39

                                The flow net is drawn in Figure Q25

                                For a soil prism 150 300m adjacent to the piling

                                hm frac14 26

                                9 500 frac14 145m

                                20 Effective stress

                                Factor of safety against lsquoheavingrsquo (Equation 310)

                                F frac14 ic

                                imfrac14 0d

                                whmfrac14 97 300

                                98 145frac14 20

                                With a filter

                                F frac14 0d thorn wwhm

                                3 frac14 eth97 300THORN thorn w98 145

                                w frac14 135 kN=m2

                                Depth of filterfrac14 13521frac14 065m (if above water level)

                                Effective stress 21

                                Chapter 4

                                Shear strength

                                41

                                frac14 295 kN=m2

                                u frac14 120 kN=m2

                                0 frac14 u frac14 295 120 frac14 175 kN=m2

                                f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                                42

                                03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                                100 452 552200 908 1108400 1810 2210800 3624 4424

                                The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                                Figure Q42

                                43

                                3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                                200 222 422400 218 618600 220 820

                                The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                                44

                                The modified shear strength parameters are

                                0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                                a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                                The coordinates of the stress point representing failure conditions in the test are

                                1

                                2eth1 2THORN frac14 1

                                2 170 frac14 85 kN=m2

                                1

                                2eth1 thorn 3THORN frac14 1

                                2eth270thorn 100THORN frac14 185 kN=m2

                                The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                                uf frac14 36 kN=m2

                                Figure Q43

                                Figure Q44

                                Shear strength 23

                                45

                                3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                46

                                03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                Figure Q45

                                24 Shear strength

                                47

                                The torque required to produce shear failure is given by

                                T frac14 dh cud

                                2thorn 2

                                Z d=2

                                0

                                2r drcur

                                frac14 cud2h

                                2thorn 4cu

                                Z d=2

                                0

                                r2dr

                                frac14 cud2h

                                2thorn d

                                3

                                6

                                Then

                                35 frac14 cu52 10

                                2thorn 53

                                6

                                103

                                cu frac14 76 kN=m3

                                400

                                0 400 800 1200 1600

                                τ (k

                                Nm

                                2 )

                                σprime (kNm2)

                                34deg

                                315deg29deg

                                (a)

                                (b)

                                0 400

                                400

                                800 1200 1600

                                Failure envelope

                                300 500

                                σprime (kNm2)

                                τ (k

                                Nm

                                2 )

                                20 (kNm2)

                                31deg

                                Figure Q46

                                Shear strength 25

                                48

                                The relevant stress values are calculated as follows

                                3 frac14 600 kN=m2

                                1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                2(1 3) 0 40 79 107 139 159

                                1

                                2(01 thorn 03) 400 411 402 389 351 326

                                1

                                2(1 thorn 3) 600 640 679 707 739 759

                                The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                Af frac14 433 200

                                319frac14 073

                                49

                                B frac14 u33

                                frac14 144

                                350 200frac14 096

                                a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                10 283 65 023

                                Figure Q48

                                26 Shear strength

                                The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                Figure Q49

                                Shear strength 27

                                Chapter 5

                                Stresses and displacements

                                51

                                Vertical stress is given by

                                z frac14 Qz2Ip frac14 5000

                                52Ip

                                Values of Ip are obtained from Table 51

                                r (m) rz Ip z (kNm2)

                                0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                10 20 0009 2

                                The variation of z with radial distance (r) is plotted in Figure Q51

                                Figure Q51

                                52

                                Below the centre load (Figure Q52)

                                r

                                zfrac14 0 for the 7500-kN load

                                Ip frac14 0478

                                r

                                zfrac14 5

                                4frac14 125 for the 10 000- and 9000-kN loads

                                Ip frac14 0045

                                Then

                                z frac14X Q

                                z2Ip

                                frac14 7500 0478

                                42thorn 10 000 0045

                                42thorn 9000 0045

                                42

                                frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                53

                                The vertical stress under a corner of a rectangular area is given by

                                z frac14 qIr

                                where values of Ir are obtained from Figure 510 In this case

                                z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                z

                                Figure Q52

                                Stresses and displacements 29

                                z (m) m n Ir z (kNm2)

                                0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                10 010 0005 5

                                z is plotted against z in Figure Q53

                                54

                                (a)

                                m frac14 125

                                12frac14 104

                                n frac14 18

                                12frac14 150

                                From Figure 510 Irfrac14 0196

                                z frac14 2 175 0196 frac14 68 kN=m2

                                Figure Q53

                                30 Stresses and displacements

                                (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                55

                                Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                Px frac14 2Q

                                1

                                m2 thorn 1frac14 2 150

                                125frac14 76 kN=m

                                Equation 517 is used to obtain the pressure distribution

                                px frac14 4Q

                                h

                                m2n

                                ethm2 thorn n2THORN2 frac14150

                                m2n

                                ethm2 thorn n2THORN2 ethkN=m2THORN

                                Figure Q54

                                Stresses and displacements 31

                                n m2n

                                (m2 thorn n2)2

                                px(kNm2)

                                0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                The pressure distribution is plotted in Figure Q55

                                56

                                H

                                Bfrac14 10

                                2frac14 5

                                L

                                Bfrac14 4

                                2frac14 2

                                D

                                Bfrac14 1

                                2frac14 05

                                Hence from Figure 515

                                131 frac14 082

                                130 frac14 094

                                Figure Q55

                                32 Stresses and displacements

                                The immediate settlement is given by Equation 528

                                si frac14 130131qB

                                Eu

                                frac14 094 082 200 2

                                45frac14 7mm

                                Stresses and displacements 33

                                Chapter 6

                                Lateral earth pressure

                                61

                                For 0 frac14 37 the active pressure coefficient is given by

                                Ka frac14 1 sin 37

                                1thorn sin 37frac14 025

                                The total active thrust (Equation 66a with c0 frac14 0) is

                                Pa frac14 1

                                2KaH

                                2 frac14 1

                                2 025 17 62 frac14 765 kN=m

                                If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                and the thrust on the wall is

                                P0 frac14 1

                                2K0H

                                2 frac14 1

                                2 040 17 62 frac14 122 kN=m

                                62

                                The active pressure coefficients for the three soil types are as follows

                                Ka1 frac141 sin 35

                                1thorn sin 35frac14 0271

                                Ka2 frac141 sin 27

                                1thorn sin 27frac14 0375

                                ffiffiffiffiffiffiffiKa2

                                p frac14 0613

                                Ka3 frac141 sin 42

                                1thorn sin 42frac14 0198

                                Distribution of active pressure (plotted in Figure Q62)

                                Depth (m) Soil Active pressure (kNm2)

                                3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                Total thrust frac14 571 kNm

                                Point of application is (4893571) m from the top of the wall ie 857m

                                Force (kN) Arm (m) Moment (kN m)

                                (1)1

                                2 0271 16 32 frac14 195 20 390

                                (2) 0271 16 3 2 frac14 260 40 1040

                                (3)1

                                2 0271 92 22 frac14 50 433 217

                                (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                (5)1

                                2 0375 102 32 frac14 172 70 1204

                                (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                (7)1

                                2 0198 112 42 frac14 177 1067 1889

                                (8)1

                                2 98 92 frac14 3969 90 35721

                                5713 48934

                                Figure Q62

                                Lateral earth pressure 35

                                63

                                (a) For u frac14 0 Ka frac14 Kp frac14 1

                                Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                frac14 245

                                At the lower end of the piling

                                pa frac14 Kaqthorn Kasatz Kaccu

                                frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                frac14 115 kN=m2

                                pp frac14 Kpsatzthorn Kpccu

                                frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                frac14 202 kN=m2

                                (b) For 0 frac14 26 and frac14 1

                                20

                                Ka frac14 035

                                Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                pfrac14 145 ethEquation 619THORN

                                Kp frac14 37

                                Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                pfrac14 47 ethEquation 624THORN

                                At the lower end of the piling

                                pa frac14 Kaqthorn Ka0z Kacc

                                0

                                frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                frac14 187 kN=m2

                                pp frac14 Kp0zthorn Kpcc

                                0

                                frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                frac14 198 kN=m2

                                36 Lateral earth pressure

                                64

                                (a) For 0 frac14 38 Ka frac14 024

                                0 frac14 20 98 frac14 102 kN=m3

                                The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                Force (kN) Arm (m) Moment (kN m)

                                (1) 024 10 66 frac14 159 33 525

                                (2)1

                                2 024 17 392 frac14 310 400 1240

                                (3) 024 17 39 27 frac14 430 135 580

                                (4)1

                                2 024 102 272 frac14 89 090 80

                                (5)1

                                2 98 272 frac14 357 090 321

                                Hfrac14 1345 MH frac14 2746

                                (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                XM frac14MV MH frac14 7790 kNm

                                Lever arm of base resultant

                                M

                                Vfrac14 779

                                488frac14 160

                                Eccentricity of base resultant

                                e frac14 200 160 frac14 040m

                                39 m

                                27 m

                                40 m

                                04 m

                                04 m

                                26 m

                                (7)

                                (9)

                                (1)(2)

                                (3)

                                (4)

                                (5)

                                (8)(6)

                                (10)

                                WT

                                10 kNm2

                                Hydrostatic

                                Figure Q64

                                Lateral earth pressure 37

                                Base pressures (Equation 627)

                                p frac14 VB

                                1 6e

                                B

                                frac14 488

                                4eth1 060THORN

                                frac14 195 kN=m2 and 49 kN=m2

                                Factor of safety against sliding (Equation 628)

                                F frac14 V tan

                                Hfrac14 488 tan 25

                                1345frac14 17

                                (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                Hfrac14 1633 kN

                                V frac14 4879 kN

                                MH frac14 3453 kNm

                                MV frac14 10536 kNm

                                The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                65

                                For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                Kp

                                Ffrac14 385

                                2

                                0 frac14 20 98 frac14 102 kN=m3

                                The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                Force (kN) Arm (m) Moment (kN m)

                                (1)1

                                2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                (2) 026 17 45 d frac14 199d d2 995d2

                                (3)1

                                2 026 102 d2 frac14 133d2 d3 044d3

                                (4)1

                                2 385

                                2 17 152 frac14 368 dthorn 05 368d 184

                                (5)385

                                2 17 15 d frac14 491d d2 2455d2

                                (6)1

                                2 385

                                2 102 d2 frac14 982d2 d3 327d3

                                38 Lateral earth pressure

                                XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                d3 thorn 516d2 283d 1724 frac14 0

                                d frac14 179m

                                Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                Over additional 20 embedded depth

                                pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                66

                                The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                Ka frac14 sin 69=sin 105

                                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                26664

                                37775

                                2

                                frac14 050

                                The total active thrust (acting at 25 above the normal) is given by Equation 616

                                Pa frac14 1

                                2 050 19 7502 frac14 267 kN=m

                                Figure Q65

                                Lateral earth pressure 39

                                Horizontal component

                                Ph frac14 267 cos 40 frac14 205 kN=m

                                Vertical component

                                Pv frac14 267 sin 40 frac14 172 kN=m

                                Consider moments about the toe of the wall (Figure Q66) (per m)

                                Force (kN) Arm (m) Moment (kN m)

                                (1)1

                                2 175 650 235 frac14 1337 258 345

                                (2) 050 650 235 frac14 764 175 134

                                (3)1

                                2 070 650 235 frac14 535 127 68

                                (4) 100 400 235 frac14 940 200 188

                                (5) 1

                                2 080 050 235 frac14 47 027 1

                                Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                Lever arm of base resultant

                                M

                                Vfrac14 795

                                525frac14 151m

                                Eccentricity of base resultant

                                e frac14 200 151 frac14 049m

                                Figure Q66

                                40 Lateral earth pressure

                                Base pressures (Equation 627)

                                p frac14 525

                                41 6 049

                                4

                                frac14 228 kN=m2 and 35 kN=m2

                                The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                67

                                For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                Force (kN) Arm (m) Moment (kNm)

                                (1)1

                                2 027 17 52 frac14 574 183 1050

                                (2) 027 17 5 3 frac14 689 500 3445

                                (3)1

                                2 027 102 32 frac14 124 550 682

                                (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                (5)1

                                2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                (7) 1

                                2 267

                                2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                2 d frac14 163d d2thorn 650 82d2 1060d

                                Tie rod force per m frac14 T 0 0

                                XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                d3 thorn 77d2 269d 1438 frac14 0

                                d frac14 467m

                                Depth of penetration frac14 12d frac14 560m

                                Lateral earth pressure 41

                                Algebraic sum of forces for d frac14 467m isX

                                F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                T frac14 905 kN=m

                                Force in each tie rod frac14 25T frac14 226 kN

                                68

                                (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                0 frac14 21 98 frac14 112 kN=m3

                                The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                uC frac14 150

                                165 15 98 frac14 134 kN=m2

                                The average seepage pressure is

                                j frac14 15

                                165 98 frac14 09 kN=m3

                                Hence

                                0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                0 j frac14 112 09 frac14 103 kN=m3

                                Figure Q67

                                42 Lateral earth pressure

                                Consider moments about the anchor point A (per m)

                                Force (kN) Arm (m) Moment (kN m)

                                (1) 10 026 150 frac14 390 60 2340

                                (2)1

                                2 026 18 452 frac14 474 15 711

                                (3) 026 18 45 105 frac14 2211 825 18240

                                (4)1

                                2 026 121 1052 frac14 1734 100 17340

                                (5)1

                                2 134 15 frac14 101 40 404

                                (6) 134 30 frac14 402 60 2412

                                (7)1

                                2 134 60 frac14 402 95 3819

                                571 4527(8) Ppm

                                115 115PPm

                                XM frac14 0

                                Ppm frac144527

                                115frac14 394 kN=m

                                Available passive resistance

                                Pp frac14 1

                                2 385 103 62 frac14 714 kN=m

                                Factor of safety

                                Fp frac14 Pp

                                Ppm

                                frac14 714

                                394frac14 18

                                Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                Figure Q68

                                Lateral earth pressure 43

                                (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                Consider moments (per m) about the tie point A

                                Force (kN) Arm (m)

                                (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                (2)1

                                2 033 18 452 frac14 601 15

                                (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                (4)1

                                2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                (5)1

                                2 134 15 frac14 101 40

                                (6) 134 30 frac14 402 60

                                (7)1

                                2 134 d frac14 67d d3thorn 75

                                (8) 1

                                2 30 103 d2 frac141545d2 2d3thorn 75

                                Moment (kN m)

                                (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                d3 thorn 827d2 466d 1518 frac14 0

                                By trial

                                d frac14 544m

                                The minimum depth of embedment required is 544m

                                69

                                For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                0 frac14 20 98 frac14 102 kN=m3

                                The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                44 Lateral earth pressure

                                uC frac14 147

                                173 26 98 frac14 216 kN=m2

                                and the average seepage pressure around the wall is

                                j frac14 26

                                173 98 frac14 15 kN=m3

                                Consider moments about the prop (A) (per m)

                                Force (kN) Arm (m) Moment (kN m)

                                (1)1

                                2 03 17 272 frac14 186 020 37

                                (2) 03 17 27 53 frac14 730 335 2445

                                (3)1

                                2 03 (102thorn 15) 532 frac14 493 423 2085

                                (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                (5)1

                                2 216 26 frac14 281 243 684

                                (6) 216 27 frac14 583 465 2712

                                (7)1

                                2 216 60 frac14 648 800 5184

                                3055(8)

                                1

                                2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                Factor of safety

                                Fr frac14 6885

                                3055frac14 225

                                Figure Q69

                                Lateral earth pressure 45

                                610

                                For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                Using the recommendations of Twine and Roscoe

                                p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                611

                                frac14 18 kN=m3 0 frac14 34

                                H frac14 350m nH frac14 335m mH frac14 185m

                                Consider a trial value of F frac14 20 Refer to Figure 635

                                0m frac14 tan1tan 34

                                20

                                frac14 186

                                Then

                                frac14 45 thorn 0m2frac14 543

                                W frac14 1

                                2 18 3502 cot 543 frac14 792 kN=m

                                Figure Q610

                                46 Lateral earth pressure

                                P frac14 1

                                2 s 3352 frac14 561s kN=m

                                U frac14 1

                                2 98 1852 cosec 543 frac14 206 kN=m

                                Equations 630 and 631 then become

                                561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                ie

                                561s 0616N 405 frac14 0

                                792 0857N thorn 563 frac14 0

                                N frac14 848

                                0857frac14 989 kN=m

                                Then

                                561s 609 405 frac14 0

                                s frac14 649

                                561frac14 116 kN=m3

                                The calculations for trial values of F of 20 15 and 10 are summarized below

                                F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                Figure Q611

                                Lateral earth pressure 47

                                612

                                For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                45 thorn 0

                                2frac14 63

                                For the retained material between the surface and a depth of 36m

                                Pa frac14 1

                                2 030 18 362 frac14 350 kN=m

                                Weight of reinforced fill between the surface and a depth of 36m is

                                Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                Eccentricity of Rv

                                e frac14 263 250 frac14 013m

                                The average vertical stress at a depth of 36m is

                                z frac14 Rv

                                L 2efrac14 324

                                474frac14 68 kN=m2

                                (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                Tensile stress in the element frac14 138 103

                                65 3frac14 71N=mm2

                                Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                The weight of ABC is

                                W frac14 1

                                2 18 52 265 frac14 124 kN=m

                                From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                48 Lateral earth pressure

                                (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                Tp frac14 032 68 120 065 frac14 170 kN

                                Tr frac14 213 420

                                418frac14 214 kN

                                Again the tensile failure and slipping limit states are satisfied for this element

                                Figure Q612

                                Lateral earth pressure 49

                                Chapter 7

                                Consolidation theory

                                71

                                Total change in thickness

                                H frac14 782 602 frac14 180mm

                                Average thickness frac14 1530thorn 180

                                2frac14 1620mm

                                Length of drainage path d frac14 1620

                                2frac14 810mm

                                Root time plot (Figure Q71a)

                                ffiffiffiffiffiffit90p frac14 33

                                t90 frac14 109min

                                cv frac14 0848d2

                                t90frac14 0848 8102

                                109 1440 365

                                106frac14 27m2=year

                                r0 frac14 782 764

                                782 602frac14 018

                                180frac14 0100

                                rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                10 119

                                9 180frac14 0735

                                rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                Log time plot (Figure Q71b)

                                t50 frac14 26min

                                cv frac14 0196d2

                                t50frac14 0196 8102

                                26 1440 365

                                106frac14 26m2=year

                                r0 frac14 782 763

                                782 602frac14 019

                                180frac14 0106

                                rp frac14 763 623

                                782 602frac14 140

                                180frac14 0778

                                rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                Figure Q71(a)

                                Figure Q71(b)

                                Final void ratio

                                e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                e

                                Hfrac14 1thorn e0

                                H0frac14 1thorn e1 thorne

                                H0

                                ie

                                e

                                180frac14 1631thorne

                                1710

                                e frac14 2936

                                1530frac14 0192

                                Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                mv frac14 1

                                1thorn e0 e0 e101 00

                                frac14 1

                                1823 0192

                                0107frac14 098m2=MN

                                k frac14 cvmvw frac14 265 098 98

                                60 1440 365 103frac14 81 1010 m=s

                                72

                                Using Equation 77 (one-dimensional method)

                                sc frac14 e0 e11thorn e0 H

                                Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                Figure Q72

                                52 Consolidation theory

                                Settlement

                                Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                318

                                Notes 5 92y 460thorn 84

                                Heave

                                Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                38

                                73

                                U frac14 f ethTvTHORN frac14 f cvt

                                d2

                                Hence if cv is constant

                                t1

                                t2frac14 d

                                21

                                d22

                                where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                d1 frac14 95mm and d2 frac14 2500mm

                                for U frac14 050 t2 frac14 t1 d22

                                d21

                                frac14 20

                                60 24 365 25002

                                952frac14 263 years

                                for U lt 060 Tv frac14

                                4U2 (Equation 724(a))

                                t030 frac14 t050 0302

                                0502

                                frac14 263 036 frac14 095 years

                                Consolidation theory 53

                                74

                                The layer is open

                                d frac14 8

                                2frac14 4m

                                Tv frac14 cvtd2frac14 24 3

                                42frac14 0450

                                ui frac14 frac14 84 kN=m2

                                The excess pore water pressure is given by Equation 721

                                ue frac14Xmfrac141mfrac140

                                2ui

                                Msin

                                Mz

                                d

                                expethM2TvTHORN

                                In this case z frac14 d

                                sinMz

                                d

                                frac14 sinM

                                where

                                M frac14

                                23

                                25

                                2

                                M sin M M2Tv exp (M2Tv)

                                2thorn1 1110 0329

                                3

                                21 9993 457 105

                                ue frac14 2 84 2

                                1 0329 ethother terms negligibleTHORN

                                frac14 352 kN=m2

                                75

                                The layer is open

                                d frac14 6

                                2frac14 3m

                                Tv frac14 cvtd2frac14 10 3

                                32frac14 0333

                                The layer thickness will be divided into six equal parts ie m frac14 6

                                54 Consolidation theory

                                For an open layer

                                Tv frac14 4n

                                m2

                                n frac14 0333 62

                                4frac14 300

                                The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                i j

                                0 1 2 3 4 5 6 7 8 9 10 11 12

                                0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                The initial and 3-year isochrones are plotted in Figure Q75

                                Area under initial isochrone frac14 180 units

                                Area under 3-year isochrone frac14 63 units

                                The average degree of consolidation is given by Equation 725Thus

                                U frac14 1 63

                                180frac14 065

                                Figure Q75

                                Consolidation theory 55

                                76

                                At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                The final consolidation settlement (one-dimensional method) is

                                sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                Corrected time t frac14 2 1

                                2

                                40

                                52

                                frac14 1615 years

                                Tv frac14 cvtd2frac14 44 1615

                                42frac14 0444

                                From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                77

                                The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                Figure Q77

                                56 Consolidation theory

                                Point m n Ir (kNm2) sc (mm)

                                13020frac14 15 20

                                20frac14 10 0194 (4) 113 124

                                260

                                20frac14 30

                                20

                                20frac14 10 0204 (2) 59 65

                                360

                                20frac14 30

                                40

                                20frac14 20 0238 (1) 35 38

                                430

                                20frac14 15

                                40

                                20frac14 20 0224 (2) 65 72

                                Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                78

                                Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                (a) Immediate settlement

                                H

                                Bfrac14 30

                                35frac14 086

                                D

                                Bfrac14 2

                                35frac14 006

                                Figure Q78

                                Consolidation theory 57

                                From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                si frac14 130131qB

                                Eufrac14 10 032 105 35

                                40frac14 30mm

                                (b) Consolidation settlement

                                Layer z (m) Dz Ic (kNm2) syod (mm)

                                1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                3150

                                Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                Now

                                H

                                Bfrac14 30

                                35frac14 086 and A frac14 065

                                from Figure 712 13 frac14 079

                                sc frac14 13sod frac14 079 315 frac14 250mm

                                Total settlement

                                s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                79

                                Without sand drains

                                Uv frac14 025

                                Tv frac14 0049 ethfrom Figure 718THORN

                                t frac14 Tvd2

                                cvfrac14 0049 82

                                cvWith sand drains

                                R frac14 0564S frac14 0564 3 frac14 169m

                                n frac14 Rrfrac14 169

                                015frac14 113

                                Tr frac14 cht

                                4R2frac14 ch

                                4 1692 0049 82

                                cvethand ch frac14 cvTHORN

                                frac14 0275

                                Ur frac14 073 (from Figure 730)

                                58 Consolidation theory

                                Using Equation 740

                                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                U frac14 080

                                710

                                Without sand drains

                                Uv frac14 090

                                Tv frac14 0848

                                t frac14 Tvd2

                                cvfrac14 0848 102

                                96frac14 88 years

                                With sand drains

                                R frac14 0564S frac14 0564 4 frac14 226m

                                n frac14 Rrfrac14 226

                                015frac14 15

                                Tr

                                Tvfrac14 chcv

                                d2

                                4R2ethsame tTHORN

                                Tr

                                Tvfrac14 140

                                96 102

                                4 2262frac14 714 eth1THORN

                                Using Equation 740

                                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                Thus

                                Uv frac14 0295 and Ur frac14 086

                                t frac14 88 00683

                                0848frac14 07 years

                                Consolidation theory 59

                                Chapter 8

                                Bearing capacity

                                81

                                (a) The ultimate bearing capacity is given by Equation 83

                                qf frac14 cNc thorn DNq thorn 1

                                2BN

                                For u frac14 0

                                Nc frac14 514 Nq frac14 1 N frac14 0

                                qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                The net ultimate bearing capacity is

                                qnf frac14 qf D frac14 540 kN=m2

                                The net foundation pressure is

                                qn frac14 q D frac14 425

                                2 eth21 1THORN frac14 192 kN=m2

                                The factor of safety (Equation 86) is

                                F frac14 qnfqnfrac14 540

                                192frac14 28

                                (b) For 0 frac14 28

                                Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                2 112 2 13

                                frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                qnf frac14 574 112 frac14 563 kN=m2

                                F frac14 563

                                192frac14 29

                                (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                82

                                For 0 frac14 38

                                Nq frac14 49 N frac14 67

                                qnf frac14 DethNq 1THORN thorn 1

                                2BN ethfrom Equation 83THORN

                                frac14 eth18 075 48THORN thorn 1

                                2 18 15 67

                                frac14 648thorn 905 frac14 1553 kN=m2

                                qn frac14 500

                                15 eth18 075THORN frac14 320 kN=m2

                                F frac14 qnfqnfrac14 1553

                                320frac14 48

                                0d frac14 tan1tan 38

                                125

                                frac14 32 therefore Nq frac14 23 and N frac14 25

                                Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                2 18 15 25

                                frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                Design load (action) Vd frac14 500 kN=m

                                The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                83

                                D

                                Bfrac14 350

                                225frac14 155

                                From Figure 85 for a square foundation

                                Nc frac14 81

                                Bearing capacity 61

                                For a rectangular foundation (L frac14 450m B frac14 225m)

                                Nc frac14 084thorn 016B

                                L

                                81 frac14 745

                                Using Equation 810

                                qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                For F frac14 3

                                qn frac14 1006

                                3frac14 335 kN=m2

                                q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                Design load frac14 405 450 225 frac14 4100 kN

                                Design undrained strength cud frac14 135

                                14frac14 96 kN=m2

                                Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                frac14 7241 kN

                                Design load Vd frac14 4100 kN

                                Rd gt Vd therefore the bearing resistance limit state is satisfied

                                84

                                For 0 frac14 40

                                Nq frac14 64 N frac14 95

                                qnf frac14 DethNq 1THORN thorn 04BN

                                (a) Water table 5m below ground level

                                qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                qn frac14 400 17 frac14 383 kN=m2

                                F frac14 2686

                                383frac14 70

                                (b) Water table 1m below ground level (ie at foundation level)

                                0 frac14 20 98 frac14 102 kN=m3

                                62 Bearing capacity

                                qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                F frac14 2040

                                383frac14 53

                                (c) Water table at ground level with upward hydraulic gradient 02

                                eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                F frac14 1296

                                392frac14 33

                                85

                                The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                Design value of 0 frac14 tan1tan 39

                                125

                                frac14 33

                                For 0 frac14 33 Nq frac14 26 and N frac14 29

                                Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                Rd gt Vd therefore the bearing resistance limit state is satisfied

                                86

                                (a) Undrained shear for u frac14 0

                                Nc frac14 514 Nq frac14 1 N frac14 0

                                qnf frac14 12cuNc

                                frac14 12 100 514 frac14 617 kN=m2

                                qn frac14 qnfFfrac14 617

                                3frac14 206 kN=m2

                                q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                Bearing capacity 63

                                Drained shear for 0 frac14 32

                                Nq frac14 23 N frac14 25

                                0 frac14 21 98 frac14 112 kN=m3

                                qnf frac14 0DethNq 1THORN thorn 040BN

                                frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                frac14 694 kN=m2

                                q frac14 694

                                3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                Design load frac14 42 227 frac14 3632 kN

                                (b) Design undrained strength cud frac14 100

                                14frac14 71 kNm2

                                Design bearing resistance Rd frac14 12cudNe area

                                frac14 12 71 514 42

                                frac14 7007 kN

                                For drained shear 0d frac14 tan1tan 32

                                125

                                frac14 26

                                Nq frac14 12 N frac14 10

                                Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                1 2 100 0175 0700qn 0182qn

                                2 6 033 0044 0176qn 0046qn

                                3 10 020 0017 0068qn 0018qn

                                0246qn

                                Diameter of equivalent circle B frac14 45m

                                H

                                Bfrac14 12

                                45frac14 27 and A frac14 042

                                13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                64 Bearing capacity

                                For sc frac14 30mm

                                qn frac14 30

                                0147frac14 204 kN=m2

                                q frac14 204thorn 21 frac14 225 kN=m2

                                Design load frac14 42 225 frac14 3600 kN

                                The design load is 3600 kN settlement being the limiting criterion

                                87

                                D

                                Bfrac14 8

                                4frac14 20

                                From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                F frac14 cuNc

                                Dfrac14 40 71

                                20 8frac14 18

                                88

                                Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                Design value of 0 frac14 tan1tan 38

                                125

                                frac14 32

                                Figure Q86

                                Bearing capacity 65

                                For 0 frac14 32 Nq frac14 23 and N frac14 25

                                Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                For B frac14 250m qn frac14 3750

                                2502 17 frac14 583 kN=m2

                                From Figure 510 m frac14 n frac14 126

                                6frac14 021

                                Ir frac14 0019

                                Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                89

                                Depth (m) N 0v (kNm2) CN N1

                                070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                Cw frac14 05thorn 0530

                                47

                                frac14 082

                                66 Bearing capacity

                                Thus

                                qa frac14 150 082 frac14 120 kN=m2

                                (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                Thus

                                qa frac14 90 15 frac14 135 kN=m2

                                (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                Ic frac14 171

                                1014frac14 0068

                                From Equation 819(a) with s frac14 25mm

                                q frac14 25

                                3507 0068frac14 150 kN=m2

                                810

                                Peak value of strain influence factor occurs at a depth of 27m and is given by

                                Izp frac14 05thorn 01130

                                16 27

                                05

                                frac14 067

                                Refer to Figure Q810

                                E frac14 25qc

                                Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                Ez (mm3MN)

                                1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                0203

                                C1 frac14 1 0500qnfrac14 1 05 12 16

                                130frac14 093

                                C2 frac14 1 ethsayTHORN

                                s frac14 C1C2qnX Iz

                                Ez frac14 093 1 130 0203 frac14 25mm

                                Bearing capacity 67

                                811

                                At pile base level

                                cu frac14 220 kN=m2

                                qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                Then

                                Qf frac14 Abqb thorn Asqs

                                frac14

                                4 32 1980

                                thorn eth 105 139 86THORN

                                frac14 13 996thorn 3941 frac14 17 937 kN

                                0 01 02 03 04 05 06 07

                                0 2 4 6 8 10 12 14

                                1

                                2

                                3

                                4

                                5

                                6

                                7

                                8

                                (1)

                                (2)

                                (3)

                                (4)

                                (5)

                                qc

                                qc

                                Iz

                                Iz

                                (MNm2)

                                z (m)

                                Figure Q810

                                68 Bearing capacity

                                Allowable load

                                ethaTHORN Qf

                                2frac14 17 937

                                2frac14 8968 kN

                                ethbTHORN Abqb

                                3thorn Asqs frac14 13 996

                                3thorn 3941 frac14 8606 kN

                                ie allowable load frac14 8600 kN

                                Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                According to the limit state method

                                Characteristic undrained strength at base level cuk frac14 220

                                150kN=m2

                                Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                150frac14 1320 kN=m2

                                Characteristic shaft resistance qsk frac14 00150

                                frac14 86

                                150frac14 57 kN=m2

                                Characteristic base and shaft resistances

                                Rbk frac14

                                4 32 1320 frac14 9330 kN

                                Rsk frac14 105 139 86

                                150frac14 2629 kN

                                For a bored pile the partial factors are b frac14 160 and s frac14 130

                                Design bearing resistance Rcd frac14 9330

                                160thorn 2629

                                130

                                frac14 5831thorn 2022

                                frac14 7850 kN

                                Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                812

                                ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                For a single pile

                                Qf frac14 Abqb thorn Asqs

                                frac14

                                4 062 1305

                                thorn eth 06 15 42THORN

                                frac14 369thorn 1187 frac14 1556 kN

                                Bearing capacity 69

                                Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                qbkfrac14 9cuk frac14 9 220

                                150frac14 1320 kN=m2

                                qskfrac14cuk frac14 040 105

                                150frac14 28 kN=m2

                                Rbkfrac14

                                4 0602 1320 frac14 373 kN

                                Rskfrac14 060 15 28 frac14 791 kN

                                Rcdfrac14 373

                                160thorn 791

                                130frac14 233thorn 608 frac14 841 kN

                                Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                q frac14 21 000

                                1762frac14 68 kN=m2

                                Immediate settlement

                                H

                                Bfrac14 15

                                176frac14 085

                                D

                                Bfrac14 13

                                176frac14 074

                                L

                                Bfrac14 1

                                Hence from Figure 515

                                130 frac14 078 and 131 frac14 041

                                70 Bearing capacity

                                Thus using Equation 528

                                si frac14 078 041 68 176

                                65frac14 6mm

                                Consolidation settlement

                                Layer z (m) Area (m2) (kNm2) mvH (mm)

                                1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                434 (sod)

                                Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                sc frac14 056 434 frac14 24mm

                                The total settlement is (6thorn 24) frac14 30mm

                                813

                                At base level N frac14 26 Then using Equation 830

                                qb frac14 40NDb

                                Bfrac14 40 26 2

                                025frac14 8320 kN=m2

                                ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                Figure Q812

                                Bearing capacity 71

                                Over the length embedded in sand

                                N frac14 21 ie18thorn 24

                                2

                                Using Equation 831

                                qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                For a single pile

                                Qf frac14 Abqb thorn Asqs

                                frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                For the pile group assuming a group efficiency of 12

                                XQf frac14 12 9 604 frac14 6523 kN

                                Then the load factor is

                                F frac14 6523

                                2000thorn 1000frac14 21

                                (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                Characteristic base resistance per unit area qbk frac14 8320

                                150frac14 5547 kNm2

                                Characteristic shaft resistance per unit area qsk frac14 42

                                150frac14 28 kNm2

                                Characteristic base and shaft resistances for a single pile

                                Rbk frac14 0252 5547 frac14 347 kN

                                Rsk frac14 4 025 2 28 frac14 56 kN

                                For a driven pile the partial factors are b frac14 s frac14 130

                                Design bearing resistance Rcd frac14 347

                                130thorn 56

                                130frac14 310 kN

                                For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                72 Bearing capacity

                                N frac14 24thorn 26thorn 34

                                3frac14 28

                                Ic frac14 171

                                2814frac14 0016 ethEquation 818THORN

                                s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                814

                                Using Equation 841

                                Tf frac14 DLcu thorn

                                4ethD2 d2THORNcuNc

                                frac14 eth 02 5 06 110THORN thorn

                                4eth022 012THORN110 9

                                frac14 207thorn 23 frac14 230 kN

                                Figure Q813

                                Bearing capacity 73

                                Chapter 9

                                Stability of slopes

                                91

                                Referring to Figure Q91

                                W frac14 417 19 frac14 792 kN=m

                                Q frac14 20 28 frac14 56 kN=m

                                Arc lengthAB frac14

                                180 73 90 frac14 115m

                                Arc length BC frac14

                                180 28 90 frac14 44m

                                The factor of safety is given by

                                F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                Depth of tension crack z0 frac14 2cu

                                frac14 2 20

                                19frac14 21m

                                Arc length BD frac14

                                180 13

                                1

                                2 90 frac14 21m

                                F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                92

                                u frac14 0

                                Depth factor D frac14 11

                                9frac14 122

                                Using Equation 92 with F frac14 10

                                Ns frac14 cu

                                FHfrac14 30

                                10 19 9frac14 0175

                                Hence from Figure 93

                                frac14 50

                                For F frac14 12

                                Ns frac14 30

                                12 19 9frac14 0146

                                frac14 27

                                93

                                Refer to Figure Q93

                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                74 m

                                214 1deg

                                213 1deg

                                39 m

                                WB

                                D

                                C

                                28 m

                                21 m

                                A

                                Q

                                Soil (1)Soil (2)

                                73deg

                                Figure Q91

                                Stability of slopes 75

                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                599 256 328 1372

                                Figure Q93

                                76 Stability of slopes

                                XW cos frac14 b

                                Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                W sin frac14 bX

                                h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                Arc length La frac14

                                180 57

                                1

                                2 326 frac14 327m

                                The factor of safety is given by

                                F frac14 c0La thorn tan0ethW cos ulTHORN

                                W sin

                                frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                frac14 091

                                According to the limit state method

                                0d frac14 tan1tan 32

                                125

                                frac14 265

                                c0 frac14 8

                                160frac14 5 kN=m2

                                Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                Design disturbing moment frac14 1075 kN=m

                                The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                94

                                F frac14 1

                                W sin

                                Xfc0bthorn ethW ubTHORN tan0g sec

                                1thorn ethtan tan0=FTHORN

                                c0 frac14 8 kN=m2

                                0 frac14 32

                                c0b frac14 8 2 frac14 16 kN=m

                                W frac14 bh frac14 21 2 h frac14 42h kN=m

                                Try F frac14 100

                                tan0

                                Ffrac14 0625

                                Stability of slopes 77

                                Values of u are as obtained in Figure Q93

                                SliceNo

                                h(m)

                                W frac14 bh(kNm)

                                W sin(kNm)

                                ub(kNm)

                                c0bthorn (W ub) tan0(kNm)

                                sec

                                1thorn (tan tan0)FProduct(kNm)

                                1 05 21 6 2 8 24 1078 262 13 55 31

                                23 33 30 1042 31

                                3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                224 92 72 0931 67

                                6 50 210 11 40 100 85 0907 777 55 231 14

                                12 58 112 90 0889 80

                                8 60 252 1812

                                80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                2154 88 116 0853 99

                                14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                1074 1091

                                F frac14 1091

                                1074frac14 102 (assumed value 100)

                                Thus

                                F frac14 101

                                95

                                F frac14 1

                                W sin

                                XfWeth1 ruTHORN tan0g sec

                                1thorn ethtan tan0THORN=F

                                0 frac14 33

                                ru frac14 020

                                W frac14 bh frac14 20 5 h frac14 100h kN=m

                                eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                Try F frac14 110

                                tan 0

                                Ffrac14 tan 33

                                110frac14 0590

                                78 Stability of slopes

                                Referring to Figure Q95

                                SliceNo

                                h(m)

                                W frac14 bh(kNm)

                                W sin(kNm)

                                W(1 ru) tan0(kNm)

                                sec

                                1thorn ( tan tan0)FProduct(kNm)

                                1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                2120 234 0892 209

                                4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                1185 1271

                                Figure Q95

                                Stability of slopes 79

                                F frac14 1271

                                1185frac14 107

                                The trial value was 110 therefore take F to be 108

                                96

                                (a) Water table at surface the factor of safety is given by Equation 912

                                F frac14 0

                                sat

                                tan0

                                tan

                                ptie 15 frac14 92

                                19

                                tan 36

                                tan

                                tan frac14 0234

                                frac14 13

                                Water table well below surface the factor of safety is given by Equation 911

                                F frac14 tan0

                                tan

                                frac14 tan 36

                                tan 13

                                frac14 31

                                (b) 0d frac14 tan1tan 36

                                125

                                frac14 30

                                Depth of potential failure surface frac14 z

                                Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                frac14 504z kN

                                Design disturbing moment per unit area Sd frac14 sat sin cos

                                frac14 19 z sin 13 cos 13

                                frac14 416z kN

                                Rd gtSd therefore the limit state for overall stability is satisfied

                                80 Stability of slopes

                                • Book Cover
                                • Title
                                • Contents
                                • Basic characteristics of soils
                                • Seepage
                                • Effective stress
                                • Shear strength
                                • Stresses and displacements
                                • Lateral earth pressure
                                • Consolidation theory
                                • Bearing capacity
                                • Stability of slopes

                                  Hence

                                  x0 frac14 20m

                                  Using Equation 234 with x0frac14 20m the coordinates of a number of points on thebasic parabola are calculated ie

                                  x frac14 20 z2

                                  80

                                  x 20 0 50 100 200 300z 0 400 748 980 1327 1600

                                  The basic parabola is plotted in Figure Q27 The upstream correction is drawn usingpersonal judgement

                                  No downstream correction is required in this case since frac14 180 If required the topflow line can be plotted back onto the natural section the x coordinates above beingdivided by the scale transformation factor The quantity of seepage can be calculatedusing Equation 233 ie

                                  q frac14 2k0x0 frac14 2 45 106 20 frac14 18 105 m3=s per m

                                  28

                                  The flow net is drawn in Figure Q28 from which Nffrac14 33 and Ndfrac14 7 The overall lossin total head is 28m Then

                                  Figure Q27

                                  Seepage 11

                                  q frac14 kh Nf

                                  Ndfrac14 45 105 28 33

                                  7

                                  frac14 59 105 m3=s per m

                                  29

                                  The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

                                  kx frac14 H1k1 thornH2k2

                                  H1 thornH2frac14 106

                                  10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

                                  kz frac14 H1 thornH2

                                  H1

                                  k1thornH2

                                  k2

                                  frac14 10

                                  5

                                  eth2 106THORN thorn5

                                  eth16 106THORNfrac14 36 106 m=s

                                  Then the scale transformation factor is given by

                                  xt frac14 xffiffiffiffiffikz

                                  pffiffiffiffiffikx

                                  p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

                                  Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

                                  Figure Q28

                                  12 Seepage

                                  k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                                  qfrac14

                                  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

                                  p 106 frac14 57 106 m=s

                                  Then the quantity of seepage is given by

                                  q frac14 k0h Nf

                                  Ndfrac14 57 106 350 56

                                  11

                                  frac14 10 105 m3=s per m

                                  Figure Q29

                                  Seepage 13

                                  Chapter 3

                                  Effective stress

                                  31

                                  Buoyant unit weight

                                  0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                                  Effective vertical stress

                                  0v frac14 5 102 frac14 51 kN=m2 or

                                  Total vertical stress

                                  v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

                                  Pore water pressure

                                  u frac14 7 98 frac14 686 kN=m2

                                  Effective vertical stress

                                  0v frac14 v u frac14 1196 686 frac14 51 kN=m2

                                  32

                                  Buoyant unit weight

                                  0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                                  Effective vertical stress

                                  0v frac14 5 102 frac14 51 kN=m2 or

                                  Total vertical stress

                                  v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

                                  Pore water pressure

                                  u frac14 205 98 frac14 2009 kN=m2

                                  Effective vertical stress

                                  0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

                                  33

                                  At top of the clay

                                  v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

                                  u frac14 2 98 frac14 196 kN=m2

                                  0v frac14 v u frac14 710 196 frac14 514 kN=m2

                                  Alternatively

                                  0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                                  0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

                                  At bottom of the clay

                                  v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

                                  u frac14 12 98 frac14 1176 kN=m2

                                  0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

                                  NB The alternative method of calculation is not applicable because of the artesiancondition

                                  Figure Q3132

                                  Effective stress 15

                                  34

                                  0 frac14 20 98 frac14 102 kN=m3

                                  At 8m depth

                                  0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                                  35

                                  0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                                  0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                                  Figure Q33

                                  Figure Q34

                                  16 Effective stress

                                  (a) Immediately after WT rise

                                  At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                                  0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                                  At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                                  0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                                  (b) Several years after WT rise

                                  At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                                  At 8m depth

                                  0v frac14 940 kN=m2 (as above)

                                  At 12m depth

                                  0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                                  Figure Q35

                                  Effective stress 17

                                  36

                                  Total weight

                                  ab frac14 210 kN

                                  Effective weight

                                  ac frac14 112 kN

                                  Resultant boundary water force

                                  be frac14 119 kN

                                  Seepage force

                                  ce frac14 34 kN

                                  Resultant body force

                                  ae frac14 99 kN eth73 to horizontalTHORN

                                  (Refer to Figure Q36)

                                  Figure Q36

                                  18 Effective stress

                                  37

                                  Situation (1)(a)

                                  frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                                  u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                  0 frac14 u frac14 694 392 frac14 302 kN=m2

                                  (b)

                                  i frac14 2

                                  4frac14 05

                                  j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                                  Situation (2)(a)

                                  frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                                  u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                  0 frac14 u frac14 498 392 frac14 106 kN=m2

                                  (b)

                                  i frac14 2

                                  4frac14 05

                                  j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                                  38

                                  The flow net is drawn in Figure Q24

                                  Loss in total head between adjacent equipotentials

                                  h frac14 550

                                  Ndfrac14 550

                                  11frac14 050m

                                  Exit hydraulic gradient

                                  ie frac14 h

                                  sfrac14 050

                                  070frac14 071

                                  Effective stress 19

                                  The critical hydraulic gradient is given by Equation 39

                                  ic frac14 0

                                  wfrac14 102

                                  98frac14 104

                                  Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                                  F frac14 iciefrac14 104

                                  071frac14 15

                                  Total head at C

                                  hC frac14 nd

                                  Ndh frac14 24

                                  11 550 frac14 120m

                                  Elevation head at C

                                  zC frac14 250m

                                  Pore water pressure at C

                                  uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                                  Therefore effective vertical stress at C

                                  0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                                  For point D

                                  hD frac14 73

                                  11 550 frac14 365m

                                  zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                                  0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                                  39

                                  The flow net is drawn in Figure Q25

                                  For a soil prism 150 300m adjacent to the piling

                                  hm frac14 26

                                  9 500 frac14 145m

                                  20 Effective stress

                                  Factor of safety against lsquoheavingrsquo (Equation 310)

                                  F frac14 ic

                                  imfrac14 0d

                                  whmfrac14 97 300

                                  98 145frac14 20

                                  With a filter

                                  F frac14 0d thorn wwhm

                                  3 frac14 eth97 300THORN thorn w98 145

                                  w frac14 135 kN=m2

                                  Depth of filterfrac14 13521frac14 065m (if above water level)

                                  Effective stress 21

                                  Chapter 4

                                  Shear strength

                                  41

                                  frac14 295 kN=m2

                                  u frac14 120 kN=m2

                                  0 frac14 u frac14 295 120 frac14 175 kN=m2

                                  f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                                  42

                                  03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                                  100 452 552200 908 1108400 1810 2210800 3624 4424

                                  The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                                  Figure Q42

                                  43

                                  3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                                  200 222 422400 218 618600 220 820

                                  The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                                  44

                                  The modified shear strength parameters are

                                  0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                                  a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                                  The coordinates of the stress point representing failure conditions in the test are

                                  1

                                  2eth1 2THORN frac14 1

                                  2 170 frac14 85 kN=m2

                                  1

                                  2eth1 thorn 3THORN frac14 1

                                  2eth270thorn 100THORN frac14 185 kN=m2

                                  The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                                  uf frac14 36 kN=m2

                                  Figure Q43

                                  Figure Q44

                                  Shear strength 23

                                  45

                                  3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                  150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                  The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                  The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                  46

                                  03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                  200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                  The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                  Figure Q45

                                  24 Shear strength

                                  47

                                  The torque required to produce shear failure is given by

                                  T frac14 dh cud

                                  2thorn 2

                                  Z d=2

                                  0

                                  2r drcur

                                  frac14 cud2h

                                  2thorn 4cu

                                  Z d=2

                                  0

                                  r2dr

                                  frac14 cud2h

                                  2thorn d

                                  3

                                  6

                                  Then

                                  35 frac14 cu52 10

                                  2thorn 53

                                  6

                                  103

                                  cu frac14 76 kN=m3

                                  400

                                  0 400 800 1200 1600

                                  τ (k

                                  Nm

                                  2 )

                                  σprime (kNm2)

                                  34deg

                                  315deg29deg

                                  (a)

                                  (b)

                                  0 400

                                  400

                                  800 1200 1600

                                  Failure envelope

                                  300 500

                                  σprime (kNm2)

                                  τ (k

                                  Nm

                                  2 )

                                  20 (kNm2)

                                  31deg

                                  Figure Q46

                                  Shear strength 25

                                  48

                                  The relevant stress values are calculated as follows

                                  3 frac14 600 kN=m2

                                  1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                  2(1 3) 0 40 79 107 139 159

                                  1

                                  2(01 thorn 03) 400 411 402 389 351 326

                                  1

                                  2(1 thorn 3) 600 640 679 707 739 759

                                  The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                  Af frac14 433 200

                                  319frac14 073

                                  49

                                  B frac14 u33

                                  frac14 144

                                  350 200frac14 096

                                  a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                  0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                  10 283 65 023

                                  Figure Q48

                                  26 Shear strength

                                  The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                  Figure Q49

                                  Shear strength 27

                                  Chapter 5

                                  Stresses and displacements

                                  51

                                  Vertical stress is given by

                                  z frac14 Qz2Ip frac14 5000

                                  52Ip

                                  Values of Ip are obtained from Table 51

                                  r (m) rz Ip z (kNm2)

                                  0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                  10 20 0009 2

                                  The variation of z with radial distance (r) is plotted in Figure Q51

                                  Figure Q51

                                  52

                                  Below the centre load (Figure Q52)

                                  r

                                  zfrac14 0 for the 7500-kN load

                                  Ip frac14 0478

                                  r

                                  zfrac14 5

                                  4frac14 125 for the 10 000- and 9000-kN loads

                                  Ip frac14 0045

                                  Then

                                  z frac14X Q

                                  z2Ip

                                  frac14 7500 0478

                                  42thorn 10 000 0045

                                  42thorn 9000 0045

                                  42

                                  frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                  53

                                  The vertical stress under a corner of a rectangular area is given by

                                  z frac14 qIr

                                  where values of Ir are obtained from Figure 510 In this case

                                  z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                  z

                                  Figure Q52

                                  Stresses and displacements 29

                                  z (m) m n Ir z (kNm2)

                                  0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                  10 010 0005 5

                                  z is plotted against z in Figure Q53

                                  54

                                  (a)

                                  m frac14 125

                                  12frac14 104

                                  n frac14 18

                                  12frac14 150

                                  From Figure 510 Irfrac14 0196

                                  z frac14 2 175 0196 frac14 68 kN=m2

                                  Figure Q53

                                  30 Stresses and displacements

                                  (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                  z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                  55

                                  Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                  Px frac14 2Q

                                  1

                                  m2 thorn 1frac14 2 150

                                  125frac14 76 kN=m

                                  Equation 517 is used to obtain the pressure distribution

                                  px frac14 4Q

                                  h

                                  m2n

                                  ethm2 thorn n2THORN2 frac14150

                                  m2n

                                  ethm2 thorn n2THORN2 ethkN=m2THORN

                                  Figure Q54

                                  Stresses and displacements 31

                                  n m2n

                                  (m2 thorn n2)2

                                  px(kNm2)

                                  0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                  The pressure distribution is plotted in Figure Q55

                                  56

                                  H

                                  Bfrac14 10

                                  2frac14 5

                                  L

                                  Bfrac14 4

                                  2frac14 2

                                  D

                                  Bfrac14 1

                                  2frac14 05

                                  Hence from Figure 515

                                  131 frac14 082

                                  130 frac14 094

                                  Figure Q55

                                  32 Stresses and displacements

                                  The immediate settlement is given by Equation 528

                                  si frac14 130131qB

                                  Eu

                                  frac14 094 082 200 2

                                  45frac14 7mm

                                  Stresses and displacements 33

                                  Chapter 6

                                  Lateral earth pressure

                                  61

                                  For 0 frac14 37 the active pressure coefficient is given by

                                  Ka frac14 1 sin 37

                                  1thorn sin 37frac14 025

                                  The total active thrust (Equation 66a with c0 frac14 0) is

                                  Pa frac14 1

                                  2KaH

                                  2 frac14 1

                                  2 025 17 62 frac14 765 kN=m

                                  If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                  K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                  and the thrust on the wall is

                                  P0 frac14 1

                                  2K0H

                                  2 frac14 1

                                  2 040 17 62 frac14 122 kN=m

                                  62

                                  The active pressure coefficients for the three soil types are as follows

                                  Ka1 frac141 sin 35

                                  1thorn sin 35frac14 0271

                                  Ka2 frac141 sin 27

                                  1thorn sin 27frac14 0375

                                  ffiffiffiffiffiffiffiKa2

                                  p frac14 0613

                                  Ka3 frac141 sin 42

                                  1thorn sin 42frac14 0198

                                  Distribution of active pressure (plotted in Figure Q62)

                                  Depth (m) Soil Active pressure (kNm2)

                                  3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                  12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                  At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                  Total thrust frac14 571 kNm

                                  Point of application is (4893571) m from the top of the wall ie 857m

                                  Force (kN) Arm (m) Moment (kN m)

                                  (1)1

                                  2 0271 16 32 frac14 195 20 390

                                  (2) 0271 16 3 2 frac14 260 40 1040

                                  (3)1

                                  2 0271 92 22 frac14 50 433 217

                                  (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                  (5)1

                                  2 0375 102 32 frac14 172 70 1204

                                  (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                  (7)1

                                  2 0198 112 42 frac14 177 1067 1889

                                  (8)1

                                  2 98 92 frac14 3969 90 35721

                                  5713 48934

                                  Figure Q62

                                  Lateral earth pressure 35

                                  63

                                  (a) For u frac14 0 Ka frac14 Kp frac14 1

                                  Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                  frac14 245

                                  At the lower end of the piling

                                  pa frac14 Kaqthorn Kasatz Kaccu

                                  frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                  frac14 115 kN=m2

                                  pp frac14 Kpsatzthorn Kpccu

                                  frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                  frac14 202 kN=m2

                                  (b) For 0 frac14 26 and frac14 1

                                  20

                                  Ka frac14 035

                                  Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                  pfrac14 145 ethEquation 619THORN

                                  Kp frac14 37

                                  Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                  pfrac14 47 ethEquation 624THORN

                                  At the lower end of the piling

                                  pa frac14 Kaqthorn Ka0z Kacc

                                  0

                                  frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                  frac14 187 kN=m2

                                  pp frac14 Kp0zthorn Kpcc

                                  0

                                  frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                  frac14 198 kN=m2

                                  36 Lateral earth pressure

                                  64

                                  (a) For 0 frac14 38 Ka frac14 024

                                  0 frac14 20 98 frac14 102 kN=m3

                                  The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                  Force (kN) Arm (m) Moment (kN m)

                                  (1) 024 10 66 frac14 159 33 525

                                  (2)1

                                  2 024 17 392 frac14 310 400 1240

                                  (3) 024 17 39 27 frac14 430 135 580

                                  (4)1

                                  2 024 102 272 frac14 89 090 80

                                  (5)1

                                  2 98 272 frac14 357 090 321

                                  Hfrac14 1345 MH frac14 2746

                                  (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                  (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                  XM frac14MV MH frac14 7790 kNm

                                  Lever arm of base resultant

                                  M

                                  Vfrac14 779

                                  488frac14 160

                                  Eccentricity of base resultant

                                  e frac14 200 160 frac14 040m

                                  39 m

                                  27 m

                                  40 m

                                  04 m

                                  04 m

                                  26 m

                                  (7)

                                  (9)

                                  (1)(2)

                                  (3)

                                  (4)

                                  (5)

                                  (8)(6)

                                  (10)

                                  WT

                                  10 kNm2

                                  Hydrostatic

                                  Figure Q64

                                  Lateral earth pressure 37

                                  Base pressures (Equation 627)

                                  p frac14 VB

                                  1 6e

                                  B

                                  frac14 488

                                  4eth1 060THORN

                                  frac14 195 kN=m2 and 49 kN=m2

                                  Factor of safety against sliding (Equation 628)

                                  F frac14 V tan

                                  Hfrac14 488 tan 25

                                  1345frac14 17

                                  (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                  Hfrac14 1633 kN

                                  V frac14 4879 kN

                                  MH frac14 3453 kNm

                                  MV frac14 10536 kNm

                                  The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                  65

                                  For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                  Kp

                                  Ffrac14 385

                                  2

                                  0 frac14 20 98 frac14 102 kN=m3

                                  The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                  Force (kN) Arm (m) Moment (kN m)

                                  (1)1

                                  2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                  (2) 026 17 45 d frac14 199d d2 995d2

                                  (3)1

                                  2 026 102 d2 frac14 133d2 d3 044d3

                                  (4)1

                                  2 385

                                  2 17 152 frac14 368 dthorn 05 368d 184

                                  (5)385

                                  2 17 15 d frac14 491d d2 2455d2

                                  (6)1

                                  2 385

                                  2 102 d2 frac14 982d2 d3 327d3

                                  38 Lateral earth pressure

                                  XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                  d3 thorn 516d2 283d 1724 frac14 0

                                  d frac14 179m

                                  Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                  Over additional 20 embedded depth

                                  pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                  Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                  66

                                  The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                  Ka frac14 sin 69=sin 105

                                  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                  pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                  26664

                                  37775

                                  2

                                  frac14 050

                                  The total active thrust (acting at 25 above the normal) is given by Equation 616

                                  Pa frac14 1

                                  2 050 19 7502 frac14 267 kN=m

                                  Figure Q65

                                  Lateral earth pressure 39

                                  Horizontal component

                                  Ph frac14 267 cos 40 frac14 205 kN=m

                                  Vertical component

                                  Pv frac14 267 sin 40 frac14 172 kN=m

                                  Consider moments about the toe of the wall (Figure Q66) (per m)

                                  Force (kN) Arm (m) Moment (kN m)

                                  (1)1

                                  2 175 650 235 frac14 1337 258 345

                                  (2) 050 650 235 frac14 764 175 134

                                  (3)1

                                  2 070 650 235 frac14 535 127 68

                                  (4) 100 400 235 frac14 940 200 188

                                  (5) 1

                                  2 080 050 235 frac14 47 027 1

                                  Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                  Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                  Lever arm of base resultant

                                  M

                                  Vfrac14 795

                                  525frac14 151m

                                  Eccentricity of base resultant

                                  e frac14 200 151 frac14 049m

                                  Figure Q66

                                  40 Lateral earth pressure

                                  Base pressures (Equation 627)

                                  p frac14 525

                                  41 6 049

                                  4

                                  frac14 228 kN=m2 and 35 kN=m2

                                  The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                  The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                  The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                  67

                                  For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                  Force (kN) Arm (m) Moment (kNm)

                                  (1)1

                                  2 027 17 52 frac14 574 183 1050

                                  (2) 027 17 5 3 frac14 689 500 3445

                                  (3)1

                                  2 027 102 32 frac14 124 550 682

                                  (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                  (5)1

                                  2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                  (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                  (7) 1

                                  2 267

                                  2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                  (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                  2 d frac14 163d d2thorn 650 82d2 1060d

                                  Tie rod force per m frac14 T 0 0

                                  XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                  d3 thorn 77d2 269d 1438 frac14 0

                                  d frac14 467m

                                  Depth of penetration frac14 12d frac14 560m

                                  Lateral earth pressure 41

                                  Algebraic sum of forces for d frac14 467m isX

                                  F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                  T frac14 905 kN=m

                                  Force in each tie rod frac14 25T frac14 226 kN

                                  68

                                  (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                  0 frac14 21 98 frac14 112 kN=m3

                                  The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                  uC frac14 150

                                  165 15 98 frac14 134 kN=m2

                                  The average seepage pressure is

                                  j frac14 15

                                  165 98 frac14 09 kN=m3

                                  Hence

                                  0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                  0 j frac14 112 09 frac14 103 kN=m3

                                  Figure Q67

                                  42 Lateral earth pressure

                                  Consider moments about the anchor point A (per m)

                                  Force (kN) Arm (m) Moment (kN m)

                                  (1) 10 026 150 frac14 390 60 2340

                                  (2)1

                                  2 026 18 452 frac14 474 15 711

                                  (3) 026 18 45 105 frac14 2211 825 18240

                                  (4)1

                                  2 026 121 1052 frac14 1734 100 17340

                                  (5)1

                                  2 134 15 frac14 101 40 404

                                  (6) 134 30 frac14 402 60 2412

                                  (7)1

                                  2 134 60 frac14 402 95 3819

                                  571 4527(8) Ppm

                                  115 115PPm

                                  XM frac14 0

                                  Ppm frac144527

                                  115frac14 394 kN=m

                                  Available passive resistance

                                  Pp frac14 1

                                  2 385 103 62 frac14 714 kN=m

                                  Factor of safety

                                  Fp frac14 Pp

                                  Ppm

                                  frac14 714

                                  394frac14 18

                                  Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                  Figure Q68

                                  Lateral earth pressure 43

                                  (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                  Consider moments (per m) about the tie point A

                                  Force (kN) Arm (m)

                                  (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                  (2)1

                                  2 033 18 452 frac14 601 15

                                  (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                  (4)1

                                  2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                  (5)1

                                  2 134 15 frac14 101 40

                                  (6) 134 30 frac14 402 60

                                  (7)1

                                  2 134 d frac14 67d d3thorn 75

                                  (8) 1

                                  2 30 103 d2 frac141545d2 2d3thorn 75

                                  Moment (kN m)

                                  (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                  XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                  d3 thorn 827d2 466d 1518 frac14 0

                                  By trial

                                  d frac14 544m

                                  The minimum depth of embedment required is 544m

                                  69

                                  For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                  0 frac14 20 98 frac14 102 kN=m3

                                  The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                  44 Lateral earth pressure

                                  uC frac14 147

                                  173 26 98 frac14 216 kN=m2

                                  and the average seepage pressure around the wall is

                                  j frac14 26

                                  173 98 frac14 15 kN=m3

                                  Consider moments about the prop (A) (per m)

                                  Force (kN) Arm (m) Moment (kN m)

                                  (1)1

                                  2 03 17 272 frac14 186 020 37

                                  (2) 03 17 27 53 frac14 730 335 2445

                                  (3)1

                                  2 03 (102thorn 15) 532 frac14 493 423 2085

                                  (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                  (5)1

                                  2 216 26 frac14 281 243 684

                                  (6) 216 27 frac14 583 465 2712

                                  (7)1

                                  2 216 60 frac14 648 800 5184

                                  3055(8)

                                  1

                                  2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                  Factor of safety

                                  Fr frac14 6885

                                  3055frac14 225

                                  Figure Q69

                                  Lateral earth pressure 45

                                  610

                                  For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                  p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                  Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                  Using the recommendations of Twine and Roscoe

                                  p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                  Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                  611

                                  frac14 18 kN=m3 0 frac14 34

                                  H frac14 350m nH frac14 335m mH frac14 185m

                                  Consider a trial value of F frac14 20 Refer to Figure 635

                                  0m frac14 tan1tan 34

                                  20

                                  frac14 186

                                  Then

                                  frac14 45 thorn 0m2frac14 543

                                  W frac14 1

                                  2 18 3502 cot 543 frac14 792 kN=m

                                  Figure Q610

                                  46 Lateral earth pressure

                                  P frac14 1

                                  2 s 3352 frac14 561s kN=m

                                  U frac14 1

                                  2 98 1852 cosec 543 frac14 206 kN=m

                                  Equations 630 and 631 then become

                                  561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                  792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                  ie

                                  561s 0616N 405 frac14 0

                                  792 0857N thorn 563 frac14 0

                                  N frac14 848

                                  0857frac14 989 kN=m

                                  Then

                                  561s 609 405 frac14 0

                                  s frac14 649

                                  561frac14 116 kN=m3

                                  The calculations for trial values of F of 20 15 and 10 are summarized below

                                  F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                  20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                  s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                  Figure Q611

                                  Lateral earth pressure 47

                                  612

                                  For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                  45 thorn 0

                                  2frac14 63

                                  For the retained material between the surface and a depth of 36m

                                  Pa frac14 1

                                  2 030 18 362 frac14 350 kN=m

                                  Weight of reinforced fill between the surface and a depth of 36m is

                                  Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                  eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                  Eccentricity of Rv

                                  e frac14 263 250 frac14 013m

                                  The average vertical stress at a depth of 36m is

                                  z frac14 Rv

                                  L 2efrac14 324

                                  474frac14 68 kN=m2

                                  (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                  Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                  Tensile stress in the element frac14 138 103

                                  65 3frac14 71N=mm2

                                  Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                  Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                  Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                  The weight of ABC is

                                  W frac14 1

                                  2 18 52 265 frac14 124 kN=m

                                  From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                  48 Lateral earth pressure

                                  (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                  Tp frac14 032 68 120 065 frac14 170 kN

                                  Tr frac14 213 420

                                  418frac14 214 kN

                                  Again the tensile failure and slipping limit states are satisfied for this element

                                  Figure Q612

                                  Lateral earth pressure 49

                                  Chapter 7

                                  Consolidation theory

                                  71

                                  Total change in thickness

                                  H frac14 782 602 frac14 180mm

                                  Average thickness frac14 1530thorn 180

                                  2frac14 1620mm

                                  Length of drainage path d frac14 1620

                                  2frac14 810mm

                                  Root time plot (Figure Q71a)

                                  ffiffiffiffiffiffit90p frac14 33

                                  t90 frac14 109min

                                  cv frac14 0848d2

                                  t90frac14 0848 8102

                                  109 1440 365

                                  106frac14 27m2=year

                                  r0 frac14 782 764

                                  782 602frac14 018

                                  180frac14 0100

                                  rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                  10 119

                                  9 180frac14 0735

                                  rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                  Log time plot (Figure Q71b)

                                  t50 frac14 26min

                                  cv frac14 0196d2

                                  t50frac14 0196 8102

                                  26 1440 365

                                  106frac14 26m2=year

                                  r0 frac14 782 763

                                  782 602frac14 019

                                  180frac14 0106

                                  rp frac14 763 623

                                  782 602frac14 140

                                  180frac14 0778

                                  rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                  Figure Q71(a)

                                  Figure Q71(b)

                                  Final void ratio

                                  e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                  e

                                  Hfrac14 1thorn e0

                                  H0frac14 1thorn e1 thorne

                                  H0

                                  ie

                                  e

                                  180frac14 1631thorne

                                  1710

                                  e frac14 2936

                                  1530frac14 0192

                                  Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                  mv frac14 1

                                  1thorn e0 e0 e101 00

                                  frac14 1

                                  1823 0192

                                  0107frac14 098m2=MN

                                  k frac14 cvmvw frac14 265 098 98

                                  60 1440 365 103frac14 81 1010 m=s

                                  72

                                  Using Equation 77 (one-dimensional method)

                                  sc frac14 e0 e11thorn e0 H

                                  Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                  Figure Q72

                                  52 Consolidation theory

                                  Settlement

                                  Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                  1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                  318

                                  Notes 5 92y 460thorn 84

                                  Heave

                                  Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                  1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                  38

                                  73

                                  U frac14 f ethTvTHORN frac14 f cvt

                                  d2

                                  Hence if cv is constant

                                  t1

                                  t2frac14 d

                                  21

                                  d22

                                  where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                  d1 frac14 95mm and d2 frac14 2500mm

                                  for U frac14 050 t2 frac14 t1 d22

                                  d21

                                  frac14 20

                                  60 24 365 25002

                                  952frac14 263 years

                                  for U lt 060 Tv frac14

                                  4U2 (Equation 724(a))

                                  t030 frac14 t050 0302

                                  0502

                                  frac14 263 036 frac14 095 years

                                  Consolidation theory 53

                                  74

                                  The layer is open

                                  d frac14 8

                                  2frac14 4m

                                  Tv frac14 cvtd2frac14 24 3

                                  42frac14 0450

                                  ui frac14 frac14 84 kN=m2

                                  The excess pore water pressure is given by Equation 721

                                  ue frac14Xmfrac141mfrac140

                                  2ui

                                  Msin

                                  Mz

                                  d

                                  expethM2TvTHORN

                                  In this case z frac14 d

                                  sinMz

                                  d

                                  frac14 sinM

                                  where

                                  M frac14

                                  23

                                  25

                                  2

                                  M sin M M2Tv exp (M2Tv)

                                  2thorn1 1110 0329

                                  3

                                  21 9993 457 105

                                  ue frac14 2 84 2

                                  1 0329 ethother terms negligibleTHORN

                                  frac14 352 kN=m2

                                  75

                                  The layer is open

                                  d frac14 6

                                  2frac14 3m

                                  Tv frac14 cvtd2frac14 10 3

                                  32frac14 0333

                                  The layer thickness will be divided into six equal parts ie m frac14 6

                                  54 Consolidation theory

                                  For an open layer

                                  Tv frac14 4n

                                  m2

                                  n frac14 0333 62

                                  4frac14 300

                                  The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                  ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                  i j

                                  0 1 2 3 4 5 6 7 8 9 10 11 12

                                  0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                  The initial and 3-year isochrones are plotted in Figure Q75

                                  Area under initial isochrone frac14 180 units

                                  Area under 3-year isochrone frac14 63 units

                                  The average degree of consolidation is given by Equation 725Thus

                                  U frac14 1 63

                                  180frac14 065

                                  Figure Q75

                                  Consolidation theory 55

                                  76

                                  At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                  0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                  The final consolidation settlement (one-dimensional method) is

                                  sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                  Corrected time t frac14 2 1

                                  2

                                  40

                                  52

                                  frac14 1615 years

                                  Tv frac14 cvtd2frac14 44 1615

                                  42frac14 0444

                                  From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                  77

                                  The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                  Figure Q77

                                  56 Consolidation theory

                                  Point m n Ir (kNm2) sc (mm)

                                  13020frac14 15 20

                                  20frac14 10 0194 (4) 113 124

                                  260

                                  20frac14 30

                                  20

                                  20frac14 10 0204 (2) 59 65

                                  360

                                  20frac14 30

                                  40

                                  20frac14 20 0238 (1) 35 38

                                  430

                                  20frac14 15

                                  40

                                  20frac14 20 0224 (2) 65 72

                                  Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                  78

                                  Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                  (a) Immediate settlement

                                  H

                                  Bfrac14 30

                                  35frac14 086

                                  D

                                  Bfrac14 2

                                  35frac14 006

                                  Figure Q78

                                  Consolidation theory 57

                                  From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                  si frac14 130131qB

                                  Eufrac14 10 032 105 35

                                  40frac14 30mm

                                  (b) Consolidation settlement

                                  Layer z (m) Dz Ic (kNm2) syod (mm)

                                  1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                  3150

                                  Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                  Now

                                  H

                                  Bfrac14 30

                                  35frac14 086 and A frac14 065

                                  from Figure 712 13 frac14 079

                                  sc frac14 13sod frac14 079 315 frac14 250mm

                                  Total settlement

                                  s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                  79

                                  Without sand drains

                                  Uv frac14 025

                                  Tv frac14 0049 ethfrom Figure 718THORN

                                  t frac14 Tvd2

                                  cvfrac14 0049 82

                                  cvWith sand drains

                                  R frac14 0564S frac14 0564 3 frac14 169m

                                  n frac14 Rrfrac14 169

                                  015frac14 113

                                  Tr frac14 cht

                                  4R2frac14 ch

                                  4 1692 0049 82

                                  cvethand ch frac14 cvTHORN

                                  frac14 0275

                                  Ur frac14 073 (from Figure 730)

                                  58 Consolidation theory

                                  Using Equation 740

                                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                  U frac14 080

                                  710

                                  Without sand drains

                                  Uv frac14 090

                                  Tv frac14 0848

                                  t frac14 Tvd2

                                  cvfrac14 0848 102

                                  96frac14 88 years

                                  With sand drains

                                  R frac14 0564S frac14 0564 4 frac14 226m

                                  n frac14 Rrfrac14 226

                                  015frac14 15

                                  Tr

                                  Tvfrac14 chcv

                                  d2

                                  4R2ethsame tTHORN

                                  Tr

                                  Tvfrac14 140

                                  96 102

                                  4 2262frac14 714 eth1THORN

                                  Using Equation 740

                                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                  An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                  Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                  040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                  Thus

                                  Uv frac14 0295 and Ur frac14 086

                                  t frac14 88 00683

                                  0848frac14 07 years

                                  Consolidation theory 59

                                  Chapter 8

                                  Bearing capacity

                                  81

                                  (a) The ultimate bearing capacity is given by Equation 83

                                  qf frac14 cNc thorn DNq thorn 1

                                  2BN

                                  For u frac14 0

                                  Nc frac14 514 Nq frac14 1 N frac14 0

                                  qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                  The net ultimate bearing capacity is

                                  qnf frac14 qf D frac14 540 kN=m2

                                  The net foundation pressure is

                                  qn frac14 q D frac14 425

                                  2 eth21 1THORN frac14 192 kN=m2

                                  The factor of safety (Equation 86) is

                                  F frac14 qnfqnfrac14 540

                                  192frac14 28

                                  (b) For 0 frac14 28

                                  Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                  qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                  2 112 2 13

                                  frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                  qnf frac14 574 112 frac14 563 kN=m2

                                  F frac14 563

                                  192frac14 29

                                  (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                  82

                                  For 0 frac14 38

                                  Nq frac14 49 N frac14 67

                                  qnf frac14 DethNq 1THORN thorn 1

                                  2BN ethfrom Equation 83THORN

                                  frac14 eth18 075 48THORN thorn 1

                                  2 18 15 67

                                  frac14 648thorn 905 frac14 1553 kN=m2

                                  qn frac14 500

                                  15 eth18 075THORN frac14 320 kN=m2

                                  F frac14 qnfqnfrac14 1553

                                  320frac14 48

                                  0d frac14 tan1tan 38

                                  125

                                  frac14 32 therefore Nq frac14 23 and N frac14 25

                                  Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                  2 18 15 25

                                  frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                  Design load (action) Vd frac14 500 kN=m

                                  The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                  83

                                  D

                                  Bfrac14 350

                                  225frac14 155

                                  From Figure 85 for a square foundation

                                  Nc frac14 81

                                  Bearing capacity 61

                                  For a rectangular foundation (L frac14 450m B frac14 225m)

                                  Nc frac14 084thorn 016B

                                  L

                                  81 frac14 745

                                  Using Equation 810

                                  qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                  For F frac14 3

                                  qn frac14 1006

                                  3frac14 335 kN=m2

                                  q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                  Design load frac14 405 450 225 frac14 4100 kN

                                  Design undrained strength cud frac14 135

                                  14frac14 96 kN=m2

                                  Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                  frac14 7241 kN

                                  Design load Vd frac14 4100 kN

                                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                                  84

                                  For 0 frac14 40

                                  Nq frac14 64 N frac14 95

                                  qnf frac14 DethNq 1THORN thorn 04BN

                                  (a) Water table 5m below ground level

                                  qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                  qn frac14 400 17 frac14 383 kN=m2

                                  F frac14 2686

                                  383frac14 70

                                  (b) Water table 1m below ground level (ie at foundation level)

                                  0 frac14 20 98 frac14 102 kN=m3

                                  62 Bearing capacity

                                  qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                  F frac14 2040

                                  383frac14 53

                                  (c) Water table at ground level with upward hydraulic gradient 02

                                  eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                  qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                  F frac14 1296

                                  392frac14 33

                                  85

                                  The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                  Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                  Design value of 0 frac14 tan1tan 39

                                  125

                                  frac14 33

                                  For 0 frac14 33 Nq frac14 26 and N frac14 29

                                  Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                                  86

                                  (a) Undrained shear for u frac14 0

                                  Nc frac14 514 Nq frac14 1 N frac14 0

                                  qnf frac14 12cuNc

                                  frac14 12 100 514 frac14 617 kN=m2

                                  qn frac14 qnfFfrac14 617

                                  3frac14 206 kN=m2

                                  q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                  Bearing capacity 63

                                  Drained shear for 0 frac14 32

                                  Nq frac14 23 N frac14 25

                                  0 frac14 21 98 frac14 112 kN=m3

                                  qnf frac14 0DethNq 1THORN thorn 040BN

                                  frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                  frac14 694 kN=m2

                                  q frac14 694

                                  3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                  Design load frac14 42 227 frac14 3632 kN

                                  (b) Design undrained strength cud frac14 100

                                  14frac14 71 kNm2

                                  Design bearing resistance Rd frac14 12cudNe area

                                  frac14 12 71 514 42

                                  frac14 7007 kN

                                  For drained shear 0d frac14 tan1tan 32

                                  125

                                  frac14 26

                                  Nq frac14 12 N frac14 10

                                  Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                  (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                  Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                  1 2 100 0175 0700qn 0182qn

                                  2 6 033 0044 0176qn 0046qn

                                  3 10 020 0017 0068qn 0018qn

                                  0246qn

                                  Diameter of equivalent circle B frac14 45m

                                  H

                                  Bfrac14 12

                                  45frac14 27 and A frac14 042

                                  13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                  64 Bearing capacity

                                  For sc frac14 30mm

                                  qn frac14 30

                                  0147frac14 204 kN=m2

                                  q frac14 204thorn 21 frac14 225 kN=m2

                                  Design load frac14 42 225 frac14 3600 kN

                                  The design load is 3600 kN settlement being the limiting criterion

                                  87

                                  D

                                  Bfrac14 8

                                  4frac14 20

                                  From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                  F frac14 cuNc

                                  Dfrac14 40 71

                                  20 8frac14 18

                                  88

                                  Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                  Design value of 0 frac14 tan1tan 38

                                  125

                                  frac14 32

                                  Figure Q86

                                  Bearing capacity 65

                                  For 0 frac14 32 Nq frac14 23 and N frac14 25

                                  Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                  The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                  Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                  For B frac14 250m qn frac14 3750

                                  2502 17 frac14 583 kN=m2

                                  From Figure 510 m frac14 n frac14 126

                                  6frac14 021

                                  Ir frac14 0019

                                  Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                  Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                  The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                  89

                                  Depth (m) N 0v (kNm2) CN N1

                                  070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                  Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                  (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                  Cw frac14 05thorn 0530

                                  47

                                  frac14 082

                                  66 Bearing capacity

                                  Thus

                                  qa frac14 150 082 frac14 120 kN=m2

                                  (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                  Thus

                                  qa frac14 90 15 frac14 135 kN=m2

                                  (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                  Ic frac14 171

                                  1014frac14 0068

                                  From Equation 819(a) with s frac14 25mm

                                  q frac14 25

                                  3507 0068frac14 150 kN=m2

                                  810

                                  Peak value of strain influence factor occurs at a depth of 27m and is given by

                                  Izp frac14 05thorn 01130

                                  16 27

                                  05

                                  frac14 067

                                  Refer to Figure Q810

                                  E frac14 25qc

                                  Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                  Ez (mm3MN)

                                  1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                  0203

                                  C1 frac14 1 0500qnfrac14 1 05 12 16

                                  130frac14 093

                                  C2 frac14 1 ethsayTHORN

                                  s frac14 C1C2qnX Iz

                                  Ez frac14 093 1 130 0203 frac14 25mm

                                  Bearing capacity 67

                                  811

                                  At pile base level

                                  cu frac14 220 kN=m2

                                  qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                  Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                  00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                  qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                  Then

                                  Qf frac14 Abqb thorn Asqs

                                  frac14

                                  4 32 1980

                                  thorn eth 105 139 86THORN

                                  frac14 13 996thorn 3941 frac14 17 937 kN

                                  0 01 02 03 04 05 06 07

                                  0 2 4 6 8 10 12 14

                                  1

                                  2

                                  3

                                  4

                                  5

                                  6

                                  7

                                  8

                                  (1)

                                  (2)

                                  (3)

                                  (4)

                                  (5)

                                  qc

                                  qc

                                  Iz

                                  Iz

                                  (MNm2)

                                  z (m)

                                  Figure Q810

                                  68 Bearing capacity

                                  Allowable load

                                  ethaTHORN Qf

                                  2frac14 17 937

                                  2frac14 8968 kN

                                  ethbTHORN Abqb

                                  3thorn Asqs frac14 13 996

                                  3thorn 3941 frac14 8606 kN

                                  ie allowable load frac14 8600 kN

                                  Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                  According to the limit state method

                                  Characteristic undrained strength at base level cuk frac14 220

                                  150kN=m2

                                  Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                  150frac14 1320 kN=m2

                                  Characteristic shaft resistance qsk frac14 00150

                                  frac14 86

                                  150frac14 57 kN=m2

                                  Characteristic base and shaft resistances

                                  Rbk frac14

                                  4 32 1320 frac14 9330 kN

                                  Rsk frac14 105 139 86

                                  150frac14 2629 kN

                                  For a bored pile the partial factors are b frac14 160 and s frac14 130

                                  Design bearing resistance Rcd frac14 9330

                                  160thorn 2629

                                  130

                                  frac14 5831thorn 2022

                                  frac14 7850 kN

                                  Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                  812

                                  ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                  qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                  For a single pile

                                  Qf frac14 Abqb thorn Asqs

                                  frac14

                                  4 062 1305

                                  thorn eth 06 15 42THORN

                                  frac14 369thorn 1187 frac14 1556 kN

                                  Bearing capacity 69

                                  Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                  eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                  (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                  (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                  qbkfrac14 9cuk frac14 9 220

                                  150frac14 1320 kN=m2

                                  qskfrac14cuk frac14 040 105

                                  150frac14 28 kN=m2

                                  Rbkfrac14

                                  4 0602 1320 frac14 373 kN

                                  Rskfrac14 060 15 28 frac14 791 kN

                                  Rcdfrac14 373

                                  160thorn 791

                                  130frac14 233thorn 608 frac14 841 kN

                                  Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                  Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                  (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                  q frac14 21 000

                                  1762frac14 68 kN=m2

                                  Immediate settlement

                                  H

                                  Bfrac14 15

                                  176frac14 085

                                  D

                                  Bfrac14 13

                                  176frac14 074

                                  L

                                  Bfrac14 1

                                  Hence from Figure 515

                                  130 frac14 078 and 131 frac14 041

                                  70 Bearing capacity

                                  Thus using Equation 528

                                  si frac14 078 041 68 176

                                  65frac14 6mm

                                  Consolidation settlement

                                  Layer z (m) Area (m2) (kNm2) mvH (mm)

                                  1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                  434 (sod)

                                  Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                  sc frac14 056 434 frac14 24mm

                                  The total settlement is (6thorn 24) frac14 30mm

                                  813

                                  At base level N frac14 26 Then using Equation 830

                                  qb frac14 40NDb

                                  Bfrac14 40 26 2

                                  025frac14 8320 kN=m2

                                  ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                  Figure Q812

                                  Bearing capacity 71

                                  Over the length embedded in sand

                                  N frac14 21 ie18thorn 24

                                  2

                                  Using Equation 831

                                  qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                  For a single pile

                                  Qf frac14 Abqb thorn Asqs

                                  frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                  For the pile group assuming a group efficiency of 12

                                  XQf frac14 12 9 604 frac14 6523 kN

                                  Then the load factor is

                                  F frac14 6523

                                  2000thorn 1000frac14 21

                                  (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                  Characteristic base resistance per unit area qbk frac14 8320

                                  150frac14 5547 kNm2

                                  Characteristic shaft resistance per unit area qsk frac14 42

                                  150frac14 28 kNm2

                                  Characteristic base and shaft resistances for a single pile

                                  Rbk frac14 0252 5547 frac14 347 kN

                                  Rsk frac14 4 025 2 28 frac14 56 kN

                                  For a driven pile the partial factors are b frac14 s frac14 130

                                  Design bearing resistance Rcd frac14 347

                                  130thorn 56

                                  130frac14 310 kN

                                  For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                  Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                  (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                  From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                  72 Bearing capacity

                                  N frac14 24thorn 26thorn 34

                                  3frac14 28

                                  Ic frac14 171

                                  2814frac14 0016 ethEquation 818THORN

                                  s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                  814

                                  Using Equation 841

                                  Tf frac14 DLcu thorn

                                  4ethD2 d2THORNcuNc

                                  frac14 eth 02 5 06 110THORN thorn

                                  4eth022 012THORN110 9

                                  frac14 207thorn 23 frac14 230 kN

                                  Figure Q813

                                  Bearing capacity 73

                                  Chapter 9

                                  Stability of slopes

                                  91

                                  Referring to Figure Q91

                                  W frac14 417 19 frac14 792 kN=m

                                  Q frac14 20 28 frac14 56 kN=m

                                  Arc lengthAB frac14

                                  180 73 90 frac14 115m

                                  Arc length BC frac14

                                  180 28 90 frac14 44m

                                  The factor of safety is given by

                                  F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                  90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                  Depth of tension crack z0 frac14 2cu

                                  frac14 2 20

                                  19frac14 21m

                                  Arc length BD frac14

                                  180 13

                                  1

                                  2 90 frac14 21m

                                  F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                  The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                  Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                  14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                  Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                  The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                  92

                                  u frac14 0

                                  Depth factor D frac14 11

                                  9frac14 122

                                  Using Equation 92 with F frac14 10

                                  Ns frac14 cu

                                  FHfrac14 30

                                  10 19 9frac14 0175

                                  Hence from Figure 93

                                  frac14 50

                                  For F frac14 12

                                  Ns frac14 30

                                  12 19 9frac14 0146

                                  frac14 27

                                  93

                                  Refer to Figure Q93

                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                  1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                  74 m

                                  214 1deg

                                  213 1deg

                                  39 m

                                  WB

                                  D

                                  C

                                  28 m

                                  21 m

                                  A

                                  Q

                                  Soil (1)Soil (2)

                                  73deg

                                  Figure Q91

                                  Stability of slopes 75

                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                  9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                  599 256 328 1372

                                  Figure Q93

                                  76 Stability of slopes

                                  XW cos frac14 b

                                  Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                  W sin frac14 bX

                                  h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                  Arc length La frac14

                                  180 57

                                  1

                                  2 326 frac14 327m

                                  The factor of safety is given by

                                  F frac14 c0La thorn tan0ethW cos ulTHORN

                                  W sin

                                  frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                  frac14 091

                                  According to the limit state method

                                  0d frac14 tan1tan 32

                                  125

                                  frac14 265

                                  c0 frac14 8

                                  160frac14 5 kN=m2

                                  Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                  Design disturbing moment frac14 1075 kN=m

                                  The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                  94

                                  F frac14 1

                                  W sin

                                  Xfc0bthorn ethW ubTHORN tan0g sec

                                  1thorn ethtan tan0=FTHORN

                                  c0 frac14 8 kN=m2

                                  0 frac14 32

                                  c0b frac14 8 2 frac14 16 kN=m

                                  W frac14 bh frac14 21 2 h frac14 42h kN=m

                                  Try F frac14 100

                                  tan0

                                  Ffrac14 0625

                                  Stability of slopes 77

                                  Values of u are as obtained in Figure Q93

                                  SliceNo

                                  h(m)

                                  W frac14 bh(kNm)

                                  W sin(kNm)

                                  ub(kNm)

                                  c0bthorn (W ub) tan0(kNm)

                                  sec

                                  1thorn (tan tan0)FProduct(kNm)

                                  1 05 21 6 2 8 24 1078 262 13 55 31

                                  23 33 30 1042 31

                                  3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                  224 92 72 0931 67

                                  6 50 210 11 40 100 85 0907 777 55 231 14

                                  12 58 112 90 0889 80

                                  8 60 252 1812

                                  80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                  10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                  2154 88 116 0853 99

                                  14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                  1074 1091

                                  F frac14 1091

                                  1074frac14 102 (assumed value 100)

                                  Thus

                                  F frac14 101

                                  95

                                  F frac14 1

                                  W sin

                                  XfWeth1 ruTHORN tan0g sec

                                  1thorn ethtan tan0THORN=F

                                  0 frac14 33

                                  ru frac14 020

                                  W frac14 bh frac14 20 5 h frac14 100h kN=m

                                  eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                  Try F frac14 110

                                  tan 0

                                  Ffrac14 tan 33

                                  110frac14 0590

                                  78 Stability of slopes

                                  Referring to Figure Q95

                                  SliceNo

                                  h(m)

                                  W frac14 bh(kNm)

                                  W sin(kNm)

                                  W(1 ru) tan0(kNm)

                                  sec

                                  1thorn ( tan tan0)FProduct(kNm)

                                  1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                  2120 234 0892 209

                                  4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                  1185 1271

                                  Figure Q95

                                  Stability of slopes 79

                                  F frac14 1271

                                  1185frac14 107

                                  The trial value was 110 therefore take F to be 108

                                  96

                                  (a) Water table at surface the factor of safety is given by Equation 912

                                  F frac14 0

                                  sat

                                  tan0

                                  tan

                                  ptie 15 frac14 92

                                  19

                                  tan 36

                                  tan

                                  tan frac14 0234

                                  frac14 13

                                  Water table well below surface the factor of safety is given by Equation 911

                                  F frac14 tan0

                                  tan

                                  frac14 tan 36

                                  tan 13

                                  frac14 31

                                  (b) 0d frac14 tan1tan 36

                                  125

                                  frac14 30

                                  Depth of potential failure surface frac14 z

                                  Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                  frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                  frac14 504z kN

                                  Design disturbing moment per unit area Sd frac14 sat sin cos

                                  frac14 19 z sin 13 cos 13

                                  frac14 416z kN

                                  Rd gtSd therefore the limit state for overall stability is satisfied

                                  80 Stability of slopes

                                  • Book Cover
                                  • Title
                                  • Contents
                                  • Basic characteristics of soils
                                  • Seepage
                                  • Effective stress
                                  • Shear strength
                                  • Stresses and displacements
                                  • Lateral earth pressure
                                  • Consolidation theory
                                  • Bearing capacity
                                  • Stability of slopes

                                    q frac14 kh Nf

                                    Ndfrac14 45 105 28 33

                                    7

                                    frac14 59 105 m3=s per m

                                    29

                                    The two isotropic soil layers each 5m thick can be considered as a single homo-geneous anisotropic layer of thickness 10m in which the coefficients of permeability inthe horizontal and vertical directions respectively are given by Equations 224 and225 ie

                                    kx frac14 H1k1 thornH2k2

                                    H1 thornH2frac14 106

                                    10feth5 20THORN thorn eth5 16THORNg frac14 90 106 m=s

                                    kz frac14 H1 thornH2

                                    H1

                                    k1thornH2

                                    k2

                                    frac14 10

                                    5

                                    eth2 106THORN thorn5

                                    eth16 106THORNfrac14 36 106 m=s

                                    Then the scale transformation factor is given by

                                    xt frac14 xffiffiffiffiffikz

                                    pffiffiffiffiffikx

                                    p frac14 xffiffiffiffiffiffiffi36pffiffiffiffiffiffiffi90p frac14 063x

                                    Thus in the transformed section the dimension 1000m becomes 630m vertical dimen-sions are unchanged The transformed section is shown in Figure Q29 and the flow netis drawn as for a single isotropic layer From the flow net Nffrac14 56 and Ndfrac14 11 Theoverall loss in total head is 350m The equivalent isotropic permeability is

                                    Figure Q28

                                    12 Seepage

                                    k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                                    qfrac14

                                    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

                                    p 106 frac14 57 106 m=s

                                    Then the quantity of seepage is given by

                                    q frac14 k0h Nf

                                    Ndfrac14 57 106 350 56

                                    11

                                    frac14 10 105 m3=s per m

                                    Figure Q29

                                    Seepage 13

                                    Chapter 3

                                    Effective stress

                                    31

                                    Buoyant unit weight

                                    0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                                    Effective vertical stress

                                    0v frac14 5 102 frac14 51 kN=m2 or

                                    Total vertical stress

                                    v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

                                    Pore water pressure

                                    u frac14 7 98 frac14 686 kN=m2

                                    Effective vertical stress

                                    0v frac14 v u frac14 1196 686 frac14 51 kN=m2

                                    32

                                    Buoyant unit weight

                                    0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                                    Effective vertical stress

                                    0v frac14 5 102 frac14 51 kN=m2 or

                                    Total vertical stress

                                    v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

                                    Pore water pressure

                                    u frac14 205 98 frac14 2009 kN=m2

                                    Effective vertical stress

                                    0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

                                    33

                                    At top of the clay

                                    v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

                                    u frac14 2 98 frac14 196 kN=m2

                                    0v frac14 v u frac14 710 196 frac14 514 kN=m2

                                    Alternatively

                                    0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                                    0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

                                    At bottom of the clay

                                    v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

                                    u frac14 12 98 frac14 1176 kN=m2

                                    0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

                                    NB The alternative method of calculation is not applicable because of the artesiancondition

                                    Figure Q3132

                                    Effective stress 15

                                    34

                                    0 frac14 20 98 frac14 102 kN=m3

                                    At 8m depth

                                    0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                                    35

                                    0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                                    0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                                    Figure Q33

                                    Figure Q34

                                    16 Effective stress

                                    (a) Immediately after WT rise

                                    At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                                    0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                                    At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                                    0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                                    (b) Several years after WT rise

                                    At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                                    At 8m depth

                                    0v frac14 940 kN=m2 (as above)

                                    At 12m depth

                                    0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                                    Figure Q35

                                    Effective stress 17

                                    36

                                    Total weight

                                    ab frac14 210 kN

                                    Effective weight

                                    ac frac14 112 kN

                                    Resultant boundary water force

                                    be frac14 119 kN

                                    Seepage force

                                    ce frac14 34 kN

                                    Resultant body force

                                    ae frac14 99 kN eth73 to horizontalTHORN

                                    (Refer to Figure Q36)

                                    Figure Q36

                                    18 Effective stress

                                    37

                                    Situation (1)(a)

                                    frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                                    u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                    0 frac14 u frac14 694 392 frac14 302 kN=m2

                                    (b)

                                    i frac14 2

                                    4frac14 05

                                    j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                                    Situation (2)(a)

                                    frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                                    u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                    0 frac14 u frac14 498 392 frac14 106 kN=m2

                                    (b)

                                    i frac14 2

                                    4frac14 05

                                    j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                                    38

                                    The flow net is drawn in Figure Q24

                                    Loss in total head between adjacent equipotentials

                                    h frac14 550

                                    Ndfrac14 550

                                    11frac14 050m

                                    Exit hydraulic gradient

                                    ie frac14 h

                                    sfrac14 050

                                    070frac14 071

                                    Effective stress 19

                                    The critical hydraulic gradient is given by Equation 39

                                    ic frac14 0

                                    wfrac14 102

                                    98frac14 104

                                    Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                                    F frac14 iciefrac14 104

                                    071frac14 15

                                    Total head at C

                                    hC frac14 nd

                                    Ndh frac14 24

                                    11 550 frac14 120m

                                    Elevation head at C

                                    zC frac14 250m

                                    Pore water pressure at C

                                    uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                                    Therefore effective vertical stress at C

                                    0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                                    For point D

                                    hD frac14 73

                                    11 550 frac14 365m

                                    zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                                    0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                                    39

                                    The flow net is drawn in Figure Q25

                                    For a soil prism 150 300m adjacent to the piling

                                    hm frac14 26

                                    9 500 frac14 145m

                                    20 Effective stress

                                    Factor of safety against lsquoheavingrsquo (Equation 310)

                                    F frac14 ic

                                    imfrac14 0d

                                    whmfrac14 97 300

                                    98 145frac14 20

                                    With a filter

                                    F frac14 0d thorn wwhm

                                    3 frac14 eth97 300THORN thorn w98 145

                                    w frac14 135 kN=m2

                                    Depth of filterfrac14 13521frac14 065m (if above water level)

                                    Effective stress 21

                                    Chapter 4

                                    Shear strength

                                    41

                                    frac14 295 kN=m2

                                    u frac14 120 kN=m2

                                    0 frac14 u frac14 295 120 frac14 175 kN=m2

                                    f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                                    42

                                    03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                                    100 452 552200 908 1108400 1810 2210800 3624 4424

                                    The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                                    Figure Q42

                                    43

                                    3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                                    200 222 422400 218 618600 220 820

                                    The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                                    44

                                    The modified shear strength parameters are

                                    0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                                    a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                                    The coordinates of the stress point representing failure conditions in the test are

                                    1

                                    2eth1 2THORN frac14 1

                                    2 170 frac14 85 kN=m2

                                    1

                                    2eth1 thorn 3THORN frac14 1

                                    2eth270thorn 100THORN frac14 185 kN=m2

                                    The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                                    uf frac14 36 kN=m2

                                    Figure Q43

                                    Figure Q44

                                    Shear strength 23

                                    45

                                    3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                    150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                    The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                    The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                    46

                                    03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                    200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                    The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                    Figure Q45

                                    24 Shear strength

                                    47

                                    The torque required to produce shear failure is given by

                                    T frac14 dh cud

                                    2thorn 2

                                    Z d=2

                                    0

                                    2r drcur

                                    frac14 cud2h

                                    2thorn 4cu

                                    Z d=2

                                    0

                                    r2dr

                                    frac14 cud2h

                                    2thorn d

                                    3

                                    6

                                    Then

                                    35 frac14 cu52 10

                                    2thorn 53

                                    6

                                    103

                                    cu frac14 76 kN=m3

                                    400

                                    0 400 800 1200 1600

                                    τ (k

                                    Nm

                                    2 )

                                    σprime (kNm2)

                                    34deg

                                    315deg29deg

                                    (a)

                                    (b)

                                    0 400

                                    400

                                    800 1200 1600

                                    Failure envelope

                                    300 500

                                    σprime (kNm2)

                                    τ (k

                                    Nm

                                    2 )

                                    20 (kNm2)

                                    31deg

                                    Figure Q46

                                    Shear strength 25

                                    48

                                    The relevant stress values are calculated as follows

                                    3 frac14 600 kN=m2

                                    1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                    2(1 3) 0 40 79 107 139 159

                                    1

                                    2(01 thorn 03) 400 411 402 389 351 326

                                    1

                                    2(1 thorn 3) 600 640 679 707 739 759

                                    The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                    Af frac14 433 200

                                    319frac14 073

                                    49

                                    B frac14 u33

                                    frac14 144

                                    350 200frac14 096

                                    a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                    0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                    10 283 65 023

                                    Figure Q48

                                    26 Shear strength

                                    The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                    Figure Q49

                                    Shear strength 27

                                    Chapter 5

                                    Stresses and displacements

                                    51

                                    Vertical stress is given by

                                    z frac14 Qz2Ip frac14 5000

                                    52Ip

                                    Values of Ip are obtained from Table 51

                                    r (m) rz Ip z (kNm2)

                                    0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                    10 20 0009 2

                                    The variation of z with radial distance (r) is plotted in Figure Q51

                                    Figure Q51

                                    52

                                    Below the centre load (Figure Q52)

                                    r

                                    zfrac14 0 for the 7500-kN load

                                    Ip frac14 0478

                                    r

                                    zfrac14 5

                                    4frac14 125 for the 10 000- and 9000-kN loads

                                    Ip frac14 0045

                                    Then

                                    z frac14X Q

                                    z2Ip

                                    frac14 7500 0478

                                    42thorn 10 000 0045

                                    42thorn 9000 0045

                                    42

                                    frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                    53

                                    The vertical stress under a corner of a rectangular area is given by

                                    z frac14 qIr

                                    where values of Ir are obtained from Figure 510 In this case

                                    z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                    z

                                    Figure Q52

                                    Stresses and displacements 29

                                    z (m) m n Ir z (kNm2)

                                    0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                    10 010 0005 5

                                    z is plotted against z in Figure Q53

                                    54

                                    (a)

                                    m frac14 125

                                    12frac14 104

                                    n frac14 18

                                    12frac14 150

                                    From Figure 510 Irfrac14 0196

                                    z frac14 2 175 0196 frac14 68 kN=m2

                                    Figure Q53

                                    30 Stresses and displacements

                                    (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                    z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                    55

                                    Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                    Px frac14 2Q

                                    1

                                    m2 thorn 1frac14 2 150

                                    125frac14 76 kN=m

                                    Equation 517 is used to obtain the pressure distribution

                                    px frac14 4Q

                                    h

                                    m2n

                                    ethm2 thorn n2THORN2 frac14150

                                    m2n

                                    ethm2 thorn n2THORN2 ethkN=m2THORN

                                    Figure Q54

                                    Stresses and displacements 31

                                    n m2n

                                    (m2 thorn n2)2

                                    px(kNm2)

                                    0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                    The pressure distribution is plotted in Figure Q55

                                    56

                                    H

                                    Bfrac14 10

                                    2frac14 5

                                    L

                                    Bfrac14 4

                                    2frac14 2

                                    D

                                    Bfrac14 1

                                    2frac14 05

                                    Hence from Figure 515

                                    131 frac14 082

                                    130 frac14 094

                                    Figure Q55

                                    32 Stresses and displacements

                                    The immediate settlement is given by Equation 528

                                    si frac14 130131qB

                                    Eu

                                    frac14 094 082 200 2

                                    45frac14 7mm

                                    Stresses and displacements 33

                                    Chapter 6

                                    Lateral earth pressure

                                    61

                                    For 0 frac14 37 the active pressure coefficient is given by

                                    Ka frac14 1 sin 37

                                    1thorn sin 37frac14 025

                                    The total active thrust (Equation 66a with c0 frac14 0) is

                                    Pa frac14 1

                                    2KaH

                                    2 frac14 1

                                    2 025 17 62 frac14 765 kN=m

                                    If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                    K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                    and the thrust on the wall is

                                    P0 frac14 1

                                    2K0H

                                    2 frac14 1

                                    2 040 17 62 frac14 122 kN=m

                                    62

                                    The active pressure coefficients for the three soil types are as follows

                                    Ka1 frac141 sin 35

                                    1thorn sin 35frac14 0271

                                    Ka2 frac141 sin 27

                                    1thorn sin 27frac14 0375

                                    ffiffiffiffiffiffiffiKa2

                                    p frac14 0613

                                    Ka3 frac141 sin 42

                                    1thorn sin 42frac14 0198

                                    Distribution of active pressure (plotted in Figure Q62)

                                    Depth (m) Soil Active pressure (kNm2)

                                    3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                    12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                    At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                    Total thrust frac14 571 kNm

                                    Point of application is (4893571) m from the top of the wall ie 857m

                                    Force (kN) Arm (m) Moment (kN m)

                                    (1)1

                                    2 0271 16 32 frac14 195 20 390

                                    (2) 0271 16 3 2 frac14 260 40 1040

                                    (3)1

                                    2 0271 92 22 frac14 50 433 217

                                    (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                    (5)1

                                    2 0375 102 32 frac14 172 70 1204

                                    (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                    (7)1

                                    2 0198 112 42 frac14 177 1067 1889

                                    (8)1

                                    2 98 92 frac14 3969 90 35721

                                    5713 48934

                                    Figure Q62

                                    Lateral earth pressure 35

                                    63

                                    (a) For u frac14 0 Ka frac14 Kp frac14 1

                                    Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                    frac14 245

                                    At the lower end of the piling

                                    pa frac14 Kaqthorn Kasatz Kaccu

                                    frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                    frac14 115 kN=m2

                                    pp frac14 Kpsatzthorn Kpccu

                                    frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                    frac14 202 kN=m2

                                    (b) For 0 frac14 26 and frac14 1

                                    20

                                    Ka frac14 035

                                    Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                    pfrac14 145 ethEquation 619THORN

                                    Kp frac14 37

                                    Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                    pfrac14 47 ethEquation 624THORN

                                    At the lower end of the piling

                                    pa frac14 Kaqthorn Ka0z Kacc

                                    0

                                    frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                    frac14 187 kN=m2

                                    pp frac14 Kp0zthorn Kpcc

                                    0

                                    frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                    frac14 198 kN=m2

                                    36 Lateral earth pressure

                                    64

                                    (a) For 0 frac14 38 Ka frac14 024

                                    0 frac14 20 98 frac14 102 kN=m3

                                    The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                    Force (kN) Arm (m) Moment (kN m)

                                    (1) 024 10 66 frac14 159 33 525

                                    (2)1

                                    2 024 17 392 frac14 310 400 1240

                                    (3) 024 17 39 27 frac14 430 135 580

                                    (4)1

                                    2 024 102 272 frac14 89 090 80

                                    (5)1

                                    2 98 272 frac14 357 090 321

                                    Hfrac14 1345 MH frac14 2746

                                    (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                    (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                    XM frac14MV MH frac14 7790 kNm

                                    Lever arm of base resultant

                                    M

                                    Vfrac14 779

                                    488frac14 160

                                    Eccentricity of base resultant

                                    e frac14 200 160 frac14 040m

                                    39 m

                                    27 m

                                    40 m

                                    04 m

                                    04 m

                                    26 m

                                    (7)

                                    (9)

                                    (1)(2)

                                    (3)

                                    (4)

                                    (5)

                                    (8)(6)

                                    (10)

                                    WT

                                    10 kNm2

                                    Hydrostatic

                                    Figure Q64

                                    Lateral earth pressure 37

                                    Base pressures (Equation 627)

                                    p frac14 VB

                                    1 6e

                                    B

                                    frac14 488

                                    4eth1 060THORN

                                    frac14 195 kN=m2 and 49 kN=m2

                                    Factor of safety against sliding (Equation 628)

                                    F frac14 V tan

                                    Hfrac14 488 tan 25

                                    1345frac14 17

                                    (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                    Hfrac14 1633 kN

                                    V frac14 4879 kN

                                    MH frac14 3453 kNm

                                    MV frac14 10536 kNm

                                    The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                    65

                                    For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                    Kp

                                    Ffrac14 385

                                    2

                                    0 frac14 20 98 frac14 102 kN=m3

                                    The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                    Force (kN) Arm (m) Moment (kN m)

                                    (1)1

                                    2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                    (2) 026 17 45 d frac14 199d d2 995d2

                                    (3)1

                                    2 026 102 d2 frac14 133d2 d3 044d3

                                    (4)1

                                    2 385

                                    2 17 152 frac14 368 dthorn 05 368d 184

                                    (5)385

                                    2 17 15 d frac14 491d d2 2455d2

                                    (6)1

                                    2 385

                                    2 102 d2 frac14 982d2 d3 327d3

                                    38 Lateral earth pressure

                                    XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                    d3 thorn 516d2 283d 1724 frac14 0

                                    d frac14 179m

                                    Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                    Over additional 20 embedded depth

                                    pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                    Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                    66

                                    The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                    Ka frac14 sin 69=sin 105

                                    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                    pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                    26664

                                    37775

                                    2

                                    frac14 050

                                    The total active thrust (acting at 25 above the normal) is given by Equation 616

                                    Pa frac14 1

                                    2 050 19 7502 frac14 267 kN=m

                                    Figure Q65

                                    Lateral earth pressure 39

                                    Horizontal component

                                    Ph frac14 267 cos 40 frac14 205 kN=m

                                    Vertical component

                                    Pv frac14 267 sin 40 frac14 172 kN=m

                                    Consider moments about the toe of the wall (Figure Q66) (per m)

                                    Force (kN) Arm (m) Moment (kN m)

                                    (1)1

                                    2 175 650 235 frac14 1337 258 345

                                    (2) 050 650 235 frac14 764 175 134

                                    (3)1

                                    2 070 650 235 frac14 535 127 68

                                    (4) 100 400 235 frac14 940 200 188

                                    (5) 1

                                    2 080 050 235 frac14 47 027 1

                                    Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                    Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                    Lever arm of base resultant

                                    M

                                    Vfrac14 795

                                    525frac14 151m

                                    Eccentricity of base resultant

                                    e frac14 200 151 frac14 049m

                                    Figure Q66

                                    40 Lateral earth pressure

                                    Base pressures (Equation 627)

                                    p frac14 525

                                    41 6 049

                                    4

                                    frac14 228 kN=m2 and 35 kN=m2

                                    The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                    The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                    The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                    67

                                    For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                    Force (kN) Arm (m) Moment (kNm)

                                    (1)1

                                    2 027 17 52 frac14 574 183 1050

                                    (2) 027 17 5 3 frac14 689 500 3445

                                    (3)1

                                    2 027 102 32 frac14 124 550 682

                                    (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                    (5)1

                                    2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                    (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                    (7) 1

                                    2 267

                                    2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                    (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                    2 d frac14 163d d2thorn 650 82d2 1060d

                                    Tie rod force per m frac14 T 0 0

                                    XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                    d3 thorn 77d2 269d 1438 frac14 0

                                    d frac14 467m

                                    Depth of penetration frac14 12d frac14 560m

                                    Lateral earth pressure 41

                                    Algebraic sum of forces for d frac14 467m isX

                                    F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                    T frac14 905 kN=m

                                    Force in each tie rod frac14 25T frac14 226 kN

                                    68

                                    (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                    0 frac14 21 98 frac14 112 kN=m3

                                    The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                    uC frac14 150

                                    165 15 98 frac14 134 kN=m2

                                    The average seepage pressure is

                                    j frac14 15

                                    165 98 frac14 09 kN=m3

                                    Hence

                                    0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                    0 j frac14 112 09 frac14 103 kN=m3

                                    Figure Q67

                                    42 Lateral earth pressure

                                    Consider moments about the anchor point A (per m)

                                    Force (kN) Arm (m) Moment (kN m)

                                    (1) 10 026 150 frac14 390 60 2340

                                    (2)1

                                    2 026 18 452 frac14 474 15 711

                                    (3) 026 18 45 105 frac14 2211 825 18240

                                    (4)1

                                    2 026 121 1052 frac14 1734 100 17340

                                    (5)1

                                    2 134 15 frac14 101 40 404

                                    (6) 134 30 frac14 402 60 2412

                                    (7)1

                                    2 134 60 frac14 402 95 3819

                                    571 4527(8) Ppm

                                    115 115PPm

                                    XM frac14 0

                                    Ppm frac144527

                                    115frac14 394 kN=m

                                    Available passive resistance

                                    Pp frac14 1

                                    2 385 103 62 frac14 714 kN=m

                                    Factor of safety

                                    Fp frac14 Pp

                                    Ppm

                                    frac14 714

                                    394frac14 18

                                    Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                    Figure Q68

                                    Lateral earth pressure 43

                                    (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                    Consider moments (per m) about the tie point A

                                    Force (kN) Arm (m)

                                    (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                    (2)1

                                    2 033 18 452 frac14 601 15

                                    (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                    (4)1

                                    2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                    (5)1

                                    2 134 15 frac14 101 40

                                    (6) 134 30 frac14 402 60

                                    (7)1

                                    2 134 d frac14 67d d3thorn 75

                                    (8) 1

                                    2 30 103 d2 frac141545d2 2d3thorn 75

                                    Moment (kN m)

                                    (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                    XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                    d3 thorn 827d2 466d 1518 frac14 0

                                    By trial

                                    d frac14 544m

                                    The minimum depth of embedment required is 544m

                                    69

                                    For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                    0 frac14 20 98 frac14 102 kN=m3

                                    The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                    44 Lateral earth pressure

                                    uC frac14 147

                                    173 26 98 frac14 216 kN=m2

                                    and the average seepage pressure around the wall is

                                    j frac14 26

                                    173 98 frac14 15 kN=m3

                                    Consider moments about the prop (A) (per m)

                                    Force (kN) Arm (m) Moment (kN m)

                                    (1)1

                                    2 03 17 272 frac14 186 020 37

                                    (2) 03 17 27 53 frac14 730 335 2445

                                    (3)1

                                    2 03 (102thorn 15) 532 frac14 493 423 2085

                                    (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                    (5)1

                                    2 216 26 frac14 281 243 684

                                    (6) 216 27 frac14 583 465 2712

                                    (7)1

                                    2 216 60 frac14 648 800 5184

                                    3055(8)

                                    1

                                    2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                    Factor of safety

                                    Fr frac14 6885

                                    3055frac14 225

                                    Figure Q69

                                    Lateral earth pressure 45

                                    610

                                    For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                    p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                    Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                    Using the recommendations of Twine and Roscoe

                                    p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                    Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                    611

                                    frac14 18 kN=m3 0 frac14 34

                                    H frac14 350m nH frac14 335m mH frac14 185m

                                    Consider a trial value of F frac14 20 Refer to Figure 635

                                    0m frac14 tan1tan 34

                                    20

                                    frac14 186

                                    Then

                                    frac14 45 thorn 0m2frac14 543

                                    W frac14 1

                                    2 18 3502 cot 543 frac14 792 kN=m

                                    Figure Q610

                                    46 Lateral earth pressure

                                    P frac14 1

                                    2 s 3352 frac14 561s kN=m

                                    U frac14 1

                                    2 98 1852 cosec 543 frac14 206 kN=m

                                    Equations 630 and 631 then become

                                    561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                    792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                    ie

                                    561s 0616N 405 frac14 0

                                    792 0857N thorn 563 frac14 0

                                    N frac14 848

                                    0857frac14 989 kN=m

                                    Then

                                    561s 609 405 frac14 0

                                    s frac14 649

                                    561frac14 116 kN=m3

                                    The calculations for trial values of F of 20 15 and 10 are summarized below

                                    F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                    20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                    s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                    Figure Q611

                                    Lateral earth pressure 47

                                    612

                                    For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                    45 thorn 0

                                    2frac14 63

                                    For the retained material between the surface and a depth of 36m

                                    Pa frac14 1

                                    2 030 18 362 frac14 350 kN=m

                                    Weight of reinforced fill between the surface and a depth of 36m is

                                    Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                    eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                    Eccentricity of Rv

                                    e frac14 263 250 frac14 013m

                                    The average vertical stress at a depth of 36m is

                                    z frac14 Rv

                                    L 2efrac14 324

                                    474frac14 68 kN=m2

                                    (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                    Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                    Tensile stress in the element frac14 138 103

                                    65 3frac14 71N=mm2

                                    Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                    Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                    Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                    The weight of ABC is

                                    W frac14 1

                                    2 18 52 265 frac14 124 kN=m

                                    From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                    48 Lateral earth pressure

                                    (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                    Tp frac14 032 68 120 065 frac14 170 kN

                                    Tr frac14 213 420

                                    418frac14 214 kN

                                    Again the tensile failure and slipping limit states are satisfied for this element

                                    Figure Q612

                                    Lateral earth pressure 49

                                    Chapter 7

                                    Consolidation theory

                                    71

                                    Total change in thickness

                                    H frac14 782 602 frac14 180mm

                                    Average thickness frac14 1530thorn 180

                                    2frac14 1620mm

                                    Length of drainage path d frac14 1620

                                    2frac14 810mm

                                    Root time plot (Figure Q71a)

                                    ffiffiffiffiffiffit90p frac14 33

                                    t90 frac14 109min

                                    cv frac14 0848d2

                                    t90frac14 0848 8102

                                    109 1440 365

                                    106frac14 27m2=year

                                    r0 frac14 782 764

                                    782 602frac14 018

                                    180frac14 0100

                                    rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                    10 119

                                    9 180frac14 0735

                                    rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                    Log time plot (Figure Q71b)

                                    t50 frac14 26min

                                    cv frac14 0196d2

                                    t50frac14 0196 8102

                                    26 1440 365

                                    106frac14 26m2=year

                                    r0 frac14 782 763

                                    782 602frac14 019

                                    180frac14 0106

                                    rp frac14 763 623

                                    782 602frac14 140

                                    180frac14 0778

                                    rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                    Figure Q71(a)

                                    Figure Q71(b)

                                    Final void ratio

                                    e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                    e

                                    Hfrac14 1thorn e0

                                    H0frac14 1thorn e1 thorne

                                    H0

                                    ie

                                    e

                                    180frac14 1631thorne

                                    1710

                                    e frac14 2936

                                    1530frac14 0192

                                    Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                    mv frac14 1

                                    1thorn e0 e0 e101 00

                                    frac14 1

                                    1823 0192

                                    0107frac14 098m2=MN

                                    k frac14 cvmvw frac14 265 098 98

                                    60 1440 365 103frac14 81 1010 m=s

                                    72

                                    Using Equation 77 (one-dimensional method)

                                    sc frac14 e0 e11thorn e0 H

                                    Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                    Figure Q72

                                    52 Consolidation theory

                                    Settlement

                                    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                    1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                    318

                                    Notes 5 92y 460thorn 84

                                    Heave

                                    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                    1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                    38

                                    73

                                    U frac14 f ethTvTHORN frac14 f cvt

                                    d2

                                    Hence if cv is constant

                                    t1

                                    t2frac14 d

                                    21

                                    d22

                                    where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                    d1 frac14 95mm and d2 frac14 2500mm

                                    for U frac14 050 t2 frac14 t1 d22

                                    d21

                                    frac14 20

                                    60 24 365 25002

                                    952frac14 263 years

                                    for U lt 060 Tv frac14

                                    4U2 (Equation 724(a))

                                    t030 frac14 t050 0302

                                    0502

                                    frac14 263 036 frac14 095 years

                                    Consolidation theory 53

                                    74

                                    The layer is open

                                    d frac14 8

                                    2frac14 4m

                                    Tv frac14 cvtd2frac14 24 3

                                    42frac14 0450

                                    ui frac14 frac14 84 kN=m2

                                    The excess pore water pressure is given by Equation 721

                                    ue frac14Xmfrac141mfrac140

                                    2ui

                                    Msin

                                    Mz

                                    d

                                    expethM2TvTHORN

                                    In this case z frac14 d

                                    sinMz

                                    d

                                    frac14 sinM

                                    where

                                    M frac14

                                    23

                                    25

                                    2

                                    M sin M M2Tv exp (M2Tv)

                                    2thorn1 1110 0329

                                    3

                                    21 9993 457 105

                                    ue frac14 2 84 2

                                    1 0329 ethother terms negligibleTHORN

                                    frac14 352 kN=m2

                                    75

                                    The layer is open

                                    d frac14 6

                                    2frac14 3m

                                    Tv frac14 cvtd2frac14 10 3

                                    32frac14 0333

                                    The layer thickness will be divided into six equal parts ie m frac14 6

                                    54 Consolidation theory

                                    For an open layer

                                    Tv frac14 4n

                                    m2

                                    n frac14 0333 62

                                    4frac14 300

                                    The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                    ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                    i j

                                    0 1 2 3 4 5 6 7 8 9 10 11 12

                                    0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                    The initial and 3-year isochrones are plotted in Figure Q75

                                    Area under initial isochrone frac14 180 units

                                    Area under 3-year isochrone frac14 63 units

                                    The average degree of consolidation is given by Equation 725Thus

                                    U frac14 1 63

                                    180frac14 065

                                    Figure Q75

                                    Consolidation theory 55

                                    76

                                    At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                    0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                    The final consolidation settlement (one-dimensional method) is

                                    sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                    Corrected time t frac14 2 1

                                    2

                                    40

                                    52

                                    frac14 1615 years

                                    Tv frac14 cvtd2frac14 44 1615

                                    42frac14 0444

                                    From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                    77

                                    The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                    Figure Q77

                                    56 Consolidation theory

                                    Point m n Ir (kNm2) sc (mm)

                                    13020frac14 15 20

                                    20frac14 10 0194 (4) 113 124

                                    260

                                    20frac14 30

                                    20

                                    20frac14 10 0204 (2) 59 65

                                    360

                                    20frac14 30

                                    40

                                    20frac14 20 0238 (1) 35 38

                                    430

                                    20frac14 15

                                    40

                                    20frac14 20 0224 (2) 65 72

                                    Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                    78

                                    Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                    (a) Immediate settlement

                                    H

                                    Bfrac14 30

                                    35frac14 086

                                    D

                                    Bfrac14 2

                                    35frac14 006

                                    Figure Q78

                                    Consolidation theory 57

                                    From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                    si frac14 130131qB

                                    Eufrac14 10 032 105 35

                                    40frac14 30mm

                                    (b) Consolidation settlement

                                    Layer z (m) Dz Ic (kNm2) syod (mm)

                                    1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                    3150

                                    Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                    Now

                                    H

                                    Bfrac14 30

                                    35frac14 086 and A frac14 065

                                    from Figure 712 13 frac14 079

                                    sc frac14 13sod frac14 079 315 frac14 250mm

                                    Total settlement

                                    s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                    79

                                    Without sand drains

                                    Uv frac14 025

                                    Tv frac14 0049 ethfrom Figure 718THORN

                                    t frac14 Tvd2

                                    cvfrac14 0049 82

                                    cvWith sand drains

                                    R frac14 0564S frac14 0564 3 frac14 169m

                                    n frac14 Rrfrac14 169

                                    015frac14 113

                                    Tr frac14 cht

                                    4R2frac14 ch

                                    4 1692 0049 82

                                    cvethand ch frac14 cvTHORN

                                    frac14 0275

                                    Ur frac14 073 (from Figure 730)

                                    58 Consolidation theory

                                    Using Equation 740

                                    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                    U frac14 080

                                    710

                                    Without sand drains

                                    Uv frac14 090

                                    Tv frac14 0848

                                    t frac14 Tvd2

                                    cvfrac14 0848 102

                                    96frac14 88 years

                                    With sand drains

                                    R frac14 0564S frac14 0564 4 frac14 226m

                                    n frac14 Rrfrac14 226

                                    015frac14 15

                                    Tr

                                    Tvfrac14 chcv

                                    d2

                                    4R2ethsame tTHORN

                                    Tr

                                    Tvfrac14 140

                                    96 102

                                    4 2262frac14 714 eth1THORN

                                    Using Equation 740

                                    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                    An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                    Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                    040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                    Thus

                                    Uv frac14 0295 and Ur frac14 086

                                    t frac14 88 00683

                                    0848frac14 07 years

                                    Consolidation theory 59

                                    Chapter 8

                                    Bearing capacity

                                    81

                                    (a) The ultimate bearing capacity is given by Equation 83

                                    qf frac14 cNc thorn DNq thorn 1

                                    2BN

                                    For u frac14 0

                                    Nc frac14 514 Nq frac14 1 N frac14 0

                                    qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                    The net ultimate bearing capacity is

                                    qnf frac14 qf D frac14 540 kN=m2

                                    The net foundation pressure is

                                    qn frac14 q D frac14 425

                                    2 eth21 1THORN frac14 192 kN=m2

                                    The factor of safety (Equation 86) is

                                    F frac14 qnfqnfrac14 540

                                    192frac14 28

                                    (b) For 0 frac14 28

                                    Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                    qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                    2 112 2 13

                                    frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                    qnf frac14 574 112 frac14 563 kN=m2

                                    F frac14 563

                                    192frac14 29

                                    (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                    82

                                    For 0 frac14 38

                                    Nq frac14 49 N frac14 67

                                    qnf frac14 DethNq 1THORN thorn 1

                                    2BN ethfrom Equation 83THORN

                                    frac14 eth18 075 48THORN thorn 1

                                    2 18 15 67

                                    frac14 648thorn 905 frac14 1553 kN=m2

                                    qn frac14 500

                                    15 eth18 075THORN frac14 320 kN=m2

                                    F frac14 qnfqnfrac14 1553

                                    320frac14 48

                                    0d frac14 tan1tan 38

                                    125

                                    frac14 32 therefore Nq frac14 23 and N frac14 25

                                    Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                    2 18 15 25

                                    frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                    Design load (action) Vd frac14 500 kN=m

                                    The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                    83

                                    D

                                    Bfrac14 350

                                    225frac14 155

                                    From Figure 85 for a square foundation

                                    Nc frac14 81

                                    Bearing capacity 61

                                    For a rectangular foundation (L frac14 450m B frac14 225m)

                                    Nc frac14 084thorn 016B

                                    L

                                    81 frac14 745

                                    Using Equation 810

                                    qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                    For F frac14 3

                                    qn frac14 1006

                                    3frac14 335 kN=m2

                                    q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                    Design load frac14 405 450 225 frac14 4100 kN

                                    Design undrained strength cud frac14 135

                                    14frac14 96 kN=m2

                                    Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                    frac14 7241 kN

                                    Design load Vd frac14 4100 kN

                                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                                    84

                                    For 0 frac14 40

                                    Nq frac14 64 N frac14 95

                                    qnf frac14 DethNq 1THORN thorn 04BN

                                    (a) Water table 5m below ground level

                                    qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                    qn frac14 400 17 frac14 383 kN=m2

                                    F frac14 2686

                                    383frac14 70

                                    (b) Water table 1m below ground level (ie at foundation level)

                                    0 frac14 20 98 frac14 102 kN=m3

                                    62 Bearing capacity

                                    qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                    F frac14 2040

                                    383frac14 53

                                    (c) Water table at ground level with upward hydraulic gradient 02

                                    eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                    qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                    F frac14 1296

                                    392frac14 33

                                    85

                                    The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                    Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                    Design value of 0 frac14 tan1tan 39

                                    125

                                    frac14 33

                                    For 0 frac14 33 Nq frac14 26 and N frac14 29

                                    Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                                    86

                                    (a) Undrained shear for u frac14 0

                                    Nc frac14 514 Nq frac14 1 N frac14 0

                                    qnf frac14 12cuNc

                                    frac14 12 100 514 frac14 617 kN=m2

                                    qn frac14 qnfFfrac14 617

                                    3frac14 206 kN=m2

                                    q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                    Bearing capacity 63

                                    Drained shear for 0 frac14 32

                                    Nq frac14 23 N frac14 25

                                    0 frac14 21 98 frac14 112 kN=m3

                                    qnf frac14 0DethNq 1THORN thorn 040BN

                                    frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                    frac14 694 kN=m2

                                    q frac14 694

                                    3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                    Design load frac14 42 227 frac14 3632 kN

                                    (b) Design undrained strength cud frac14 100

                                    14frac14 71 kNm2

                                    Design bearing resistance Rd frac14 12cudNe area

                                    frac14 12 71 514 42

                                    frac14 7007 kN

                                    For drained shear 0d frac14 tan1tan 32

                                    125

                                    frac14 26

                                    Nq frac14 12 N frac14 10

                                    Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                    (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                    Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                    1 2 100 0175 0700qn 0182qn

                                    2 6 033 0044 0176qn 0046qn

                                    3 10 020 0017 0068qn 0018qn

                                    0246qn

                                    Diameter of equivalent circle B frac14 45m

                                    H

                                    Bfrac14 12

                                    45frac14 27 and A frac14 042

                                    13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                    64 Bearing capacity

                                    For sc frac14 30mm

                                    qn frac14 30

                                    0147frac14 204 kN=m2

                                    q frac14 204thorn 21 frac14 225 kN=m2

                                    Design load frac14 42 225 frac14 3600 kN

                                    The design load is 3600 kN settlement being the limiting criterion

                                    87

                                    D

                                    Bfrac14 8

                                    4frac14 20

                                    From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                    F frac14 cuNc

                                    Dfrac14 40 71

                                    20 8frac14 18

                                    88

                                    Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                    Design value of 0 frac14 tan1tan 38

                                    125

                                    frac14 32

                                    Figure Q86

                                    Bearing capacity 65

                                    For 0 frac14 32 Nq frac14 23 and N frac14 25

                                    Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                    The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                    Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                    For B frac14 250m qn frac14 3750

                                    2502 17 frac14 583 kN=m2

                                    From Figure 510 m frac14 n frac14 126

                                    6frac14 021

                                    Ir frac14 0019

                                    Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                    Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                    The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                    89

                                    Depth (m) N 0v (kNm2) CN N1

                                    070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                    Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                    (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                    Cw frac14 05thorn 0530

                                    47

                                    frac14 082

                                    66 Bearing capacity

                                    Thus

                                    qa frac14 150 082 frac14 120 kN=m2

                                    (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                    Thus

                                    qa frac14 90 15 frac14 135 kN=m2

                                    (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                    Ic frac14 171

                                    1014frac14 0068

                                    From Equation 819(a) with s frac14 25mm

                                    q frac14 25

                                    3507 0068frac14 150 kN=m2

                                    810

                                    Peak value of strain influence factor occurs at a depth of 27m and is given by

                                    Izp frac14 05thorn 01130

                                    16 27

                                    05

                                    frac14 067

                                    Refer to Figure Q810

                                    E frac14 25qc

                                    Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                    Ez (mm3MN)

                                    1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                    0203

                                    C1 frac14 1 0500qnfrac14 1 05 12 16

                                    130frac14 093

                                    C2 frac14 1 ethsayTHORN

                                    s frac14 C1C2qnX Iz

                                    Ez frac14 093 1 130 0203 frac14 25mm

                                    Bearing capacity 67

                                    811

                                    At pile base level

                                    cu frac14 220 kN=m2

                                    qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                    Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                    00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                    qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                    Then

                                    Qf frac14 Abqb thorn Asqs

                                    frac14

                                    4 32 1980

                                    thorn eth 105 139 86THORN

                                    frac14 13 996thorn 3941 frac14 17 937 kN

                                    0 01 02 03 04 05 06 07

                                    0 2 4 6 8 10 12 14

                                    1

                                    2

                                    3

                                    4

                                    5

                                    6

                                    7

                                    8

                                    (1)

                                    (2)

                                    (3)

                                    (4)

                                    (5)

                                    qc

                                    qc

                                    Iz

                                    Iz

                                    (MNm2)

                                    z (m)

                                    Figure Q810

                                    68 Bearing capacity

                                    Allowable load

                                    ethaTHORN Qf

                                    2frac14 17 937

                                    2frac14 8968 kN

                                    ethbTHORN Abqb

                                    3thorn Asqs frac14 13 996

                                    3thorn 3941 frac14 8606 kN

                                    ie allowable load frac14 8600 kN

                                    Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                    According to the limit state method

                                    Characteristic undrained strength at base level cuk frac14 220

                                    150kN=m2

                                    Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                    150frac14 1320 kN=m2

                                    Characteristic shaft resistance qsk frac14 00150

                                    frac14 86

                                    150frac14 57 kN=m2

                                    Characteristic base and shaft resistances

                                    Rbk frac14

                                    4 32 1320 frac14 9330 kN

                                    Rsk frac14 105 139 86

                                    150frac14 2629 kN

                                    For a bored pile the partial factors are b frac14 160 and s frac14 130

                                    Design bearing resistance Rcd frac14 9330

                                    160thorn 2629

                                    130

                                    frac14 5831thorn 2022

                                    frac14 7850 kN

                                    Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                    812

                                    ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                    qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                    For a single pile

                                    Qf frac14 Abqb thorn Asqs

                                    frac14

                                    4 062 1305

                                    thorn eth 06 15 42THORN

                                    frac14 369thorn 1187 frac14 1556 kN

                                    Bearing capacity 69

                                    Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                    eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                    (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                    (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                    qbkfrac14 9cuk frac14 9 220

                                    150frac14 1320 kN=m2

                                    qskfrac14cuk frac14 040 105

                                    150frac14 28 kN=m2

                                    Rbkfrac14

                                    4 0602 1320 frac14 373 kN

                                    Rskfrac14 060 15 28 frac14 791 kN

                                    Rcdfrac14 373

                                    160thorn 791

                                    130frac14 233thorn 608 frac14 841 kN

                                    Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                    Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                    (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                    q frac14 21 000

                                    1762frac14 68 kN=m2

                                    Immediate settlement

                                    H

                                    Bfrac14 15

                                    176frac14 085

                                    D

                                    Bfrac14 13

                                    176frac14 074

                                    L

                                    Bfrac14 1

                                    Hence from Figure 515

                                    130 frac14 078 and 131 frac14 041

                                    70 Bearing capacity

                                    Thus using Equation 528

                                    si frac14 078 041 68 176

                                    65frac14 6mm

                                    Consolidation settlement

                                    Layer z (m) Area (m2) (kNm2) mvH (mm)

                                    1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                    434 (sod)

                                    Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                    sc frac14 056 434 frac14 24mm

                                    The total settlement is (6thorn 24) frac14 30mm

                                    813

                                    At base level N frac14 26 Then using Equation 830

                                    qb frac14 40NDb

                                    Bfrac14 40 26 2

                                    025frac14 8320 kN=m2

                                    ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                    Figure Q812

                                    Bearing capacity 71

                                    Over the length embedded in sand

                                    N frac14 21 ie18thorn 24

                                    2

                                    Using Equation 831

                                    qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                    For a single pile

                                    Qf frac14 Abqb thorn Asqs

                                    frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                    For the pile group assuming a group efficiency of 12

                                    XQf frac14 12 9 604 frac14 6523 kN

                                    Then the load factor is

                                    F frac14 6523

                                    2000thorn 1000frac14 21

                                    (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                    Characteristic base resistance per unit area qbk frac14 8320

                                    150frac14 5547 kNm2

                                    Characteristic shaft resistance per unit area qsk frac14 42

                                    150frac14 28 kNm2

                                    Characteristic base and shaft resistances for a single pile

                                    Rbk frac14 0252 5547 frac14 347 kN

                                    Rsk frac14 4 025 2 28 frac14 56 kN

                                    For a driven pile the partial factors are b frac14 s frac14 130

                                    Design bearing resistance Rcd frac14 347

                                    130thorn 56

                                    130frac14 310 kN

                                    For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                    Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                    (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                    From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                    72 Bearing capacity

                                    N frac14 24thorn 26thorn 34

                                    3frac14 28

                                    Ic frac14 171

                                    2814frac14 0016 ethEquation 818THORN

                                    s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                    814

                                    Using Equation 841

                                    Tf frac14 DLcu thorn

                                    4ethD2 d2THORNcuNc

                                    frac14 eth 02 5 06 110THORN thorn

                                    4eth022 012THORN110 9

                                    frac14 207thorn 23 frac14 230 kN

                                    Figure Q813

                                    Bearing capacity 73

                                    Chapter 9

                                    Stability of slopes

                                    91

                                    Referring to Figure Q91

                                    W frac14 417 19 frac14 792 kN=m

                                    Q frac14 20 28 frac14 56 kN=m

                                    Arc lengthAB frac14

                                    180 73 90 frac14 115m

                                    Arc length BC frac14

                                    180 28 90 frac14 44m

                                    The factor of safety is given by

                                    F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                    90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                    Depth of tension crack z0 frac14 2cu

                                    frac14 2 20

                                    19frac14 21m

                                    Arc length BD frac14

                                    180 13

                                    1

                                    2 90 frac14 21m

                                    F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                    The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                    Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                    14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                    Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                    The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                    92

                                    u frac14 0

                                    Depth factor D frac14 11

                                    9frac14 122

                                    Using Equation 92 with F frac14 10

                                    Ns frac14 cu

                                    FHfrac14 30

                                    10 19 9frac14 0175

                                    Hence from Figure 93

                                    frac14 50

                                    For F frac14 12

                                    Ns frac14 30

                                    12 19 9frac14 0146

                                    frac14 27

                                    93

                                    Refer to Figure Q93

                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                    1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                    74 m

                                    214 1deg

                                    213 1deg

                                    39 m

                                    WB

                                    D

                                    C

                                    28 m

                                    21 m

                                    A

                                    Q

                                    Soil (1)Soil (2)

                                    73deg

                                    Figure Q91

                                    Stability of slopes 75

                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                    9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                    599 256 328 1372

                                    Figure Q93

                                    76 Stability of slopes

                                    XW cos frac14 b

                                    Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                    W sin frac14 bX

                                    h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                    Arc length La frac14

                                    180 57

                                    1

                                    2 326 frac14 327m

                                    The factor of safety is given by

                                    F frac14 c0La thorn tan0ethW cos ulTHORN

                                    W sin

                                    frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                    frac14 091

                                    According to the limit state method

                                    0d frac14 tan1tan 32

                                    125

                                    frac14 265

                                    c0 frac14 8

                                    160frac14 5 kN=m2

                                    Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                    Design disturbing moment frac14 1075 kN=m

                                    The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                    94

                                    F frac14 1

                                    W sin

                                    Xfc0bthorn ethW ubTHORN tan0g sec

                                    1thorn ethtan tan0=FTHORN

                                    c0 frac14 8 kN=m2

                                    0 frac14 32

                                    c0b frac14 8 2 frac14 16 kN=m

                                    W frac14 bh frac14 21 2 h frac14 42h kN=m

                                    Try F frac14 100

                                    tan0

                                    Ffrac14 0625

                                    Stability of slopes 77

                                    Values of u are as obtained in Figure Q93

                                    SliceNo

                                    h(m)

                                    W frac14 bh(kNm)

                                    W sin(kNm)

                                    ub(kNm)

                                    c0bthorn (W ub) tan0(kNm)

                                    sec

                                    1thorn (tan tan0)FProduct(kNm)

                                    1 05 21 6 2 8 24 1078 262 13 55 31

                                    23 33 30 1042 31

                                    3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                    224 92 72 0931 67

                                    6 50 210 11 40 100 85 0907 777 55 231 14

                                    12 58 112 90 0889 80

                                    8 60 252 1812

                                    80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                    10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                    2154 88 116 0853 99

                                    14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                    1074 1091

                                    F frac14 1091

                                    1074frac14 102 (assumed value 100)

                                    Thus

                                    F frac14 101

                                    95

                                    F frac14 1

                                    W sin

                                    XfWeth1 ruTHORN tan0g sec

                                    1thorn ethtan tan0THORN=F

                                    0 frac14 33

                                    ru frac14 020

                                    W frac14 bh frac14 20 5 h frac14 100h kN=m

                                    eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                    Try F frac14 110

                                    tan 0

                                    Ffrac14 tan 33

                                    110frac14 0590

                                    78 Stability of slopes

                                    Referring to Figure Q95

                                    SliceNo

                                    h(m)

                                    W frac14 bh(kNm)

                                    W sin(kNm)

                                    W(1 ru) tan0(kNm)

                                    sec

                                    1thorn ( tan tan0)FProduct(kNm)

                                    1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                    2120 234 0892 209

                                    4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                    1185 1271

                                    Figure Q95

                                    Stability of slopes 79

                                    F frac14 1271

                                    1185frac14 107

                                    The trial value was 110 therefore take F to be 108

                                    96

                                    (a) Water table at surface the factor of safety is given by Equation 912

                                    F frac14 0

                                    sat

                                    tan0

                                    tan

                                    ptie 15 frac14 92

                                    19

                                    tan 36

                                    tan

                                    tan frac14 0234

                                    frac14 13

                                    Water table well below surface the factor of safety is given by Equation 911

                                    F frac14 tan0

                                    tan

                                    frac14 tan 36

                                    tan 13

                                    frac14 31

                                    (b) 0d frac14 tan1tan 36

                                    125

                                    frac14 30

                                    Depth of potential failure surface frac14 z

                                    Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                    frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                    frac14 504z kN

                                    Design disturbing moment per unit area Sd frac14 sat sin cos

                                    frac14 19 z sin 13 cos 13

                                    frac14 416z kN

                                    Rd gtSd therefore the limit state for overall stability is satisfied

                                    80 Stability of slopes

                                    • Book Cover
                                    • Title
                                    • Contents
                                    • Basic characteristics of soils
                                    • Seepage
                                    • Effective stress
                                    • Shear strength
                                    • Stresses and displacements
                                    • Lateral earth pressure
                                    • Consolidation theory
                                    • Bearing capacity
                                    • Stability of slopes

                                      k0 frac14ffiffiffiffiffiffiffiffiffiffiffiffiffiethkxkzTHORN

                                      qfrac14

                                      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth90 36THORN

                                      p 106 frac14 57 106 m=s

                                      Then the quantity of seepage is given by

                                      q frac14 k0h Nf

                                      Ndfrac14 57 106 350 56

                                      11

                                      frac14 10 105 m3=s per m

                                      Figure Q29

                                      Seepage 13

                                      Chapter 3

                                      Effective stress

                                      31

                                      Buoyant unit weight

                                      0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                                      Effective vertical stress

                                      0v frac14 5 102 frac14 51 kN=m2 or

                                      Total vertical stress

                                      v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

                                      Pore water pressure

                                      u frac14 7 98 frac14 686 kN=m2

                                      Effective vertical stress

                                      0v frac14 v u frac14 1196 686 frac14 51 kN=m2

                                      32

                                      Buoyant unit weight

                                      0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                                      Effective vertical stress

                                      0v frac14 5 102 frac14 51 kN=m2 or

                                      Total vertical stress

                                      v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

                                      Pore water pressure

                                      u frac14 205 98 frac14 2009 kN=m2

                                      Effective vertical stress

                                      0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

                                      33

                                      At top of the clay

                                      v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

                                      u frac14 2 98 frac14 196 kN=m2

                                      0v frac14 v u frac14 710 196 frac14 514 kN=m2

                                      Alternatively

                                      0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                                      0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

                                      At bottom of the clay

                                      v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

                                      u frac14 12 98 frac14 1176 kN=m2

                                      0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

                                      NB The alternative method of calculation is not applicable because of the artesiancondition

                                      Figure Q3132

                                      Effective stress 15

                                      34

                                      0 frac14 20 98 frac14 102 kN=m3

                                      At 8m depth

                                      0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                                      35

                                      0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                                      0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                                      Figure Q33

                                      Figure Q34

                                      16 Effective stress

                                      (a) Immediately after WT rise

                                      At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                                      0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                                      At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                                      0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                                      (b) Several years after WT rise

                                      At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                                      At 8m depth

                                      0v frac14 940 kN=m2 (as above)

                                      At 12m depth

                                      0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                                      Figure Q35

                                      Effective stress 17

                                      36

                                      Total weight

                                      ab frac14 210 kN

                                      Effective weight

                                      ac frac14 112 kN

                                      Resultant boundary water force

                                      be frac14 119 kN

                                      Seepage force

                                      ce frac14 34 kN

                                      Resultant body force

                                      ae frac14 99 kN eth73 to horizontalTHORN

                                      (Refer to Figure Q36)

                                      Figure Q36

                                      18 Effective stress

                                      37

                                      Situation (1)(a)

                                      frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                                      u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                      0 frac14 u frac14 694 392 frac14 302 kN=m2

                                      (b)

                                      i frac14 2

                                      4frac14 05

                                      j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                                      Situation (2)(a)

                                      frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                                      u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                      0 frac14 u frac14 498 392 frac14 106 kN=m2

                                      (b)

                                      i frac14 2

                                      4frac14 05

                                      j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                                      38

                                      The flow net is drawn in Figure Q24

                                      Loss in total head between adjacent equipotentials

                                      h frac14 550

                                      Ndfrac14 550

                                      11frac14 050m

                                      Exit hydraulic gradient

                                      ie frac14 h

                                      sfrac14 050

                                      070frac14 071

                                      Effective stress 19

                                      The critical hydraulic gradient is given by Equation 39

                                      ic frac14 0

                                      wfrac14 102

                                      98frac14 104

                                      Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                                      F frac14 iciefrac14 104

                                      071frac14 15

                                      Total head at C

                                      hC frac14 nd

                                      Ndh frac14 24

                                      11 550 frac14 120m

                                      Elevation head at C

                                      zC frac14 250m

                                      Pore water pressure at C

                                      uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                                      Therefore effective vertical stress at C

                                      0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                                      For point D

                                      hD frac14 73

                                      11 550 frac14 365m

                                      zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                                      0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                                      39

                                      The flow net is drawn in Figure Q25

                                      For a soil prism 150 300m adjacent to the piling

                                      hm frac14 26

                                      9 500 frac14 145m

                                      20 Effective stress

                                      Factor of safety against lsquoheavingrsquo (Equation 310)

                                      F frac14 ic

                                      imfrac14 0d

                                      whmfrac14 97 300

                                      98 145frac14 20

                                      With a filter

                                      F frac14 0d thorn wwhm

                                      3 frac14 eth97 300THORN thorn w98 145

                                      w frac14 135 kN=m2

                                      Depth of filterfrac14 13521frac14 065m (if above water level)

                                      Effective stress 21

                                      Chapter 4

                                      Shear strength

                                      41

                                      frac14 295 kN=m2

                                      u frac14 120 kN=m2

                                      0 frac14 u frac14 295 120 frac14 175 kN=m2

                                      f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                                      42

                                      03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                                      100 452 552200 908 1108400 1810 2210800 3624 4424

                                      The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                                      Figure Q42

                                      43

                                      3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                                      200 222 422400 218 618600 220 820

                                      The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                                      44

                                      The modified shear strength parameters are

                                      0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                                      a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                                      The coordinates of the stress point representing failure conditions in the test are

                                      1

                                      2eth1 2THORN frac14 1

                                      2 170 frac14 85 kN=m2

                                      1

                                      2eth1 thorn 3THORN frac14 1

                                      2eth270thorn 100THORN frac14 185 kN=m2

                                      The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                                      uf frac14 36 kN=m2

                                      Figure Q43

                                      Figure Q44

                                      Shear strength 23

                                      45

                                      3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                      150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                      The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                      The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                      46

                                      03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                      200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                      The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                      Figure Q45

                                      24 Shear strength

                                      47

                                      The torque required to produce shear failure is given by

                                      T frac14 dh cud

                                      2thorn 2

                                      Z d=2

                                      0

                                      2r drcur

                                      frac14 cud2h

                                      2thorn 4cu

                                      Z d=2

                                      0

                                      r2dr

                                      frac14 cud2h

                                      2thorn d

                                      3

                                      6

                                      Then

                                      35 frac14 cu52 10

                                      2thorn 53

                                      6

                                      103

                                      cu frac14 76 kN=m3

                                      400

                                      0 400 800 1200 1600

                                      τ (k

                                      Nm

                                      2 )

                                      σprime (kNm2)

                                      34deg

                                      315deg29deg

                                      (a)

                                      (b)

                                      0 400

                                      400

                                      800 1200 1600

                                      Failure envelope

                                      300 500

                                      σprime (kNm2)

                                      τ (k

                                      Nm

                                      2 )

                                      20 (kNm2)

                                      31deg

                                      Figure Q46

                                      Shear strength 25

                                      48

                                      The relevant stress values are calculated as follows

                                      3 frac14 600 kN=m2

                                      1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                      2(1 3) 0 40 79 107 139 159

                                      1

                                      2(01 thorn 03) 400 411 402 389 351 326

                                      1

                                      2(1 thorn 3) 600 640 679 707 739 759

                                      The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                      Af frac14 433 200

                                      319frac14 073

                                      49

                                      B frac14 u33

                                      frac14 144

                                      350 200frac14 096

                                      a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                      0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                      10 283 65 023

                                      Figure Q48

                                      26 Shear strength

                                      The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                      Figure Q49

                                      Shear strength 27

                                      Chapter 5

                                      Stresses and displacements

                                      51

                                      Vertical stress is given by

                                      z frac14 Qz2Ip frac14 5000

                                      52Ip

                                      Values of Ip are obtained from Table 51

                                      r (m) rz Ip z (kNm2)

                                      0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                      10 20 0009 2

                                      The variation of z with radial distance (r) is plotted in Figure Q51

                                      Figure Q51

                                      52

                                      Below the centre load (Figure Q52)

                                      r

                                      zfrac14 0 for the 7500-kN load

                                      Ip frac14 0478

                                      r

                                      zfrac14 5

                                      4frac14 125 for the 10 000- and 9000-kN loads

                                      Ip frac14 0045

                                      Then

                                      z frac14X Q

                                      z2Ip

                                      frac14 7500 0478

                                      42thorn 10 000 0045

                                      42thorn 9000 0045

                                      42

                                      frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                      53

                                      The vertical stress under a corner of a rectangular area is given by

                                      z frac14 qIr

                                      where values of Ir are obtained from Figure 510 In this case

                                      z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                      z

                                      Figure Q52

                                      Stresses and displacements 29

                                      z (m) m n Ir z (kNm2)

                                      0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                      10 010 0005 5

                                      z is plotted against z in Figure Q53

                                      54

                                      (a)

                                      m frac14 125

                                      12frac14 104

                                      n frac14 18

                                      12frac14 150

                                      From Figure 510 Irfrac14 0196

                                      z frac14 2 175 0196 frac14 68 kN=m2

                                      Figure Q53

                                      30 Stresses and displacements

                                      (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                      z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                      55

                                      Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                      Px frac14 2Q

                                      1

                                      m2 thorn 1frac14 2 150

                                      125frac14 76 kN=m

                                      Equation 517 is used to obtain the pressure distribution

                                      px frac14 4Q

                                      h

                                      m2n

                                      ethm2 thorn n2THORN2 frac14150

                                      m2n

                                      ethm2 thorn n2THORN2 ethkN=m2THORN

                                      Figure Q54

                                      Stresses and displacements 31

                                      n m2n

                                      (m2 thorn n2)2

                                      px(kNm2)

                                      0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                      The pressure distribution is plotted in Figure Q55

                                      56

                                      H

                                      Bfrac14 10

                                      2frac14 5

                                      L

                                      Bfrac14 4

                                      2frac14 2

                                      D

                                      Bfrac14 1

                                      2frac14 05

                                      Hence from Figure 515

                                      131 frac14 082

                                      130 frac14 094

                                      Figure Q55

                                      32 Stresses and displacements

                                      The immediate settlement is given by Equation 528

                                      si frac14 130131qB

                                      Eu

                                      frac14 094 082 200 2

                                      45frac14 7mm

                                      Stresses and displacements 33

                                      Chapter 6

                                      Lateral earth pressure

                                      61

                                      For 0 frac14 37 the active pressure coefficient is given by

                                      Ka frac14 1 sin 37

                                      1thorn sin 37frac14 025

                                      The total active thrust (Equation 66a with c0 frac14 0) is

                                      Pa frac14 1

                                      2KaH

                                      2 frac14 1

                                      2 025 17 62 frac14 765 kN=m

                                      If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                      K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                      and the thrust on the wall is

                                      P0 frac14 1

                                      2K0H

                                      2 frac14 1

                                      2 040 17 62 frac14 122 kN=m

                                      62

                                      The active pressure coefficients for the three soil types are as follows

                                      Ka1 frac141 sin 35

                                      1thorn sin 35frac14 0271

                                      Ka2 frac141 sin 27

                                      1thorn sin 27frac14 0375

                                      ffiffiffiffiffiffiffiKa2

                                      p frac14 0613

                                      Ka3 frac141 sin 42

                                      1thorn sin 42frac14 0198

                                      Distribution of active pressure (plotted in Figure Q62)

                                      Depth (m) Soil Active pressure (kNm2)

                                      3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                      12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                      At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                      Total thrust frac14 571 kNm

                                      Point of application is (4893571) m from the top of the wall ie 857m

                                      Force (kN) Arm (m) Moment (kN m)

                                      (1)1

                                      2 0271 16 32 frac14 195 20 390

                                      (2) 0271 16 3 2 frac14 260 40 1040

                                      (3)1

                                      2 0271 92 22 frac14 50 433 217

                                      (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                      (5)1

                                      2 0375 102 32 frac14 172 70 1204

                                      (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                      (7)1

                                      2 0198 112 42 frac14 177 1067 1889

                                      (8)1

                                      2 98 92 frac14 3969 90 35721

                                      5713 48934

                                      Figure Q62

                                      Lateral earth pressure 35

                                      63

                                      (a) For u frac14 0 Ka frac14 Kp frac14 1

                                      Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                      frac14 245

                                      At the lower end of the piling

                                      pa frac14 Kaqthorn Kasatz Kaccu

                                      frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                      frac14 115 kN=m2

                                      pp frac14 Kpsatzthorn Kpccu

                                      frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                      frac14 202 kN=m2

                                      (b) For 0 frac14 26 and frac14 1

                                      20

                                      Ka frac14 035

                                      Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                      pfrac14 145 ethEquation 619THORN

                                      Kp frac14 37

                                      Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                      pfrac14 47 ethEquation 624THORN

                                      At the lower end of the piling

                                      pa frac14 Kaqthorn Ka0z Kacc

                                      0

                                      frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                      frac14 187 kN=m2

                                      pp frac14 Kp0zthorn Kpcc

                                      0

                                      frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                      frac14 198 kN=m2

                                      36 Lateral earth pressure

                                      64

                                      (a) For 0 frac14 38 Ka frac14 024

                                      0 frac14 20 98 frac14 102 kN=m3

                                      The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                      Force (kN) Arm (m) Moment (kN m)

                                      (1) 024 10 66 frac14 159 33 525

                                      (2)1

                                      2 024 17 392 frac14 310 400 1240

                                      (3) 024 17 39 27 frac14 430 135 580

                                      (4)1

                                      2 024 102 272 frac14 89 090 80

                                      (5)1

                                      2 98 272 frac14 357 090 321

                                      Hfrac14 1345 MH frac14 2746

                                      (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                      (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                      XM frac14MV MH frac14 7790 kNm

                                      Lever arm of base resultant

                                      M

                                      Vfrac14 779

                                      488frac14 160

                                      Eccentricity of base resultant

                                      e frac14 200 160 frac14 040m

                                      39 m

                                      27 m

                                      40 m

                                      04 m

                                      04 m

                                      26 m

                                      (7)

                                      (9)

                                      (1)(2)

                                      (3)

                                      (4)

                                      (5)

                                      (8)(6)

                                      (10)

                                      WT

                                      10 kNm2

                                      Hydrostatic

                                      Figure Q64

                                      Lateral earth pressure 37

                                      Base pressures (Equation 627)

                                      p frac14 VB

                                      1 6e

                                      B

                                      frac14 488

                                      4eth1 060THORN

                                      frac14 195 kN=m2 and 49 kN=m2

                                      Factor of safety against sliding (Equation 628)

                                      F frac14 V tan

                                      Hfrac14 488 tan 25

                                      1345frac14 17

                                      (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                      Hfrac14 1633 kN

                                      V frac14 4879 kN

                                      MH frac14 3453 kNm

                                      MV frac14 10536 kNm

                                      The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                      65

                                      For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                      Kp

                                      Ffrac14 385

                                      2

                                      0 frac14 20 98 frac14 102 kN=m3

                                      The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                      Force (kN) Arm (m) Moment (kN m)

                                      (1)1

                                      2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                      (2) 026 17 45 d frac14 199d d2 995d2

                                      (3)1

                                      2 026 102 d2 frac14 133d2 d3 044d3

                                      (4)1

                                      2 385

                                      2 17 152 frac14 368 dthorn 05 368d 184

                                      (5)385

                                      2 17 15 d frac14 491d d2 2455d2

                                      (6)1

                                      2 385

                                      2 102 d2 frac14 982d2 d3 327d3

                                      38 Lateral earth pressure

                                      XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                      d3 thorn 516d2 283d 1724 frac14 0

                                      d frac14 179m

                                      Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                      Over additional 20 embedded depth

                                      pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                      Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                      66

                                      The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                      Ka frac14 sin 69=sin 105

                                      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                      pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                      26664

                                      37775

                                      2

                                      frac14 050

                                      The total active thrust (acting at 25 above the normal) is given by Equation 616

                                      Pa frac14 1

                                      2 050 19 7502 frac14 267 kN=m

                                      Figure Q65

                                      Lateral earth pressure 39

                                      Horizontal component

                                      Ph frac14 267 cos 40 frac14 205 kN=m

                                      Vertical component

                                      Pv frac14 267 sin 40 frac14 172 kN=m

                                      Consider moments about the toe of the wall (Figure Q66) (per m)

                                      Force (kN) Arm (m) Moment (kN m)

                                      (1)1

                                      2 175 650 235 frac14 1337 258 345

                                      (2) 050 650 235 frac14 764 175 134

                                      (3)1

                                      2 070 650 235 frac14 535 127 68

                                      (4) 100 400 235 frac14 940 200 188

                                      (5) 1

                                      2 080 050 235 frac14 47 027 1

                                      Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                      Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                      Lever arm of base resultant

                                      M

                                      Vfrac14 795

                                      525frac14 151m

                                      Eccentricity of base resultant

                                      e frac14 200 151 frac14 049m

                                      Figure Q66

                                      40 Lateral earth pressure

                                      Base pressures (Equation 627)

                                      p frac14 525

                                      41 6 049

                                      4

                                      frac14 228 kN=m2 and 35 kN=m2

                                      The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                      The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                      The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                      67

                                      For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                      Force (kN) Arm (m) Moment (kNm)

                                      (1)1

                                      2 027 17 52 frac14 574 183 1050

                                      (2) 027 17 5 3 frac14 689 500 3445

                                      (3)1

                                      2 027 102 32 frac14 124 550 682

                                      (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                      (5)1

                                      2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                      (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                      (7) 1

                                      2 267

                                      2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                      (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                      2 d frac14 163d d2thorn 650 82d2 1060d

                                      Tie rod force per m frac14 T 0 0

                                      XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                      d3 thorn 77d2 269d 1438 frac14 0

                                      d frac14 467m

                                      Depth of penetration frac14 12d frac14 560m

                                      Lateral earth pressure 41

                                      Algebraic sum of forces for d frac14 467m isX

                                      F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                      T frac14 905 kN=m

                                      Force in each tie rod frac14 25T frac14 226 kN

                                      68

                                      (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                      0 frac14 21 98 frac14 112 kN=m3

                                      The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                      uC frac14 150

                                      165 15 98 frac14 134 kN=m2

                                      The average seepage pressure is

                                      j frac14 15

                                      165 98 frac14 09 kN=m3

                                      Hence

                                      0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                      0 j frac14 112 09 frac14 103 kN=m3

                                      Figure Q67

                                      42 Lateral earth pressure

                                      Consider moments about the anchor point A (per m)

                                      Force (kN) Arm (m) Moment (kN m)

                                      (1) 10 026 150 frac14 390 60 2340

                                      (2)1

                                      2 026 18 452 frac14 474 15 711

                                      (3) 026 18 45 105 frac14 2211 825 18240

                                      (4)1

                                      2 026 121 1052 frac14 1734 100 17340

                                      (5)1

                                      2 134 15 frac14 101 40 404

                                      (6) 134 30 frac14 402 60 2412

                                      (7)1

                                      2 134 60 frac14 402 95 3819

                                      571 4527(8) Ppm

                                      115 115PPm

                                      XM frac14 0

                                      Ppm frac144527

                                      115frac14 394 kN=m

                                      Available passive resistance

                                      Pp frac14 1

                                      2 385 103 62 frac14 714 kN=m

                                      Factor of safety

                                      Fp frac14 Pp

                                      Ppm

                                      frac14 714

                                      394frac14 18

                                      Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                      Figure Q68

                                      Lateral earth pressure 43

                                      (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                      Consider moments (per m) about the tie point A

                                      Force (kN) Arm (m)

                                      (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                      (2)1

                                      2 033 18 452 frac14 601 15

                                      (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                      (4)1

                                      2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                      (5)1

                                      2 134 15 frac14 101 40

                                      (6) 134 30 frac14 402 60

                                      (7)1

                                      2 134 d frac14 67d d3thorn 75

                                      (8) 1

                                      2 30 103 d2 frac141545d2 2d3thorn 75

                                      Moment (kN m)

                                      (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                      XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                      d3 thorn 827d2 466d 1518 frac14 0

                                      By trial

                                      d frac14 544m

                                      The minimum depth of embedment required is 544m

                                      69

                                      For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                      0 frac14 20 98 frac14 102 kN=m3

                                      The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                      44 Lateral earth pressure

                                      uC frac14 147

                                      173 26 98 frac14 216 kN=m2

                                      and the average seepage pressure around the wall is

                                      j frac14 26

                                      173 98 frac14 15 kN=m3

                                      Consider moments about the prop (A) (per m)

                                      Force (kN) Arm (m) Moment (kN m)

                                      (1)1

                                      2 03 17 272 frac14 186 020 37

                                      (2) 03 17 27 53 frac14 730 335 2445

                                      (3)1

                                      2 03 (102thorn 15) 532 frac14 493 423 2085

                                      (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                      (5)1

                                      2 216 26 frac14 281 243 684

                                      (6) 216 27 frac14 583 465 2712

                                      (7)1

                                      2 216 60 frac14 648 800 5184

                                      3055(8)

                                      1

                                      2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                      Factor of safety

                                      Fr frac14 6885

                                      3055frac14 225

                                      Figure Q69

                                      Lateral earth pressure 45

                                      610

                                      For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                      p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                      Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                      Using the recommendations of Twine and Roscoe

                                      p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                      Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                      611

                                      frac14 18 kN=m3 0 frac14 34

                                      H frac14 350m nH frac14 335m mH frac14 185m

                                      Consider a trial value of F frac14 20 Refer to Figure 635

                                      0m frac14 tan1tan 34

                                      20

                                      frac14 186

                                      Then

                                      frac14 45 thorn 0m2frac14 543

                                      W frac14 1

                                      2 18 3502 cot 543 frac14 792 kN=m

                                      Figure Q610

                                      46 Lateral earth pressure

                                      P frac14 1

                                      2 s 3352 frac14 561s kN=m

                                      U frac14 1

                                      2 98 1852 cosec 543 frac14 206 kN=m

                                      Equations 630 and 631 then become

                                      561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                      792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                      ie

                                      561s 0616N 405 frac14 0

                                      792 0857N thorn 563 frac14 0

                                      N frac14 848

                                      0857frac14 989 kN=m

                                      Then

                                      561s 609 405 frac14 0

                                      s frac14 649

                                      561frac14 116 kN=m3

                                      The calculations for trial values of F of 20 15 and 10 are summarized below

                                      F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                      20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                      s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                      Figure Q611

                                      Lateral earth pressure 47

                                      612

                                      For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                      45 thorn 0

                                      2frac14 63

                                      For the retained material between the surface and a depth of 36m

                                      Pa frac14 1

                                      2 030 18 362 frac14 350 kN=m

                                      Weight of reinforced fill between the surface and a depth of 36m is

                                      Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                      eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                      Eccentricity of Rv

                                      e frac14 263 250 frac14 013m

                                      The average vertical stress at a depth of 36m is

                                      z frac14 Rv

                                      L 2efrac14 324

                                      474frac14 68 kN=m2

                                      (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                      Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                      Tensile stress in the element frac14 138 103

                                      65 3frac14 71N=mm2

                                      Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                      Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                      Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                      The weight of ABC is

                                      W frac14 1

                                      2 18 52 265 frac14 124 kN=m

                                      From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                      48 Lateral earth pressure

                                      (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                      Tp frac14 032 68 120 065 frac14 170 kN

                                      Tr frac14 213 420

                                      418frac14 214 kN

                                      Again the tensile failure and slipping limit states are satisfied for this element

                                      Figure Q612

                                      Lateral earth pressure 49

                                      Chapter 7

                                      Consolidation theory

                                      71

                                      Total change in thickness

                                      H frac14 782 602 frac14 180mm

                                      Average thickness frac14 1530thorn 180

                                      2frac14 1620mm

                                      Length of drainage path d frac14 1620

                                      2frac14 810mm

                                      Root time plot (Figure Q71a)

                                      ffiffiffiffiffiffit90p frac14 33

                                      t90 frac14 109min

                                      cv frac14 0848d2

                                      t90frac14 0848 8102

                                      109 1440 365

                                      106frac14 27m2=year

                                      r0 frac14 782 764

                                      782 602frac14 018

                                      180frac14 0100

                                      rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                      10 119

                                      9 180frac14 0735

                                      rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                      Log time plot (Figure Q71b)

                                      t50 frac14 26min

                                      cv frac14 0196d2

                                      t50frac14 0196 8102

                                      26 1440 365

                                      106frac14 26m2=year

                                      r0 frac14 782 763

                                      782 602frac14 019

                                      180frac14 0106

                                      rp frac14 763 623

                                      782 602frac14 140

                                      180frac14 0778

                                      rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                      Figure Q71(a)

                                      Figure Q71(b)

                                      Final void ratio

                                      e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                      e

                                      Hfrac14 1thorn e0

                                      H0frac14 1thorn e1 thorne

                                      H0

                                      ie

                                      e

                                      180frac14 1631thorne

                                      1710

                                      e frac14 2936

                                      1530frac14 0192

                                      Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                      mv frac14 1

                                      1thorn e0 e0 e101 00

                                      frac14 1

                                      1823 0192

                                      0107frac14 098m2=MN

                                      k frac14 cvmvw frac14 265 098 98

                                      60 1440 365 103frac14 81 1010 m=s

                                      72

                                      Using Equation 77 (one-dimensional method)

                                      sc frac14 e0 e11thorn e0 H

                                      Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                      Figure Q72

                                      52 Consolidation theory

                                      Settlement

                                      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                      1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                      318

                                      Notes 5 92y 460thorn 84

                                      Heave

                                      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                      1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                      38

                                      73

                                      U frac14 f ethTvTHORN frac14 f cvt

                                      d2

                                      Hence if cv is constant

                                      t1

                                      t2frac14 d

                                      21

                                      d22

                                      where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                      d1 frac14 95mm and d2 frac14 2500mm

                                      for U frac14 050 t2 frac14 t1 d22

                                      d21

                                      frac14 20

                                      60 24 365 25002

                                      952frac14 263 years

                                      for U lt 060 Tv frac14

                                      4U2 (Equation 724(a))

                                      t030 frac14 t050 0302

                                      0502

                                      frac14 263 036 frac14 095 years

                                      Consolidation theory 53

                                      74

                                      The layer is open

                                      d frac14 8

                                      2frac14 4m

                                      Tv frac14 cvtd2frac14 24 3

                                      42frac14 0450

                                      ui frac14 frac14 84 kN=m2

                                      The excess pore water pressure is given by Equation 721

                                      ue frac14Xmfrac141mfrac140

                                      2ui

                                      Msin

                                      Mz

                                      d

                                      expethM2TvTHORN

                                      In this case z frac14 d

                                      sinMz

                                      d

                                      frac14 sinM

                                      where

                                      M frac14

                                      23

                                      25

                                      2

                                      M sin M M2Tv exp (M2Tv)

                                      2thorn1 1110 0329

                                      3

                                      21 9993 457 105

                                      ue frac14 2 84 2

                                      1 0329 ethother terms negligibleTHORN

                                      frac14 352 kN=m2

                                      75

                                      The layer is open

                                      d frac14 6

                                      2frac14 3m

                                      Tv frac14 cvtd2frac14 10 3

                                      32frac14 0333

                                      The layer thickness will be divided into six equal parts ie m frac14 6

                                      54 Consolidation theory

                                      For an open layer

                                      Tv frac14 4n

                                      m2

                                      n frac14 0333 62

                                      4frac14 300

                                      The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                      ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                      i j

                                      0 1 2 3 4 5 6 7 8 9 10 11 12

                                      0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                      The initial and 3-year isochrones are plotted in Figure Q75

                                      Area under initial isochrone frac14 180 units

                                      Area under 3-year isochrone frac14 63 units

                                      The average degree of consolidation is given by Equation 725Thus

                                      U frac14 1 63

                                      180frac14 065

                                      Figure Q75

                                      Consolidation theory 55

                                      76

                                      At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                      0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                      The final consolidation settlement (one-dimensional method) is

                                      sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                      Corrected time t frac14 2 1

                                      2

                                      40

                                      52

                                      frac14 1615 years

                                      Tv frac14 cvtd2frac14 44 1615

                                      42frac14 0444

                                      From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                      77

                                      The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                      Figure Q77

                                      56 Consolidation theory

                                      Point m n Ir (kNm2) sc (mm)

                                      13020frac14 15 20

                                      20frac14 10 0194 (4) 113 124

                                      260

                                      20frac14 30

                                      20

                                      20frac14 10 0204 (2) 59 65

                                      360

                                      20frac14 30

                                      40

                                      20frac14 20 0238 (1) 35 38

                                      430

                                      20frac14 15

                                      40

                                      20frac14 20 0224 (2) 65 72

                                      Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                      78

                                      Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                      (a) Immediate settlement

                                      H

                                      Bfrac14 30

                                      35frac14 086

                                      D

                                      Bfrac14 2

                                      35frac14 006

                                      Figure Q78

                                      Consolidation theory 57

                                      From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                      si frac14 130131qB

                                      Eufrac14 10 032 105 35

                                      40frac14 30mm

                                      (b) Consolidation settlement

                                      Layer z (m) Dz Ic (kNm2) syod (mm)

                                      1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                      3150

                                      Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                      Now

                                      H

                                      Bfrac14 30

                                      35frac14 086 and A frac14 065

                                      from Figure 712 13 frac14 079

                                      sc frac14 13sod frac14 079 315 frac14 250mm

                                      Total settlement

                                      s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                      79

                                      Without sand drains

                                      Uv frac14 025

                                      Tv frac14 0049 ethfrom Figure 718THORN

                                      t frac14 Tvd2

                                      cvfrac14 0049 82

                                      cvWith sand drains

                                      R frac14 0564S frac14 0564 3 frac14 169m

                                      n frac14 Rrfrac14 169

                                      015frac14 113

                                      Tr frac14 cht

                                      4R2frac14 ch

                                      4 1692 0049 82

                                      cvethand ch frac14 cvTHORN

                                      frac14 0275

                                      Ur frac14 073 (from Figure 730)

                                      58 Consolidation theory

                                      Using Equation 740

                                      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                      U frac14 080

                                      710

                                      Without sand drains

                                      Uv frac14 090

                                      Tv frac14 0848

                                      t frac14 Tvd2

                                      cvfrac14 0848 102

                                      96frac14 88 years

                                      With sand drains

                                      R frac14 0564S frac14 0564 4 frac14 226m

                                      n frac14 Rrfrac14 226

                                      015frac14 15

                                      Tr

                                      Tvfrac14 chcv

                                      d2

                                      4R2ethsame tTHORN

                                      Tr

                                      Tvfrac14 140

                                      96 102

                                      4 2262frac14 714 eth1THORN

                                      Using Equation 740

                                      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                      An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                      Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                      040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                      Thus

                                      Uv frac14 0295 and Ur frac14 086

                                      t frac14 88 00683

                                      0848frac14 07 years

                                      Consolidation theory 59

                                      Chapter 8

                                      Bearing capacity

                                      81

                                      (a) The ultimate bearing capacity is given by Equation 83

                                      qf frac14 cNc thorn DNq thorn 1

                                      2BN

                                      For u frac14 0

                                      Nc frac14 514 Nq frac14 1 N frac14 0

                                      qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                      The net ultimate bearing capacity is

                                      qnf frac14 qf D frac14 540 kN=m2

                                      The net foundation pressure is

                                      qn frac14 q D frac14 425

                                      2 eth21 1THORN frac14 192 kN=m2

                                      The factor of safety (Equation 86) is

                                      F frac14 qnfqnfrac14 540

                                      192frac14 28

                                      (b) For 0 frac14 28

                                      Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                      qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                      2 112 2 13

                                      frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                      qnf frac14 574 112 frac14 563 kN=m2

                                      F frac14 563

                                      192frac14 29

                                      (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                      82

                                      For 0 frac14 38

                                      Nq frac14 49 N frac14 67

                                      qnf frac14 DethNq 1THORN thorn 1

                                      2BN ethfrom Equation 83THORN

                                      frac14 eth18 075 48THORN thorn 1

                                      2 18 15 67

                                      frac14 648thorn 905 frac14 1553 kN=m2

                                      qn frac14 500

                                      15 eth18 075THORN frac14 320 kN=m2

                                      F frac14 qnfqnfrac14 1553

                                      320frac14 48

                                      0d frac14 tan1tan 38

                                      125

                                      frac14 32 therefore Nq frac14 23 and N frac14 25

                                      Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                      2 18 15 25

                                      frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                      Design load (action) Vd frac14 500 kN=m

                                      The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                      83

                                      D

                                      Bfrac14 350

                                      225frac14 155

                                      From Figure 85 for a square foundation

                                      Nc frac14 81

                                      Bearing capacity 61

                                      For a rectangular foundation (L frac14 450m B frac14 225m)

                                      Nc frac14 084thorn 016B

                                      L

                                      81 frac14 745

                                      Using Equation 810

                                      qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                      For F frac14 3

                                      qn frac14 1006

                                      3frac14 335 kN=m2

                                      q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                      Design load frac14 405 450 225 frac14 4100 kN

                                      Design undrained strength cud frac14 135

                                      14frac14 96 kN=m2

                                      Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                      frac14 7241 kN

                                      Design load Vd frac14 4100 kN

                                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                                      84

                                      For 0 frac14 40

                                      Nq frac14 64 N frac14 95

                                      qnf frac14 DethNq 1THORN thorn 04BN

                                      (a) Water table 5m below ground level

                                      qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                      qn frac14 400 17 frac14 383 kN=m2

                                      F frac14 2686

                                      383frac14 70

                                      (b) Water table 1m below ground level (ie at foundation level)

                                      0 frac14 20 98 frac14 102 kN=m3

                                      62 Bearing capacity

                                      qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                      F frac14 2040

                                      383frac14 53

                                      (c) Water table at ground level with upward hydraulic gradient 02

                                      eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                      qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                      F frac14 1296

                                      392frac14 33

                                      85

                                      The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                      Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                      Design value of 0 frac14 tan1tan 39

                                      125

                                      frac14 33

                                      For 0 frac14 33 Nq frac14 26 and N frac14 29

                                      Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                                      86

                                      (a) Undrained shear for u frac14 0

                                      Nc frac14 514 Nq frac14 1 N frac14 0

                                      qnf frac14 12cuNc

                                      frac14 12 100 514 frac14 617 kN=m2

                                      qn frac14 qnfFfrac14 617

                                      3frac14 206 kN=m2

                                      q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                      Bearing capacity 63

                                      Drained shear for 0 frac14 32

                                      Nq frac14 23 N frac14 25

                                      0 frac14 21 98 frac14 112 kN=m3

                                      qnf frac14 0DethNq 1THORN thorn 040BN

                                      frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                      frac14 694 kN=m2

                                      q frac14 694

                                      3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                      Design load frac14 42 227 frac14 3632 kN

                                      (b) Design undrained strength cud frac14 100

                                      14frac14 71 kNm2

                                      Design bearing resistance Rd frac14 12cudNe area

                                      frac14 12 71 514 42

                                      frac14 7007 kN

                                      For drained shear 0d frac14 tan1tan 32

                                      125

                                      frac14 26

                                      Nq frac14 12 N frac14 10

                                      Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                      (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                      Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                      1 2 100 0175 0700qn 0182qn

                                      2 6 033 0044 0176qn 0046qn

                                      3 10 020 0017 0068qn 0018qn

                                      0246qn

                                      Diameter of equivalent circle B frac14 45m

                                      H

                                      Bfrac14 12

                                      45frac14 27 and A frac14 042

                                      13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                      64 Bearing capacity

                                      For sc frac14 30mm

                                      qn frac14 30

                                      0147frac14 204 kN=m2

                                      q frac14 204thorn 21 frac14 225 kN=m2

                                      Design load frac14 42 225 frac14 3600 kN

                                      The design load is 3600 kN settlement being the limiting criterion

                                      87

                                      D

                                      Bfrac14 8

                                      4frac14 20

                                      From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                      F frac14 cuNc

                                      Dfrac14 40 71

                                      20 8frac14 18

                                      88

                                      Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                      Design value of 0 frac14 tan1tan 38

                                      125

                                      frac14 32

                                      Figure Q86

                                      Bearing capacity 65

                                      For 0 frac14 32 Nq frac14 23 and N frac14 25

                                      Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                      The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                      Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                      For B frac14 250m qn frac14 3750

                                      2502 17 frac14 583 kN=m2

                                      From Figure 510 m frac14 n frac14 126

                                      6frac14 021

                                      Ir frac14 0019

                                      Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                      Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                      The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                      89

                                      Depth (m) N 0v (kNm2) CN N1

                                      070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                      Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                      (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                      Cw frac14 05thorn 0530

                                      47

                                      frac14 082

                                      66 Bearing capacity

                                      Thus

                                      qa frac14 150 082 frac14 120 kN=m2

                                      (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                      Thus

                                      qa frac14 90 15 frac14 135 kN=m2

                                      (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                      Ic frac14 171

                                      1014frac14 0068

                                      From Equation 819(a) with s frac14 25mm

                                      q frac14 25

                                      3507 0068frac14 150 kN=m2

                                      810

                                      Peak value of strain influence factor occurs at a depth of 27m and is given by

                                      Izp frac14 05thorn 01130

                                      16 27

                                      05

                                      frac14 067

                                      Refer to Figure Q810

                                      E frac14 25qc

                                      Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                      Ez (mm3MN)

                                      1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                      0203

                                      C1 frac14 1 0500qnfrac14 1 05 12 16

                                      130frac14 093

                                      C2 frac14 1 ethsayTHORN

                                      s frac14 C1C2qnX Iz

                                      Ez frac14 093 1 130 0203 frac14 25mm

                                      Bearing capacity 67

                                      811

                                      At pile base level

                                      cu frac14 220 kN=m2

                                      qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                      Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                      00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                      qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                      Then

                                      Qf frac14 Abqb thorn Asqs

                                      frac14

                                      4 32 1980

                                      thorn eth 105 139 86THORN

                                      frac14 13 996thorn 3941 frac14 17 937 kN

                                      0 01 02 03 04 05 06 07

                                      0 2 4 6 8 10 12 14

                                      1

                                      2

                                      3

                                      4

                                      5

                                      6

                                      7

                                      8

                                      (1)

                                      (2)

                                      (3)

                                      (4)

                                      (5)

                                      qc

                                      qc

                                      Iz

                                      Iz

                                      (MNm2)

                                      z (m)

                                      Figure Q810

                                      68 Bearing capacity

                                      Allowable load

                                      ethaTHORN Qf

                                      2frac14 17 937

                                      2frac14 8968 kN

                                      ethbTHORN Abqb

                                      3thorn Asqs frac14 13 996

                                      3thorn 3941 frac14 8606 kN

                                      ie allowable load frac14 8600 kN

                                      Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                      According to the limit state method

                                      Characteristic undrained strength at base level cuk frac14 220

                                      150kN=m2

                                      Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                      150frac14 1320 kN=m2

                                      Characteristic shaft resistance qsk frac14 00150

                                      frac14 86

                                      150frac14 57 kN=m2

                                      Characteristic base and shaft resistances

                                      Rbk frac14

                                      4 32 1320 frac14 9330 kN

                                      Rsk frac14 105 139 86

                                      150frac14 2629 kN

                                      For a bored pile the partial factors are b frac14 160 and s frac14 130

                                      Design bearing resistance Rcd frac14 9330

                                      160thorn 2629

                                      130

                                      frac14 5831thorn 2022

                                      frac14 7850 kN

                                      Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                      812

                                      ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                      qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                      For a single pile

                                      Qf frac14 Abqb thorn Asqs

                                      frac14

                                      4 062 1305

                                      thorn eth 06 15 42THORN

                                      frac14 369thorn 1187 frac14 1556 kN

                                      Bearing capacity 69

                                      Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                      eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                      (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                      (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                      qbkfrac14 9cuk frac14 9 220

                                      150frac14 1320 kN=m2

                                      qskfrac14cuk frac14 040 105

                                      150frac14 28 kN=m2

                                      Rbkfrac14

                                      4 0602 1320 frac14 373 kN

                                      Rskfrac14 060 15 28 frac14 791 kN

                                      Rcdfrac14 373

                                      160thorn 791

                                      130frac14 233thorn 608 frac14 841 kN

                                      Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                      Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                      (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                      q frac14 21 000

                                      1762frac14 68 kN=m2

                                      Immediate settlement

                                      H

                                      Bfrac14 15

                                      176frac14 085

                                      D

                                      Bfrac14 13

                                      176frac14 074

                                      L

                                      Bfrac14 1

                                      Hence from Figure 515

                                      130 frac14 078 and 131 frac14 041

                                      70 Bearing capacity

                                      Thus using Equation 528

                                      si frac14 078 041 68 176

                                      65frac14 6mm

                                      Consolidation settlement

                                      Layer z (m) Area (m2) (kNm2) mvH (mm)

                                      1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                      434 (sod)

                                      Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                      sc frac14 056 434 frac14 24mm

                                      The total settlement is (6thorn 24) frac14 30mm

                                      813

                                      At base level N frac14 26 Then using Equation 830

                                      qb frac14 40NDb

                                      Bfrac14 40 26 2

                                      025frac14 8320 kN=m2

                                      ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                      Figure Q812

                                      Bearing capacity 71

                                      Over the length embedded in sand

                                      N frac14 21 ie18thorn 24

                                      2

                                      Using Equation 831

                                      qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                      For a single pile

                                      Qf frac14 Abqb thorn Asqs

                                      frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                      For the pile group assuming a group efficiency of 12

                                      XQf frac14 12 9 604 frac14 6523 kN

                                      Then the load factor is

                                      F frac14 6523

                                      2000thorn 1000frac14 21

                                      (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                      Characteristic base resistance per unit area qbk frac14 8320

                                      150frac14 5547 kNm2

                                      Characteristic shaft resistance per unit area qsk frac14 42

                                      150frac14 28 kNm2

                                      Characteristic base and shaft resistances for a single pile

                                      Rbk frac14 0252 5547 frac14 347 kN

                                      Rsk frac14 4 025 2 28 frac14 56 kN

                                      For a driven pile the partial factors are b frac14 s frac14 130

                                      Design bearing resistance Rcd frac14 347

                                      130thorn 56

                                      130frac14 310 kN

                                      For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                      Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                      (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                      From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                      72 Bearing capacity

                                      N frac14 24thorn 26thorn 34

                                      3frac14 28

                                      Ic frac14 171

                                      2814frac14 0016 ethEquation 818THORN

                                      s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                      814

                                      Using Equation 841

                                      Tf frac14 DLcu thorn

                                      4ethD2 d2THORNcuNc

                                      frac14 eth 02 5 06 110THORN thorn

                                      4eth022 012THORN110 9

                                      frac14 207thorn 23 frac14 230 kN

                                      Figure Q813

                                      Bearing capacity 73

                                      Chapter 9

                                      Stability of slopes

                                      91

                                      Referring to Figure Q91

                                      W frac14 417 19 frac14 792 kN=m

                                      Q frac14 20 28 frac14 56 kN=m

                                      Arc lengthAB frac14

                                      180 73 90 frac14 115m

                                      Arc length BC frac14

                                      180 28 90 frac14 44m

                                      The factor of safety is given by

                                      F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                      90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                      Depth of tension crack z0 frac14 2cu

                                      frac14 2 20

                                      19frac14 21m

                                      Arc length BD frac14

                                      180 13

                                      1

                                      2 90 frac14 21m

                                      F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                      The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                      Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                      14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                      Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                      The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                      92

                                      u frac14 0

                                      Depth factor D frac14 11

                                      9frac14 122

                                      Using Equation 92 with F frac14 10

                                      Ns frac14 cu

                                      FHfrac14 30

                                      10 19 9frac14 0175

                                      Hence from Figure 93

                                      frac14 50

                                      For F frac14 12

                                      Ns frac14 30

                                      12 19 9frac14 0146

                                      frac14 27

                                      93

                                      Refer to Figure Q93

                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                      1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                      74 m

                                      214 1deg

                                      213 1deg

                                      39 m

                                      WB

                                      D

                                      C

                                      28 m

                                      21 m

                                      A

                                      Q

                                      Soil (1)Soil (2)

                                      73deg

                                      Figure Q91

                                      Stability of slopes 75

                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                      9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                      599 256 328 1372

                                      Figure Q93

                                      76 Stability of slopes

                                      XW cos frac14 b

                                      Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                      W sin frac14 bX

                                      h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                      Arc length La frac14

                                      180 57

                                      1

                                      2 326 frac14 327m

                                      The factor of safety is given by

                                      F frac14 c0La thorn tan0ethW cos ulTHORN

                                      W sin

                                      frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                      frac14 091

                                      According to the limit state method

                                      0d frac14 tan1tan 32

                                      125

                                      frac14 265

                                      c0 frac14 8

                                      160frac14 5 kN=m2

                                      Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                      Design disturbing moment frac14 1075 kN=m

                                      The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                      94

                                      F frac14 1

                                      W sin

                                      Xfc0bthorn ethW ubTHORN tan0g sec

                                      1thorn ethtan tan0=FTHORN

                                      c0 frac14 8 kN=m2

                                      0 frac14 32

                                      c0b frac14 8 2 frac14 16 kN=m

                                      W frac14 bh frac14 21 2 h frac14 42h kN=m

                                      Try F frac14 100

                                      tan0

                                      Ffrac14 0625

                                      Stability of slopes 77

                                      Values of u are as obtained in Figure Q93

                                      SliceNo

                                      h(m)

                                      W frac14 bh(kNm)

                                      W sin(kNm)

                                      ub(kNm)

                                      c0bthorn (W ub) tan0(kNm)

                                      sec

                                      1thorn (tan tan0)FProduct(kNm)

                                      1 05 21 6 2 8 24 1078 262 13 55 31

                                      23 33 30 1042 31

                                      3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                      224 92 72 0931 67

                                      6 50 210 11 40 100 85 0907 777 55 231 14

                                      12 58 112 90 0889 80

                                      8 60 252 1812

                                      80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                      10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                      2154 88 116 0853 99

                                      14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                      1074 1091

                                      F frac14 1091

                                      1074frac14 102 (assumed value 100)

                                      Thus

                                      F frac14 101

                                      95

                                      F frac14 1

                                      W sin

                                      XfWeth1 ruTHORN tan0g sec

                                      1thorn ethtan tan0THORN=F

                                      0 frac14 33

                                      ru frac14 020

                                      W frac14 bh frac14 20 5 h frac14 100h kN=m

                                      eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                      Try F frac14 110

                                      tan 0

                                      Ffrac14 tan 33

                                      110frac14 0590

                                      78 Stability of slopes

                                      Referring to Figure Q95

                                      SliceNo

                                      h(m)

                                      W frac14 bh(kNm)

                                      W sin(kNm)

                                      W(1 ru) tan0(kNm)

                                      sec

                                      1thorn ( tan tan0)FProduct(kNm)

                                      1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                      2120 234 0892 209

                                      4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                      1185 1271

                                      Figure Q95

                                      Stability of slopes 79

                                      F frac14 1271

                                      1185frac14 107

                                      The trial value was 110 therefore take F to be 108

                                      96

                                      (a) Water table at surface the factor of safety is given by Equation 912

                                      F frac14 0

                                      sat

                                      tan0

                                      tan

                                      ptie 15 frac14 92

                                      19

                                      tan 36

                                      tan

                                      tan frac14 0234

                                      frac14 13

                                      Water table well below surface the factor of safety is given by Equation 911

                                      F frac14 tan0

                                      tan

                                      frac14 tan 36

                                      tan 13

                                      frac14 31

                                      (b) 0d frac14 tan1tan 36

                                      125

                                      frac14 30

                                      Depth of potential failure surface frac14 z

                                      Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                      frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                      frac14 504z kN

                                      Design disturbing moment per unit area Sd frac14 sat sin cos

                                      frac14 19 z sin 13 cos 13

                                      frac14 416z kN

                                      Rd gtSd therefore the limit state for overall stability is satisfied

                                      80 Stability of slopes

                                      • Book Cover
                                      • Title
                                      • Contents
                                      • Basic characteristics of soils
                                      • Seepage
                                      • Effective stress
                                      • Shear strength
                                      • Stresses and displacements
                                      • Lateral earth pressure
                                      • Consolidation theory
                                      • Bearing capacity
                                      • Stability of slopes

                                        Chapter 3

                                        Effective stress

                                        31

                                        Buoyant unit weight

                                        0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                                        Effective vertical stress

                                        0v frac14 5 102 frac14 51 kN=m2 or

                                        Total vertical stress

                                        v frac14 eth2 98THORN thorn eth5 20THORN frac14 1196 kN=m2

                                        Pore water pressure

                                        u frac14 7 98 frac14 686 kN=m2

                                        Effective vertical stress

                                        0v frac14 v u frac14 1196 686 frac14 51 kN=m2

                                        32

                                        Buoyant unit weight

                                        0 frac14 sat w frac14 20 98 frac14 102 kN=m3

                                        Effective vertical stress

                                        0v frac14 5 102 frac14 51 kN=m2 or

                                        Total vertical stress

                                        v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

                                        Pore water pressure

                                        u frac14 205 98 frac14 2009 kN=m2

                                        Effective vertical stress

                                        0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

                                        33

                                        At top of the clay

                                        v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

                                        u frac14 2 98 frac14 196 kN=m2

                                        0v frac14 v u frac14 710 196 frac14 514 kN=m2

                                        Alternatively

                                        0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                                        0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

                                        At bottom of the clay

                                        v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

                                        u frac14 12 98 frac14 1176 kN=m2

                                        0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

                                        NB The alternative method of calculation is not applicable because of the artesiancondition

                                        Figure Q3132

                                        Effective stress 15

                                        34

                                        0 frac14 20 98 frac14 102 kN=m3

                                        At 8m depth

                                        0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                                        35

                                        0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                                        0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                                        Figure Q33

                                        Figure Q34

                                        16 Effective stress

                                        (a) Immediately after WT rise

                                        At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                                        0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                                        At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                                        0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                                        (b) Several years after WT rise

                                        At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                                        At 8m depth

                                        0v frac14 940 kN=m2 (as above)

                                        At 12m depth

                                        0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                                        Figure Q35

                                        Effective stress 17

                                        36

                                        Total weight

                                        ab frac14 210 kN

                                        Effective weight

                                        ac frac14 112 kN

                                        Resultant boundary water force

                                        be frac14 119 kN

                                        Seepage force

                                        ce frac14 34 kN

                                        Resultant body force

                                        ae frac14 99 kN eth73 to horizontalTHORN

                                        (Refer to Figure Q36)

                                        Figure Q36

                                        18 Effective stress

                                        37

                                        Situation (1)(a)

                                        frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                                        u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                        0 frac14 u frac14 694 392 frac14 302 kN=m2

                                        (b)

                                        i frac14 2

                                        4frac14 05

                                        j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                                        Situation (2)(a)

                                        frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                                        u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                        0 frac14 u frac14 498 392 frac14 106 kN=m2

                                        (b)

                                        i frac14 2

                                        4frac14 05

                                        j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                                        38

                                        The flow net is drawn in Figure Q24

                                        Loss in total head between adjacent equipotentials

                                        h frac14 550

                                        Ndfrac14 550

                                        11frac14 050m

                                        Exit hydraulic gradient

                                        ie frac14 h

                                        sfrac14 050

                                        070frac14 071

                                        Effective stress 19

                                        The critical hydraulic gradient is given by Equation 39

                                        ic frac14 0

                                        wfrac14 102

                                        98frac14 104

                                        Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                                        F frac14 iciefrac14 104

                                        071frac14 15

                                        Total head at C

                                        hC frac14 nd

                                        Ndh frac14 24

                                        11 550 frac14 120m

                                        Elevation head at C

                                        zC frac14 250m

                                        Pore water pressure at C

                                        uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                                        Therefore effective vertical stress at C

                                        0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                                        For point D

                                        hD frac14 73

                                        11 550 frac14 365m

                                        zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                                        0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                                        39

                                        The flow net is drawn in Figure Q25

                                        For a soil prism 150 300m adjacent to the piling

                                        hm frac14 26

                                        9 500 frac14 145m

                                        20 Effective stress

                                        Factor of safety against lsquoheavingrsquo (Equation 310)

                                        F frac14 ic

                                        imfrac14 0d

                                        whmfrac14 97 300

                                        98 145frac14 20

                                        With a filter

                                        F frac14 0d thorn wwhm

                                        3 frac14 eth97 300THORN thorn w98 145

                                        w frac14 135 kN=m2

                                        Depth of filterfrac14 13521frac14 065m (if above water level)

                                        Effective stress 21

                                        Chapter 4

                                        Shear strength

                                        41

                                        frac14 295 kN=m2

                                        u frac14 120 kN=m2

                                        0 frac14 u frac14 295 120 frac14 175 kN=m2

                                        f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                                        42

                                        03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                                        100 452 552200 908 1108400 1810 2210800 3624 4424

                                        The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                                        Figure Q42

                                        43

                                        3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                                        200 222 422400 218 618600 220 820

                                        The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                                        44

                                        The modified shear strength parameters are

                                        0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                                        a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                                        The coordinates of the stress point representing failure conditions in the test are

                                        1

                                        2eth1 2THORN frac14 1

                                        2 170 frac14 85 kN=m2

                                        1

                                        2eth1 thorn 3THORN frac14 1

                                        2eth270thorn 100THORN frac14 185 kN=m2

                                        The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                                        uf frac14 36 kN=m2

                                        Figure Q43

                                        Figure Q44

                                        Shear strength 23

                                        45

                                        3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                        150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                        The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                        The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                        46

                                        03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                        200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                        The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                        Figure Q45

                                        24 Shear strength

                                        47

                                        The torque required to produce shear failure is given by

                                        T frac14 dh cud

                                        2thorn 2

                                        Z d=2

                                        0

                                        2r drcur

                                        frac14 cud2h

                                        2thorn 4cu

                                        Z d=2

                                        0

                                        r2dr

                                        frac14 cud2h

                                        2thorn d

                                        3

                                        6

                                        Then

                                        35 frac14 cu52 10

                                        2thorn 53

                                        6

                                        103

                                        cu frac14 76 kN=m3

                                        400

                                        0 400 800 1200 1600

                                        τ (k

                                        Nm

                                        2 )

                                        σprime (kNm2)

                                        34deg

                                        315deg29deg

                                        (a)

                                        (b)

                                        0 400

                                        400

                                        800 1200 1600

                                        Failure envelope

                                        300 500

                                        σprime (kNm2)

                                        τ (k

                                        Nm

                                        2 )

                                        20 (kNm2)

                                        31deg

                                        Figure Q46

                                        Shear strength 25

                                        48

                                        The relevant stress values are calculated as follows

                                        3 frac14 600 kN=m2

                                        1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                        2(1 3) 0 40 79 107 139 159

                                        1

                                        2(01 thorn 03) 400 411 402 389 351 326

                                        1

                                        2(1 thorn 3) 600 640 679 707 739 759

                                        The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                        Af frac14 433 200

                                        319frac14 073

                                        49

                                        B frac14 u33

                                        frac14 144

                                        350 200frac14 096

                                        a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                        0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                        10 283 65 023

                                        Figure Q48

                                        26 Shear strength

                                        The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                        Figure Q49

                                        Shear strength 27

                                        Chapter 5

                                        Stresses and displacements

                                        51

                                        Vertical stress is given by

                                        z frac14 Qz2Ip frac14 5000

                                        52Ip

                                        Values of Ip are obtained from Table 51

                                        r (m) rz Ip z (kNm2)

                                        0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                        10 20 0009 2

                                        The variation of z with radial distance (r) is plotted in Figure Q51

                                        Figure Q51

                                        52

                                        Below the centre load (Figure Q52)

                                        r

                                        zfrac14 0 for the 7500-kN load

                                        Ip frac14 0478

                                        r

                                        zfrac14 5

                                        4frac14 125 for the 10 000- and 9000-kN loads

                                        Ip frac14 0045

                                        Then

                                        z frac14X Q

                                        z2Ip

                                        frac14 7500 0478

                                        42thorn 10 000 0045

                                        42thorn 9000 0045

                                        42

                                        frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                        53

                                        The vertical stress under a corner of a rectangular area is given by

                                        z frac14 qIr

                                        where values of Ir are obtained from Figure 510 In this case

                                        z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                        z

                                        Figure Q52

                                        Stresses and displacements 29

                                        z (m) m n Ir z (kNm2)

                                        0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                        10 010 0005 5

                                        z is plotted against z in Figure Q53

                                        54

                                        (a)

                                        m frac14 125

                                        12frac14 104

                                        n frac14 18

                                        12frac14 150

                                        From Figure 510 Irfrac14 0196

                                        z frac14 2 175 0196 frac14 68 kN=m2

                                        Figure Q53

                                        30 Stresses and displacements

                                        (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                        z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                        55

                                        Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                        Px frac14 2Q

                                        1

                                        m2 thorn 1frac14 2 150

                                        125frac14 76 kN=m

                                        Equation 517 is used to obtain the pressure distribution

                                        px frac14 4Q

                                        h

                                        m2n

                                        ethm2 thorn n2THORN2 frac14150

                                        m2n

                                        ethm2 thorn n2THORN2 ethkN=m2THORN

                                        Figure Q54

                                        Stresses and displacements 31

                                        n m2n

                                        (m2 thorn n2)2

                                        px(kNm2)

                                        0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                        The pressure distribution is plotted in Figure Q55

                                        56

                                        H

                                        Bfrac14 10

                                        2frac14 5

                                        L

                                        Bfrac14 4

                                        2frac14 2

                                        D

                                        Bfrac14 1

                                        2frac14 05

                                        Hence from Figure 515

                                        131 frac14 082

                                        130 frac14 094

                                        Figure Q55

                                        32 Stresses and displacements

                                        The immediate settlement is given by Equation 528

                                        si frac14 130131qB

                                        Eu

                                        frac14 094 082 200 2

                                        45frac14 7mm

                                        Stresses and displacements 33

                                        Chapter 6

                                        Lateral earth pressure

                                        61

                                        For 0 frac14 37 the active pressure coefficient is given by

                                        Ka frac14 1 sin 37

                                        1thorn sin 37frac14 025

                                        The total active thrust (Equation 66a with c0 frac14 0) is

                                        Pa frac14 1

                                        2KaH

                                        2 frac14 1

                                        2 025 17 62 frac14 765 kN=m

                                        If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                        K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                        and the thrust on the wall is

                                        P0 frac14 1

                                        2K0H

                                        2 frac14 1

                                        2 040 17 62 frac14 122 kN=m

                                        62

                                        The active pressure coefficients for the three soil types are as follows

                                        Ka1 frac141 sin 35

                                        1thorn sin 35frac14 0271

                                        Ka2 frac141 sin 27

                                        1thorn sin 27frac14 0375

                                        ffiffiffiffiffiffiffiKa2

                                        p frac14 0613

                                        Ka3 frac141 sin 42

                                        1thorn sin 42frac14 0198

                                        Distribution of active pressure (plotted in Figure Q62)

                                        Depth (m) Soil Active pressure (kNm2)

                                        3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                        12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                        At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                        Total thrust frac14 571 kNm

                                        Point of application is (4893571) m from the top of the wall ie 857m

                                        Force (kN) Arm (m) Moment (kN m)

                                        (1)1

                                        2 0271 16 32 frac14 195 20 390

                                        (2) 0271 16 3 2 frac14 260 40 1040

                                        (3)1

                                        2 0271 92 22 frac14 50 433 217

                                        (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                        (5)1

                                        2 0375 102 32 frac14 172 70 1204

                                        (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                        (7)1

                                        2 0198 112 42 frac14 177 1067 1889

                                        (8)1

                                        2 98 92 frac14 3969 90 35721

                                        5713 48934

                                        Figure Q62

                                        Lateral earth pressure 35

                                        63

                                        (a) For u frac14 0 Ka frac14 Kp frac14 1

                                        Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                        frac14 245

                                        At the lower end of the piling

                                        pa frac14 Kaqthorn Kasatz Kaccu

                                        frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                        frac14 115 kN=m2

                                        pp frac14 Kpsatzthorn Kpccu

                                        frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                        frac14 202 kN=m2

                                        (b) For 0 frac14 26 and frac14 1

                                        20

                                        Ka frac14 035

                                        Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                        pfrac14 145 ethEquation 619THORN

                                        Kp frac14 37

                                        Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                        pfrac14 47 ethEquation 624THORN

                                        At the lower end of the piling

                                        pa frac14 Kaqthorn Ka0z Kacc

                                        0

                                        frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                        frac14 187 kN=m2

                                        pp frac14 Kp0zthorn Kpcc

                                        0

                                        frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                        frac14 198 kN=m2

                                        36 Lateral earth pressure

                                        64

                                        (a) For 0 frac14 38 Ka frac14 024

                                        0 frac14 20 98 frac14 102 kN=m3

                                        The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                        Force (kN) Arm (m) Moment (kN m)

                                        (1) 024 10 66 frac14 159 33 525

                                        (2)1

                                        2 024 17 392 frac14 310 400 1240

                                        (3) 024 17 39 27 frac14 430 135 580

                                        (4)1

                                        2 024 102 272 frac14 89 090 80

                                        (5)1

                                        2 98 272 frac14 357 090 321

                                        Hfrac14 1345 MH frac14 2746

                                        (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                        (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                        XM frac14MV MH frac14 7790 kNm

                                        Lever arm of base resultant

                                        M

                                        Vfrac14 779

                                        488frac14 160

                                        Eccentricity of base resultant

                                        e frac14 200 160 frac14 040m

                                        39 m

                                        27 m

                                        40 m

                                        04 m

                                        04 m

                                        26 m

                                        (7)

                                        (9)

                                        (1)(2)

                                        (3)

                                        (4)

                                        (5)

                                        (8)(6)

                                        (10)

                                        WT

                                        10 kNm2

                                        Hydrostatic

                                        Figure Q64

                                        Lateral earth pressure 37

                                        Base pressures (Equation 627)

                                        p frac14 VB

                                        1 6e

                                        B

                                        frac14 488

                                        4eth1 060THORN

                                        frac14 195 kN=m2 and 49 kN=m2

                                        Factor of safety against sliding (Equation 628)

                                        F frac14 V tan

                                        Hfrac14 488 tan 25

                                        1345frac14 17

                                        (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                        Hfrac14 1633 kN

                                        V frac14 4879 kN

                                        MH frac14 3453 kNm

                                        MV frac14 10536 kNm

                                        The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                        65

                                        For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                        Kp

                                        Ffrac14 385

                                        2

                                        0 frac14 20 98 frac14 102 kN=m3

                                        The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                        Force (kN) Arm (m) Moment (kN m)

                                        (1)1

                                        2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                        (2) 026 17 45 d frac14 199d d2 995d2

                                        (3)1

                                        2 026 102 d2 frac14 133d2 d3 044d3

                                        (4)1

                                        2 385

                                        2 17 152 frac14 368 dthorn 05 368d 184

                                        (5)385

                                        2 17 15 d frac14 491d d2 2455d2

                                        (6)1

                                        2 385

                                        2 102 d2 frac14 982d2 d3 327d3

                                        38 Lateral earth pressure

                                        XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                        d3 thorn 516d2 283d 1724 frac14 0

                                        d frac14 179m

                                        Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                        Over additional 20 embedded depth

                                        pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                        Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                        66

                                        The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                        Ka frac14 sin 69=sin 105

                                        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                        pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                        26664

                                        37775

                                        2

                                        frac14 050

                                        The total active thrust (acting at 25 above the normal) is given by Equation 616

                                        Pa frac14 1

                                        2 050 19 7502 frac14 267 kN=m

                                        Figure Q65

                                        Lateral earth pressure 39

                                        Horizontal component

                                        Ph frac14 267 cos 40 frac14 205 kN=m

                                        Vertical component

                                        Pv frac14 267 sin 40 frac14 172 kN=m

                                        Consider moments about the toe of the wall (Figure Q66) (per m)

                                        Force (kN) Arm (m) Moment (kN m)

                                        (1)1

                                        2 175 650 235 frac14 1337 258 345

                                        (2) 050 650 235 frac14 764 175 134

                                        (3)1

                                        2 070 650 235 frac14 535 127 68

                                        (4) 100 400 235 frac14 940 200 188

                                        (5) 1

                                        2 080 050 235 frac14 47 027 1

                                        Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                        Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                        Lever arm of base resultant

                                        M

                                        Vfrac14 795

                                        525frac14 151m

                                        Eccentricity of base resultant

                                        e frac14 200 151 frac14 049m

                                        Figure Q66

                                        40 Lateral earth pressure

                                        Base pressures (Equation 627)

                                        p frac14 525

                                        41 6 049

                                        4

                                        frac14 228 kN=m2 and 35 kN=m2

                                        The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                        The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                        The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                        67

                                        For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                        Force (kN) Arm (m) Moment (kNm)

                                        (1)1

                                        2 027 17 52 frac14 574 183 1050

                                        (2) 027 17 5 3 frac14 689 500 3445

                                        (3)1

                                        2 027 102 32 frac14 124 550 682

                                        (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                        (5)1

                                        2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                        (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                        (7) 1

                                        2 267

                                        2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                        (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                        2 d frac14 163d d2thorn 650 82d2 1060d

                                        Tie rod force per m frac14 T 0 0

                                        XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                        d3 thorn 77d2 269d 1438 frac14 0

                                        d frac14 467m

                                        Depth of penetration frac14 12d frac14 560m

                                        Lateral earth pressure 41

                                        Algebraic sum of forces for d frac14 467m isX

                                        F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                        T frac14 905 kN=m

                                        Force in each tie rod frac14 25T frac14 226 kN

                                        68

                                        (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                        0 frac14 21 98 frac14 112 kN=m3

                                        The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                        uC frac14 150

                                        165 15 98 frac14 134 kN=m2

                                        The average seepage pressure is

                                        j frac14 15

                                        165 98 frac14 09 kN=m3

                                        Hence

                                        0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                        0 j frac14 112 09 frac14 103 kN=m3

                                        Figure Q67

                                        42 Lateral earth pressure

                                        Consider moments about the anchor point A (per m)

                                        Force (kN) Arm (m) Moment (kN m)

                                        (1) 10 026 150 frac14 390 60 2340

                                        (2)1

                                        2 026 18 452 frac14 474 15 711

                                        (3) 026 18 45 105 frac14 2211 825 18240

                                        (4)1

                                        2 026 121 1052 frac14 1734 100 17340

                                        (5)1

                                        2 134 15 frac14 101 40 404

                                        (6) 134 30 frac14 402 60 2412

                                        (7)1

                                        2 134 60 frac14 402 95 3819

                                        571 4527(8) Ppm

                                        115 115PPm

                                        XM frac14 0

                                        Ppm frac144527

                                        115frac14 394 kN=m

                                        Available passive resistance

                                        Pp frac14 1

                                        2 385 103 62 frac14 714 kN=m

                                        Factor of safety

                                        Fp frac14 Pp

                                        Ppm

                                        frac14 714

                                        394frac14 18

                                        Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                        Figure Q68

                                        Lateral earth pressure 43

                                        (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                        Consider moments (per m) about the tie point A

                                        Force (kN) Arm (m)

                                        (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                        (2)1

                                        2 033 18 452 frac14 601 15

                                        (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                        (4)1

                                        2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                        (5)1

                                        2 134 15 frac14 101 40

                                        (6) 134 30 frac14 402 60

                                        (7)1

                                        2 134 d frac14 67d d3thorn 75

                                        (8) 1

                                        2 30 103 d2 frac141545d2 2d3thorn 75

                                        Moment (kN m)

                                        (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                        XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                        d3 thorn 827d2 466d 1518 frac14 0

                                        By trial

                                        d frac14 544m

                                        The minimum depth of embedment required is 544m

                                        69

                                        For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                        0 frac14 20 98 frac14 102 kN=m3

                                        The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                        44 Lateral earth pressure

                                        uC frac14 147

                                        173 26 98 frac14 216 kN=m2

                                        and the average seepage pressure around the wall is

                                        j frac14 26

                                        173 98 frac14 15 kN=m3

                                        Consider moments about the prop (A) (per m)

                                        Force (kN) Arm (m) Moment (kN m)

                                        (1)1

                                        2 03 17 272 frac14 186 020 37

                                        (2) 03 17 27 53 frac14 730 335 2445

                                        (3)1

                                        2 03 (102thorn 15) 532 frac14 493 423 2085

                                        (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                        (5)1

                                        2 216 26 frac14 281 243 684

                                        (6) 216 27 frac14 583 465 2712

                                        (7)1

                                        2 216 60 frac14 648 800 5184

                                        3055(8)

                                        1

                                        2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                        Factor of safety

                                        Fr frac14 6885

                                        3055frac14 225

                                        Figure Q69

                                        Lateral earth pressure 45

                                        610

                                        For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                        p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                        Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                        Using the recommendations of Twine and Roscoe

                                        p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                        Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                        611

                                        frac14 18 kN=m3 0 frac14 34

                                        H frac14 350m nH frac14 335m mH frac14 185m

                                        Consider a trial value of F frac14 20 Refer to Figure 635

                                        0m frac14 tan1tan 34

                                        20

                                        frac14 186

                                        Then

                                        frac14 45 thorn 0m2frac14 543

                                        W frac14 1

                                        2 18 3502 cot 543 frac14 792 kN=m

                                        Figure Q610

                                        46 Lateral earth pressure

                                        P frac14 1

                                        2 s 3352 frac14 561s kN=m

                                        U frac14 1

                                        2 98 1852 cosec 543 frac14 206 kN=m

                                        Equations 630 and 631 then become

                                        561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                        792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                        ie

                                        561s 0616N 405 frac14 0

                                        792 0857N thorn 563 frac14 0

                                        N frac14 848

                                        0857frac14 989 kN=m

                                        Then

                                        561s 609 405 frac14 0

                                        s frac14 649

                                        561frac14 116 kN=m3

                                        The calculations for trial values of F of 20 15 and 10 are summarized below

                                        F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                        20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                        s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                        Figure Q611

                                        Lateral earth pressure 47

                                        612

                                        For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                        45 thorn 0

                                        2frac14 63

                                        For the retained material between the surface and a depth of 36m

                                        Pa frac14 1

                                        2 030 18 362 frac14 350 kN=m

                                        Weight of reinforced fill between the surface and a depth of 36m is

                                        Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                        eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                        Eccentricity of Rv

                                        e frac14 263 250 frac14 013m

                                        The average vertical stress at a depth of 36m is

                                        z frac14 Rv

                                        L 2efrac14 324

                                        474frac14 68 kN=m2

                                        (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                        Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                        Tensile stress in the element frac14 138 103

                                        65 3frac14 71N=mm2

                                        Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                        Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                        Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                        The weight of ABC is

                                        W frac14 1

                                        2 18 52 265 frac14 124 kN=m

                                        From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                        48 Lateral earth pressure

                                        (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                        Tp frac14 032 68 120 065 frac14 170 kN

                                        Tr frac14 213 420

                                        418frac14 214 kN

                                        Again the tensile failure and slipping limit states are satisfied for this element

                                        Figure Q612

                                        Lateral earth pressure 49

                                        Chapter 7

                                        Consolidation theory

                                        71

                                        Total change in thickness

                                        H frac14 782 602 frac14 180mm

                                        Average thickness frac14 1530thorn 180

                                        2frac14 1620mm

                                        Length of drainage path d frac14 1620

                                        2frac14 810mm

                                        Root time plot (Figure Q71a)

                                        ffiffiffiffiffiffit90p frac14 33

                                        t90 frac14 109min

                                        cv frac14 0848d2

                                        t90frac14 0848 8102

                                        109 1440 365

                                        106frac14 27m2=year

                                        r0 frac14 782 764

                                        782 602frac14 018

                                        180frac14 0100

                                        rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                        10 119

                                        9 180frac14 0735

                                        rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                        Log time plot (Figure Q71b)

                                        t50 frac14 26min

                                        cv frac14 0196d2

                                        t50frac14 0196 8102

                                        26 1440 365

                                        106frac14 26m2=year

                                        r0 frac14 782 763

                                        782 602frac14 019

                                        180frac14 0106

                                        rp frac14 763 623

                                        782 602frac14 140

                                        180frac14 0778

                                        rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                        Figure Q71(a)

                                        Figure Q71(b)

                                        Final void ratio

                                        e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                        e

                                        Hfrac14 1thorn e0

                                        H0frac14 1thorn e1 thorne

                                        H0

                                        ie

                                        e

                                        180frac14 1631thorne

                                        1710

                                        e frac14 2936

                                        1530frac14 0192

                                        Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                        mv frac14 1

                                        1thorn e0 e0 e101 00

                                        frac14 1

                                        1823 0192

                                        0107frac14 098m2=MN

                                        k frac14 cvmvw frac14 265 098 98

                                        60 1440 365 103frac14 81 1010 m=s

                                        72

                                        Using Equation 77 (one-dimensional method)

                                        sc frac14 e0 e11thorn e0 H

                                        Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                        Figure Q72

                                        52 Consolidation theory

                                        Settlement

                                        Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                        1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                        318

                                        Notes 5 92y 460thorn 84

                                        Heave

                                        Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                        1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                        38

                                        73

                                        U frac14 f ethTvTHORN frac14 f cvt

                                        d2

                                        Hence if cv is constant

                                        t1

                                        t2frac14 d

                                        21

                                        d22

                                        where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                        d1 frac14 95mm and d2 frac14 2500mm

                                        for U frac14 050 t2 frac14 t1 d22

                                        d21

                                        frac14 20

                                        60 24 365 25002

                                        952frac14 263 years

                                        for U lt 060 Tv frac14

                                        4U2 (Equation 724(a))

                                        t030 frac14 t050 0302

                                        0502

                                        frac14 263 036 frac14 095 years

                                        Consolidation theory 53

                                        74

                                        The layer is open

                                        d frac14 8

                                        2frac14 4m

                                        Tv frac14 cvtd2frac14 24 3

                                        42frac14 0450

                                        ui frac14 frac14 84 kN=m2

                                        The excess pore water pressure is given by Equation 721

                                        ue frac14Xmfrac141mfrac140

                                        2ui

                                        Msin

                                        Mz

                                        d

                                        expethM2TvTHORN

                                        In this case z frac14 d

                                        sinMz

                                        d

                                        frac14 sinM

                                        where

                                        M frac14

                                        23

                                        25

                                        2

                                        M sin M M2Tv exp (M2Tv)

                                        2thorn1 1110 0329

                                        3

                                        21 9993 457 105

                                        ue frac14 2 84 2

                                        1 0329 ethother terms negligibleTHORN

                                        frac14 352 kN=m2

                                        75

                                        The layer is open

                                        d frac14 6

                                        2frac14 3m

                                        Tv frac14 cvtd2frac14 10 3

                                        32frac14 0333

                                        The layer thickness will be divided into six equal parts ie m frac14 6

                                        54 Consolidation theory

                                        For an open layer

                                        Tv frac14 4n

                                        m2

                                        n frac14 0333 62

                                        4frac14 300

                                        The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                        ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                        i j

                                        0 1 2 3 4 5 6 7 8 9 10 11 12

                                        0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                        The initial and 3-year isochrones are plotted in Figure Q75

                                        Area under initial isochrone frac14 180 units

                                        Area under 3-year isochrone frac14 63 units

                                        The average degree of consolidation is given by Equation 725Thus

                                        U frac14 1 63

                                        180frac14 065

                                        Figure Q75

                                        Consolidation theory 55

                                        76

                                        At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                        0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                        The final consolidation settlement (one-dimensional method) is

                                        sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                        Corrected time t frac14 2 1

                                        2

                                        40

                                        52

                                        frac14 1615 years

                                        Tv frac14 cvtd2frac14 44 1615

                                        42frac14 0444

                                        From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                        77

                                        The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                        Figure Q77

                                        56 Consolidation theory

                                        Point m n Ir (kNm2) sc (mm)

                                        13020frac14 15 20

                                        20frac14 10 0194 (4) 113 124

                                        260

                                        20frac14 30

                                        20

                                        20frac14 10 0204 (2) 59 65

                                        360

                                        20frac14 30

                                        40

                                        20frac14 20 0238 (1) 35 38

                                        430

                                        20frac14 15

                                        40

                                        20frac14 20 0224 (2) 65 72

                                        Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                        78

                                        Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                        (a) Immediate settlement

                                        H

                                        Bfrac14 30

                                        35frac14 086

                                        D

                                        Bfrac14 2

                                        35frac14 006

                                        Figure Q78

                                        Consolidation theory 57

                                        From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                        si frac14 130131qB

                                        Eufrac14 10 032 105 35

                                        40frac14 30mm

                                        (b) Consolidation settlement

                                        Layer z (m) Dz Ic (kNm2) syod (mm)

                                        1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                        3150

                                        Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                        Now

                                        H

                                        Bfrac14 30

                                        35frac14 086 and A frac14 065

                                        from Figure 712 13 frac14 079

                                        sc frac14 13sod frac14 079 315 frac14 250mm

                                        Total settlement

                                        s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                        79

                                        Without sand drains

                                        Uv frac14 025

                                        Tv frac14 0049 ethfrom Figure 718THORN

                                        t frac14 Tvd2

                                        cvfrac14 0049 82

                                        cvWith sand drains

                                        R frac14 0564S frac14 0564 3 frac14 169m

                                        n frac14 Rrfrac14 169

                                        015frac14 113

                                        Tr frac14 cht

                                        4R2frac14 ch

                                        4 1692 0049 82

                                        cvethand ch frac14 cvTHORN

                                        frac14 0275

                                        Ur frac14 073 (from Figure 730)

                                        58 Consolidation theory

                                        Using Equation 740

                                        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                        U frac14 080

                                        710

                                        Without sand drains

                                        Uv frac14 090

                                        Tv frac14 0848

                                        t frac14 Tvd2

                                        cvfrac14 0848 102

                                        96frac14 88 years

                                        With sand drains

                                        R frac14 0564S frac14 0564 4 frac14 226m

                                        n frac14 Rrfrac14 226

                                        015frac14 15

                                        Tr

                                        Tvfrac14 chcv

                                        d2

                                        4R2ethsame tTHORN

                                        Tr

                                        Tvfrac14 140

                                        96 102

                                        4 2262frac14 714 eth1THORN

                                        Using Equation 740

                                        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                        An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                        Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                        040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                        Thus

                                        Uv frac14 0295 and Ur frac14 086

                                        t frac14 88 00683

                                        0848frac14 07 years

                                        Consolidation theory 59

                                        Chapter 8

                                        Bearing capacity

                                        81

                                        (a) The ultimate bearing capacity is given by Equation 83

                                        qf frac14 cNc thorn DNq thorn 1

                                        2BN

                                        For u frac14 0

                                        Nc frac14 514 Nq frac14 1 N frac14 0

                                        qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                        The net ultimate bearing capacity is

                                        qnf frac14 qf D frac14 540 kN=m2

                                        The net foundation pressure is

                                        qn frac14 q D frac14 425

                                        2 eth21 1THORN frac14 192 kN=m2

                                        The factor of safety (Equation 86) is

                                        F frac14 qnfqnfrac14 540

                                        192frac14 28

                                        (b) For 0 frac14 28

                                        Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                        qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                        2 112 2 13

                                        frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                        qnf frac14 574 112 frac14 563 kN=m2

                                        F frac14 563

                                        192frac14 29

                                        (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                        82

                                        For 0 frac14 38

                                        Nq frac14 49 N frac14 67

                                        qnf frac14 DethNq 1THORN thorn 1

                                        2BN ethfrom Equation 83THORN

                                        frac14 eth18 075 48THORN thorn 1

                                        2 18 15 67

                                        frac14 648thorn 905 frac14 1553 kN=m2

                                        qn frac14 500

                                        15 eth18 075THORN frac14 320 kN=m2

                                        F frac14 qnfqnfrac14 1553

                                        320frac14 48

                                        0d frac14 tan1tan 38

                                        125

                                        frac14 32 therefore Nq frac14 23 and N frac14 25

                                        Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                        2 18 15 25

                                        frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                        Design load (action) Vd frac14 500 kN=m

                                        The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                        83

                                        D

                                        Bfrac14 350

                                        225frac14 155

                                        From Figure 85 for a square foundation

                                        Nc frac14 81

                                        Bearing capacity 61

                                        For a rectangular foundation (L frac14 450m B frac14 225m)

                                        Nc frac14 084thorn 016B

                                        L

                                        81 frac14 745

                                        Using Equation 810

                                        qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                        For F frac14 3

                                        qn frac14 1006

                                        3frac14 335 kN=m2

                                        q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                        Design load frac14 405 450 225 frac14 4100 kN

                                        Design undrained strength cud frac14 135

                                        14frac14 96 kN=m2

                                        Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                        frac14 7241 kN

                                        Design load Vd frac14 4100 kN

                                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                                        84

                                        For 0 frac14 40

                                        Nq frac14 64 N frac14 95

                                        qnf frac14 DethNq 1THORN thorn 04BN

                                        (a) Water table 5m below ground level

                                        qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                        qn frac14 400 17 frac14 383 kN=m2

                                        F frac14 2686

                                        383frac14 70

                                        (b) Water table 1m below ground level (ie at foundation level)

                                        0 frac14 20 98 frac14 102 kN=m3

                                        62 Bearing capacity

                                        qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                        F frac14 2040

                                        383frac14 53

                                        (c) Water table at ground level with upward hydraulic gradient 02

                                        eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                        qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                        F frac14 1296

                                        392frac14 33

                                        85

                                        The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                        Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                        Design value of 0 frac14 tan1tan 39

                                        125

                                        frac14 33

                                        For 0 frac14 33 Nq frac14 26 and N frac14 29

                                        Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                                        86

                                        (a) Undrained shear for u frac14 0

                                        Nc frac14 514 Nq frac14 1 N frac14 0

                                        qnf frac14 12cuNc

                                        frac14 12 100 514 frac14 617 kN=m2

                                        qn frac14 qnfFfrac14 617

                                        3frac14 206 kN=m2

                                        q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                        Bearing capacity 63

                                        Drained shear for 0 frac14 32

                                        Nq frac14 23 N frac14 25

                                        0 frac14 21 98 frac14 112 kN=m3

                                        qnf frac14 0DethNq 1THORN thorn 040BN

                                        frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                        frac14 694 kN=m2

                                        q frac14 694

                                        3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                        Design load frac14 42 227 frac14 3632 kN

                                        (b) Design undrained strength cud frac14 100

                                        14frac14 71 kNm2

                                        Design bearing resistance Rd frac14 12cudNe area

                                        frac14 12 71 514 42

                                        frac14 7007 kN

                                        For drained shear 0d frac14 tan1tan 32

                                        125

                                        frac14 26

                                        Nq frac14 12 N frac14 10

                                        Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                        (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                        Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                        1 2 100 0175 0700qn 0182qn

                                        2 6 033 0044 0176qn 0046qn

                                        3 10 020 0017 0068qn 0018qn

                                        0246qn

                                        Diameter of equivalent circle B frac14 45m

                                        H

                                        Bfrac14 12

                                        45frac14 27 and A frac14 042

                                        13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                        64 Bearing capacity

                                        For sc frac14 30mm

                                        qn frac14 30

                                        0147frac14 204 kN=m2

                                        q frac14 204thorn 21 frac14 225 kN=m2

                                        Design load frac14 42 225 frac14 3600 kN

                                        The design load is 3600 kN settlement being the limiting criterion

                                        87

                                        D

                                        Bfrac14 8

                                        4frac14 20

                                        From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                        F frac14 cuNc

                                        Dfrac14 40 71

                                        20 8frac14 18

                                        88

                                        Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                        Design value of 0 frac14 tan1tan 38

                                        125

                                        frac14 32

                                        Figure Q86

                                        Bearing capacity 65

                                        For 0 frac14 32 Nq frac14 23 and N frac14 25

                                        Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                        The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                        Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                        For B frac14 250m qn frac14 3750

                                        2502 17 frac14 583 kN=m2

                                        From Figure 510 m frac14 n frac14 126

                                        6frac14 021

                                        Ir frac14 0019

                                        Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                        Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                        The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                        89

                                        Depth (m) N 0v (kNm2) CN N1

                                        070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                        Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                        (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                        Cw frac14 05thorn 0530

                                        47

                                        frac14 082

                                        66 Bearing capacity

                                        Thus

                                        qa frac14 150 082 frac14 120 kN=m2

                                        (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                        Thus

                                        qa frac14 90 15 frac14 135 kN=m2

                                        (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                        Ic frac14 171

                                        1014frac14 0068

                                        From Equation 819(a) with s frac14 25mm

                                        q frac14 25

                                        3507 0068frac14 150 kN=m2

                                        810

                                        Peak value of strain influence factor occurs at a depth of 27m and is given by

                                        Izp frac14 05thorn 01130

                                        16 27

                                        05

                                        frac14 067

                                        Refer to Figure Q810

                                        E frac14 25qc

                                        Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                        Ez (mm3MN)

                                        1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                        0203

                                        C1 frac14 1 0500qnfrac14 1 05 12 16

                                        130frac14 093

                                        C2 frac14 1 ethsayTHORN

                                        s frac14 C1C2qnX Iz

                                        Ez frac14 093 1 130 0203 frac14 25mm

                                        Bearing capacity 67

                                        811

                                        At pile base level

                                        cu frac14 220 kN=m2

                                        qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                        Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                        00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                        qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                        Then

                                        Qf frac14 Abqb thorn Asqs

                                        frac14

                                        4 32 1980

                                        thorn eth 105 139 86THORN

                                        frac14 13 996thorn 3941 frac14 17 937 kN

                                        0 01 02 03 04 05 06 07

                                        0 2 4 6 8 10 12 14

                                        1

                                        2

                                        3

                                        4

                                        5

                                        6

                                        7

                                        8

                                        (1)

                                        (2)

                                        (3)

                                        (4)

                                        (5)

                                        qc

                                        qc

                                        Iz

                                        Iz

                                        (MNm2)

                                        z (m)

                                        Figure Q810

                                        68 Bearing capacity

                                        Allowable load

                                        ethaTHORN Qf

                                        2frac14 17 937

                                        2frac14 8968 kN

                                        ethbTHORN Abqb

                                        3thorn Asqs frac14 13 996

                                        3thorn 3941 frac14 8606 kN

                                        ie allowable load frac14 8600 kN

                                        Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                        According to the limit state method

                                        Characteristic undrained strength at base level cuk frac14 220

                                        150kN=m2

                                        Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                        150frac14 1320 kN=m2

                                        Characteristic shaft resistance qsk frac14 00150

                                        frac14 86

                                        150frac14 57 kN=m2

                                        Characteristic base and shaft resistances

                                        Rbk frac14

                                        4 32 1320 frac14 9330 kN

                                        Rsk frac14 105 139 86

                                        150frac14 2629 kN

                                        For a bored pile the partial factors are b frac14 160 and s frac14 130

                                        Design bearing resistance Rcd frac14 9330

                                        160thorn 2629

                                        130

                                        frac14 5831thorn 2022

                                        frac14 7850 kN

                                        Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                        812

                                        ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                        qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                        For a single pile

                                        Qf frac14 Abqb thorn Asqs

                                        frac14

                                        4 062 1305

                                        thorn eth 06 15 42THORN

                                        frac14 369thorn 1187 frac14 1556 kN

                                        Bearing capacity 69

                                        Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                        eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                        (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                        (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                        qbkfrac14 9cuk frac14 9 220

                                        150frac14 1320 kN=m2

                                        qskfrac14cuk frac14 040 105

                                        150frac14 28 kN=m2

                                        Rbkfrac14

                                        4 0602 1320 frac14 373 kN

                                        Rskfrac14 060 15 28 frac14 791 kN

                                        Rcdfrac14 373

                                        160thorn 791

                                        130frac14 233thorn 608 frac14 841 kN

                                        Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                        Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                        (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                        q frac14 21 000

                                        1762frac14 68 kN=m2

                                        Immediate settlement

                                        H

                                        Bfrac14 15

                                        176frac14 085

                                        D

                                        Bfrac14 13

                                        176frac14 074

                                        L

                                        Bfrac14 1

                                        Hence from Figure 515

                                        130 frac14 078 and 131 frac14 041

                                        70 Bearing capacity

                                        Thus using Equation 528

                                        si frac14 078 041 68 176

                                        65frac14 6mm

                                        Consolidation settlement

                                        Layer z (m) Area (m2) (kNm2) mvH (mm)

                                        1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                        434 (sod)

                                        Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                        sc frac14 056 434 frac14 24mm

                                        The total settlement is (6thorn 24) frac14 30mm

                                        813

                                        At base level N frac14 26 Then using Equation 830

                                        qb frac14 40NDb

                                        Bfrac14 40 26 2

                                        025frac14 8320 kN=m2

                                        ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                        Figure Q812

                                        Bearing capacity 71

                                        Over the length embedded in sand

                                        N frac14 21 ie18thorn 24

                                        2

                                        Using Equation 831

                                        qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                        For a single pile

                                        Qf frac14 Abqb thorn Asqs

                                        frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                        For the pile group assuming a group efficiency of 12

                                        XQf frac14 12 9 604 frac14 6523 kN

                                        Then the load factor is

                                        F frac14 6523

                                        2000thorn 1000frac14 21

                                        (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                        Characteristic base resistance per unit area qbk frac14 8320

                                        150frac14 5547 kNm2

                                        Characteristic shaft resistance per unit area qsk frac14 42

                                        150frac14 28 kNm2

                                        Characteristic base and shaft resistances for a single pile

                                        Rbk frac14 0252 5547 frac14 347 kN

                                        Rsk frac14 4 025 2 28 frac14 56 kN

                                        For a driven pile the partial factors are b frac14 s frac14 130

                                        Design bearing resistance Rcd frac14 347

                                        130thorn 56

                                        130frac14 310 kN

                                        For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                        Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                        (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                        From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                        72 Bearing capacity

                                        N frac14 24thorn 26thorn 34

                                        3frac14 28

                                        Ic frac14 171

                                        2814frac14 0016 ethEquation 818THORN

                                        s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                        814

                                        Using Equation 841

                                        Tf frac14 DLcu thorn

                                        4ethD2 d2THORNcuNc

                                        frac14 eth 02 5 06 110THORN thorn

                                        4eth022 012THORN110 9

                                        frac14 207thorn 23 frac14 230 kN

                                        Figure Q813

                                        Bearing capacity 73

                                        Chapter 9

                                        Stability of slopes

                                        91

                                        Referring to Figure Q91

                                        W frac14 417 19 frac14 792 kN=m

                                        Q frac14 20 28 frac14 56 kN=m

                                        Arc lengthAB frac14

                                        180 73 90 frac14 115m

                                        Arc length BC frac14

                                        180 28 90 frac14 44m

                                        The factor of safety is given by

                                        F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                        90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                        Depth of tension crack z0 frac14 2cu

                                        frac14 2 20

                                        19frac14 21m

                                        Arc length BD frac14

                                        180 13

                                        1

                                        2 90 frac14 21m

                                        F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                        The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                        Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                        14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                        Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                        The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                        92

                                        u frac14 0

                                        Depth factor D frac14 11

                                        9frac14 122

                                        Using Equation 92 with F frac14 10

                                        Ns frac14 cu

                                        FHfrac14 30

                                        10 19 9frac14 0175

                                        Hence from Figure 93

                                        frac14 50

                                        For F frac14 12

                                        Ns frac14 30

                                        12 19 9frac14 0146

                                        frac14 27

                                        93

                                        Refer to Figure Q93

                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                        1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                        74 m

                                        214 1deg

                                        213 1deg

                                        39 m

                                        WB

                                        D

                                        C

                                        28 m

                                        21 m

                                        A

                                        Q

                                        Soil (1)Soil (2)

                                        73deg

                                        Figure Q91

                                        Stability of slopes 75

                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                        9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                        599 256 328 1372

                                        Figure Q93

                                        76 Stability of slopes

                                        XW cos frac14 b

                                        Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                        W sin frac14 bX

                                        h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                        Arc length La frac14

                                        180 57

                                        1

                                        2 326 frac14 327m

                                        The factor of safety is given by

                                        F frac14 c0La thorn tan0ethW cos ulTHORN

                                        W sin

                                        frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                        frac14 091

                                        According to the limit state method

                                        0d frac14 tan1tan 32

                                        125

                                        frac14 265

                                        c0 frac14 8

                                        160frac14 5 kN=m2

                                        Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                        Design disturbing moment frac14 1075 kN=m

                                        The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                        94

                                        F frac14 1

                                        W sin

                                        Xfc0bthorn ethW ubTHORN tan0g sec

                                        1thorn ethtan tan0=FTHORN

                                        c0 frac14 8 kN=m2

                                        0 frac14 32

                                        c0b frac14 8 2 frac14 16 kN=m

                                        W frac14 bh frac14 21 2 h frac14 42h kN=m

                                        Try F frac14 100

                                        tan0

                                        Ffrac14 0625

                                        Stability of slopes 77

                                        Values of u are as obtained in Figure Q93

                                        SliceNo

                                        h(m)

                                        W frac14 bh(kNm)

                                        W sin(kNm)

                                        ub(kNm)

                                        c0bthorn (W ub) tan0(kNm)

                                        sec

                                        1thorn (tan tan0)FProduct(kNm)

                                        1 05 21 6 2 8 24 1078 262 13 55 31

                                        23 33 30 1042 31

                                        3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                        224 92 72 0931 67

                                        6 50 210 11 40 100 85 0907 777 55 231 14

                                        12 58 112 90 0889 80

                                        8 60 252 1812

                                        80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                        10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                        2154 88 116 0853 99

                                        14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                        1074 1091

                                        F frac14 1091

                                        1074frac14 102 (assumed value 100)

                                        Thus

                                        F frac14 101

                                        95

                                        F frac14 1

                                        W sin

                                        XfWeth1 ruTHORN tan0g sec

                                        1thorn ethtan tan0THORN=F

                                        0 frac14 33

                                        ru frac14 020

                                        W frac14 bh frac14 20 5 h frac14 100h kN=m

                                        eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                        Try F frac14 110

                                        tan 0

                                        Ffrac14 tan 33

                                        110frac14 0590

                                        78 Stability of slopes

                                        Referring to Figure Q95

                                        SliceNo

                                        h(m)

                                        W frac14 bh(kNm)

                                        W sin(kNm)

                                        W(1 ru) tan0(kNm)

                                        sec

                                        1thorn ( tan tan0)FProduct(kNm)

                                        1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                        2120 234 0892 209

                                        4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                        1185 1271

                                        Figure Q95

                                        Stability of slopes 79

                                        F frac14 1271

                                        1185frac14 107

                                        The trial value was 110 therefore take F to be 108

                                        96

                                        (a) Water table at surface the factor of safety is given by Equation 912

                                        F frac14 0

                                        sat

                                        tan0

                                        tan

                                        ptie 15 frac14 92

                                        19

                                        tan 36

                                        tan

                                        tan frac14 0234

                                        frac14 13

                                        Water table well below surface the factor of safety is given by Equation 911

                                        F frac14 tan0

                                        tan

                                        frac14 tan 36

                                        tan 13

                                        frac14 31

                                        (b) 0d frac14 tan1tan 36

                                        125

                                        frac14 30

                                        Depth of potential failure surface frac14 z

                                        Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                        frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                        frac14 504z kN

                                        Design disturbing moment per unit area Sd frac14 sat sin cos

                                        frac14 19 z sin 13 cos 13

                                        frac14 416z kN

                                        Rd gtSd therefore the limit state for overall stability is satisfied

                                        80 Stability of slopes

                                        • Book Cover
                                        • Title
                                        • Contents
                                        • Basic characteristics of soils
                                        • Seepage
                                        • Effective stress
                                        • Shear strength
                                        • Stresses and displacements
                                        • Lateral earth pressure
                                        • Consolidation theory
                                        • Bearing capacity
                                        • Stability of slopes

                                          Total vertical stress

                                          v frac14 eth200 98THORN thorn eth5 20THORN frac14 2060 kN=m2

                                          Pore water pressure

                                          u frac14 205 98 frac14 2009 kN=m2

                                          Effective vertical stress

                                          0v frac14 v u frac14 2060 2009 frac14 51 kN=m2

                                          33

                                          At top of the clay

                                          v frac14 eth2 165THORN thorn eth2 19THORN frac14 710 kN=m2

                                          u frac14 2 98 frac14 196 kN=m2

                                          0v frac14 v u frac14 710 196 frac14 514 kN=m2

                                          Alternatively

                                          0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                                          0v frac14 eth2 165THORN thorn eth2 92THORN frac14 514 kN=m2

                                          At bottom of the clay

                                          v frac14 eth2 165THORN thorn eth2 19THORN thorn eth4 20THORN frac14 1510 kN=m2

                                          u frac14 12 98 frac14 1176 kN=m2

                                          0v frac14 v u frac14 1510 1176 frac14 334 kN=m2

                                          NB The alternative method of calculation is not applicable because of the artesiancondition

                                          Figure Q3132

                                          Effective stress 15

                                          34

                                          0 frac14 20 98 frac14 102 kN=m3

                                          At 8m depth

                                          0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                                          35

                                          0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                                          0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                                          Figure Q33

                                          Figure Q34

                                          16 Effective stress

                                          (a) Immediately after WT rise

                                          At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                                          0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                                          At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                                          0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                                          (b) Several years after WT rise

                                          At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                                          At 8m depth

                                          0v frac14 940 kN=m2 (as above)

                                          At 12m depth

                                          0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                                          Figure Q35

                                          Effective stress 17

                                          36

                                          Total weight

                                          ab frac14 210 kN

                                          Effective weight

                                          ac frac14 112 kN

                                          Resultant boundary water force

                                          be frac14 119 kN

                                          Seepage force

                                          ce frac14 34 kN

                                          Resultant body force

                                          ae frac14 99 kN eth73 to horizontalTHORN

                                          (Refer to Figure Q36)

                                          Figure Q36

                                          18 Effective stress

                                          37

                                          Situation (1)(a)

                                          frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                                          u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                          0 frac14 u frac14 694 392 frac14 302 kN=m2

                                          (b)

                                          i frac14 2

                                          4frac14 05

                                          j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                                          Situation (2)(a)

                                          frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                                          u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                          0 frac14 u frac14 498 392 frac14 106 kN=m2

                                          (b)

                                          i frac14 2

                                          4frac14 05

                                          j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                                          38

                                          The flow net is drawn in Figure Q24

                                          Loss in total head between adjacent equipotentials

                                          h frac14 550

                                          Ndfrac14 550

                                          11frac14 050m

                                          Exit hydraulic gradient

                                          ie frac14 h

                                          sfrac14 050

                                          070frac14 071

                                          Effective stress 19

                                          The critical hydraulic gradient is given by Equation 39

                                          ic frac14 0

                                          wfrac14 102

                                          98frac14 104

                                          Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                                          F frac14 iciefrac14 104

                                          071frac14 15

                                          Total head at C

                                          hC frac14 nd

                                          Ndh frac14 24

                                          11 550 frac14 120m

                                          Elevation head at C

                                          zC frac14 250m

                                          Pore water pressure at C

                                          uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                                          Therefore effective vertical stress at C

                                          0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                                          For point D

                                          hD frac14 73

                                          11 550 frac14 365m

                                          zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                                          0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                                          39

                                          The flow net is drawn in Figure Q25

                                          For a soil prism 150 300m adjacent to the piling

                                          hm frac14 26

                                          9 500 frac14 145m

                                          20 Effective stress

                                          Factor of safety against lsquoheavingrsquo (Equation 310)

                                          F frac14 ic

                                          imfrac14 0d

                                          whmfrac14 97 300

                                          98 145frac14 20

                                          With a filter

                                          F frac14 0d thorn wwhm

                                          3 frac14 eth97 300THORN thorn w98 145

                                          w frac14 135 kN=m2

                                          Depth of filterfrac14 13521frac14 065m (if above water level)

                                          Effective stress 21

                                          Chapter 4

                                          Shear strength

                                          41

                                          frac14 295 kN=m2

                                          u frac14 120 kN=m2

                                          0 frac14 u frac14 295 120 frac14 175 kN=m2

                                          f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                                          42

                                          03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                                          100 452 552200 908 1108400 1810 2210800 3624 4424

                                          The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                                          Figure Q42

                                          43

                                          3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                                          200 222 422400 218 618600 220 820

                                          The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                                          44

                                          The modified shear strength parameters are

                                          0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                                          a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                                          The coordinates of the stress point representing failure conditions in the test are

                                          1

                                          2eth1 2THORN frac14 1

                                          2 170 frac14 85 kN=m2

                                          1

                                          2eth1 thorn 3THORN frac14 1

                                          2eth270thorn 100THORN frac14 185 kN=m2

                                          The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                                          uf frac14 36 kN=m2

                                          Figure Q43

                                          Figure Q44

                                          Shear strength 23

                                          45

                                          3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                          150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                          The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                          The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                          46

                                          03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                          200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                          The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                          Figure Q45

                                          24 Shear strength

                                          47

                                          The torque required to produce shear failure is given by

                                          T frac14 dh cud

                                          2thorn 2

                                          Z d=2

                                          0

                                          2r drcur

                                          frac14 cud2h

                                          2thorn 4cu

                                          Z d=2

                                          0

                                          r2dr

                                          frac14 cud2h

                                          2thorn d

                                          3

                                          6

                                          Then

                                          35 frac14 cu52 10

                                          2thorn 53

                                          6

                                          103

                                          cu frac14 76 kN=m3

                                          400

                                          0 400 800 1200 1600

                                          τ (k

                                          Nm

                                          2 )

                                          σprime (kNm2)

                                          34deg

                                          315deg29deg

                                          (a)

                                          (b)

                                          0 400

                                          400

                                          800 1200 1600

                                          Failure envelope

                                          300 500

                                          σprime (kNm2)

                                          τ (k

                                          Nm

                                          2 )

                                          20 (kNm2)

                                          31deg

                                          Figure Q46

                                          Shear strength 25

                                          48

                                          The relevant stress values are calculated as follows

                                          3 frac14 600 kN=m2

                                          1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                          2(1 3) 0 40 79 107 139 159

                                          1

                                          2(01 thorn 03) 400 411 402 389 351 326

                                          1

                                          2(1 thorn 3) 600 640 679 707 739 759

                                          The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                          Af frac14 433 200

                                          319frac14 073

                                          49

                                          B frac14 u33

                                          frac14 144

                                          350 200frac14 096

                                          a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                          0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                          10 283 65 023

                                          Figure Q48

                                          26 Shear strength

                                          The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                          Figure Q49

                                          Shear strength 27

                                          Chapter 5

                                          Stresses and displacements

                                          51

                                          Vertical stress is given by

                                          z frac14 Qz2Ip frac14 5000

                                          52Ip

                                          Values of Ip are obtained from Table 51

                                          r (m) rz Ip z (kNm2)

                                          0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                          10 20 0009 2

                                          The variation of z with radial distance (r) is plotted in Figure Q51

                                          Figure Q51

                                          52

                                          Below the centre load (Figure Q52)

                                          r

                                          zfrac14 0 for the 7500-kN load

                                          Ip frac14 0478

                                          r

                                          zfrac14 5

                                          4frac14 125 for the 10 000- and 9000-kN loads

                                          Ip frac14 0045

                                          Then

                                          z frac14X Q

                                          z2Ip

                                          frac14 7500 0478

                                          42thorn 10 000 0045

                                          42thorn 9000 0045

                                          42

                                          frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                          53

                                          The vertical stress under a corner of a rectangular area is given by

                                          z frac14 qIr

                                          where values of Ir are obtained from Figure 510 In this case

                                          z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                          z

                                          Figure Q52

                                          Stresses and displacements 29

                                          z (m) m n Ir z (kNm2)

                                          0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                          10 010 0005 5

                                          z is plotted against z in Figure Q53

                                          54

                                          (a)

                                          m frac14 125

                                          12frac14 104

                                          n frac14 18

                                          12frac14 150

                                          From Figure 510 Irfrac14 0196

                                          z frac14 2 175 0196 frac14 68 kN=m2

                                          Figure Q53

                                          30 Stresses and displacements

                                          (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                          z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                          55

                                          Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                          Px frac14 2Q

                                          1

                                          m2 thorn 1frac14 2 150

                                          125frac14 76 kN=m

                                          Equation 517 is used to obtain the pressure distribution

                                          px frac14 4Q

                                          h

                                          m2n

                                          ethm2 thorn n2THORN2 frac14150

                                          m2n

                                          ethm2 thorn n2THORN2 ethkN=m2THORN

                                          Figure Q54

                                          Stresses and displacements 31

                                          n m2n

                                          (m2 thorn n2)2

                                          px(kNm2)

                                          0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                          The pressure distribution is plotted in Figure Q55

                                          56

                                          H

                                          Bfrac14 10

                                          2frac14 5

                                          L

                                          Bfrac14 4

                                          2frac14 2

                                          D

                                          Bfrac14 1

                                          2frac14 05

                                          Hence from Figure 515

                                          131 frac14 082

                                          130 frac14 094

                                          Figure Q55

                                          32 Stresses and displacements

                                          The immediate settlement is given by Equation 528

                                          si frac14 130131qB

                                          Eu

                                          frac14 094 082 200 2

                                          45frac14 7mm

                                          Stresses and displacements 33

                                          Chapter 6

                                          Lateral earth pressure

                                          61

                                          For 0 frac14 37 the active pressure coefficient is given by

                                          Ka frac14 1 sin 37

                                          1thorn sin 37frac14 025

                                          The total active thrust (Equation 66a with c0 frac14 0) is

                                          Pa frac14 1

                                          2KaH

                                          2 frac14 1

                                          2 025 17 62 frac14 765 kN=m

                                          If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                          K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                          and the thrust on the wall is

                                          P0 frac14 1

                                          2K0H

                                          2 frac14 1

                                          2 040 17 62 frac14 122 kN=m

                                          62

                                          The active pressure coefficients for the three soil types are as follows

                                          Ka1 frac141 sin 35

                                          1thorn sin 35frac14 0271

                                          Ka2 frac141 sin 27

                                          1thorn sin 27frac14 0375

                                          ffiffiffiffiffiffiffiKa2

                                          p frac14 0613

                                          Ka3 frac141 sin 42

                                          1thorn sin 42frac14 0198

                                          Distribution of active pressure (plotted in Figure Q62)

                                          Depth (m) Soil Active pressure (kNm2)

                                          3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                          12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                          At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                          Total thrust frac14 571 kNm

                                          Point of application is (4893571) m from the top of the wall ie 857m

                                          Force (kN) Arm (m) Moment (kN m)

                                          (1)1

                                          2 0271 16 32 frac14 195 20 390

                                          (2) 0271 16 3 2 frac14 260 40 1040

                                          (3)1

                                          2 0271 92 22 frac14 50 433 217

                                          (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                          (5)1

                                          2 0375 102 32 frac14 172 70 1204

                                          (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                          (7)1

                                          2 0198 112 42 frac14 177 1067 1889

                                          (8)1

                                          2 98 92 frac14 3969 90 35721

                                          5713 48934

                                          Figure Q62

                                          Lateral earth pressure 35

                                          63

                                          (a) For u frac14 0 Ka frac14 Kp frac14 1

                                          Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                          frac14 245

                                          At the lower end of the piling

                                          pa frac14 Kaqthorn Kasatz Kaccu

                                          frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                          frac14 115 kN=m2

                                          pp frac14 Kpsatzthorn Kpccu

                                          frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                          frac14 202 kN=m2

                                          (b) For 0 frac14 26 and frac14 1

                                          20

                                          Ka frac14 035

                                          Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                          pfrac14 145 ethEquation 619THORN

                                          Kp frac14 37

                                          Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                          pfrac14 47 ethEquation 624THORN

                                          At the lower end of the piling

                                          pa frac14 Kaqthorn Ka0z Kacc

                                          0

                                          frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                          frac14 187 kN=m2

                                          pp frac14 Kp0zthorn Kpcc

                                          0

                                          frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                          frac14 198 kN=m2

                                          36 Lateral earth pressure

                                          64

                                          (a) For 0 frac14 38 Ka frac14 024

                                          0 frac14 20 98 frac14 102 kN=m3

                                          The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                          Force (kN) Arm (m) Moment (kN m)

                                          (1) 024 10 66 frac14 159 33 525

                                          (2)1

                                          2 024 17 392 frac14 310 400 1240

                                          (3) 024 17 39 27 frac14 430 135 580

                                          (4)1

                                          2 024 102 272 frac14 89 090 80

                                          (5)1

                                          2 98 272 frac14 357 090 321

                                          Hfrac14 1345 MH frac14 2746

                                          (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                          (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                          XM frac14MV MH frac14 7790 kNm

                                          Lever arm of base resultant

                                          M

                                          Vfrac14 779

                                          488frac14 160

                                          Eccentricity of base resultant

                                          e frac14 200 160 frac14 040m

                                          39 m

                                          27 m

                                          40 m

                                          04 m

                                          04 m

                                          26 m

                                          (7)

                                          (9)

                                          (1)(2)

                                          (3)

                                          (4)

                                          (5)

                                          (8)(6)

                                          (10)

                                          WT

                                          10 kNm2

                                          Hydrostatic

                                          Figure Q64

                                          Lateral earth pressure 37

                                          Base pressures (Equation 627)

                                          p frac14 VB

                                          1 6e

                                          B

                                          frac14 488

                                          4eth1 060THORN

                                          frac14 195 kN=m2 and 49 kN=m2

                                          Factor of safety against sliding (Equation 628)

                                          F frac14 V tan

                                          Hfrac14 488 tan 25

                                          1345frac14 17

                                          (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                          Hfrac14 1633 kN

                                          V frac14 4879 kN

                                          MH frac14 3453 kNm

                                          MV frac14 10536 kNm

                                          The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                          65

                                          For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                          Kp

                                          Ffrac14 385

                                          2

                                          0 frac14 20 98 frac14 102 kN=m3

                                          The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                          Force (kN) Arm (m) Moment (kN m)

                                          (1)1

                                          2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                          (2) 026 17 45 d frac14 199d d2 995d2

                                          (3)1

                                          2 026 102 d2 frac14 133d2 d3 044d3

                                          (4)1

                                          2 385

                                          2 17 152 frac14 368 dthorn 05 368d 184

                                          (5)385

                                          2 17 15 d frac14 491d d2 2455d2

                                          (6)1

                                          2 385

                                          2 102 d2 frac14 982d2 d3 327d3

                                          38 Lateral earth pressure

                                          XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                          d3 thorn 516d2 283d 1724 frac14 0

                                          d frac14 179m

                                          Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                          Over additional 20 embedded depth

                                          pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                          Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                          66

                                          The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                          Ka frac14 sin 69=sin 105

                                          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                          pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                          26664

                                          37775

                                          2

                                          frac14 050

                                          The total active thrust (acting at 25 above the normal) is given by Equation 616

                                          Pa frac14 1

                                          2 050 19 7502 frac14 267 kN=m

                                          Figure Q65

                                          Lateral earth pressure 39

                                          Horizontal component

                                          Ph frac14 267 cos 40 frac14 205 kN=m

                                          Vertical component

                                          Pv frac14 267 sin 40 frac14 172 kN=m

                                          Consider moments about the toe of the wall (Figure Q66) (per m)

                                          Force (kN) Arm (m) Moment (kN m)

                                          (1)1

                                          2 175 650 235 frac14 1337 258 345

                                          (2) 050 650 235 frac14 764 175 134

                                          (3)1

                                          2 070 650 235 frac14 535 127 68

                                          (4) 100 400 235 frac14 940 200 188

                                          (5) 1

                                          2 080 050 235 frac14 47 027 1

                                          Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                          Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                          Lever arm of base resultant

                                          M

                                          Vfrac14 795

                                          525frac14 151m

                                          Eccentricity of base resultant

                                          e frac14 200 151 frac14 049m

                                          Figure Q66

                                          40 Lateral earth pressure

                                          Base pressures (Equation 627)

                                          p frac14 525

                                          41 6 049

                                          4

                                          frac14 228 kN=m2 and 35 kN=m2

                                          The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                          The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                          The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                          67

                                          For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                          Force (kN) Arm (m) Moment (kNm)

                                          (1)1

                                          2 027 17 52 frac14 574 183 1050

                                          (2) 027 17 5 3 frac14 689 500 3445

                                          (3)1

                                          2 027 102 32 frac14 124 550 682

                                          (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                          (5)1

                                          2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                          (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                          (7) 1

                                          2 267

                                          2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                          (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                          2 d frac14 163d d2thorn 650 82d2 1060d

                                          Tie rod force per m frac14 T 0 0

                                          XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                          d3 thorn 77d2 269d 1438 frac14 0

                                          d frac14 467m

                                          Depth of penetration frac14 12d frac14 560m

                                          Lateral earth pressure 41

                                          Algebraic sum of forces for d frac14 467m isX

                                          F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                          T frac14 905 kN=m

                                          Force in each tie rod frac14 25T frac14 226 kN

                                          68

                                          (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                          0 frac14 21 98 frac14 112 kN=m3

                                          The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                          uC frac14 150

                                          165 15 98 frac14 134 kN=m2

                                          The average seepage pressure is

                                          j frac14 15

                                          165 98 frac14 09 kN=m3

                                          Hence

                                          0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                          0 j frac14 112 09 frac14 103 kN=m3

                                          Figure Q67

                                          42 Lateral earth pressure

                                          Consider moments about the anchor point A (per m)

                                          Force (kN) Arm (m) Moment (kN m)

                                          (1) 10 026 150 frac14 390 60 2340

                                          (2)1

                                          2 026 18 452 frac14 474 15 711

                                          (3) 026 18 45 105 frac14 2211 825 18240

                                          (4)1

                                          2 026 121 1052 frac14 1734 100 17340

                                          (5)1

                                          2 134 15 frac14 101 40 404

                                          (6) 134 30 frac14 402 60 2412

                                          (7)1

                                          2 134 60 frac14 402 95 3819

                                          571 4527(8) Ppm

                                          115 115PPm

                                          XM frac14 0

                                          Ppm frac144527

                                          115frac14 394 kN=m

                                          Available passive resistance

                                          Pp frac14 1

                                          2 385 103 62 frac14 714 kN=m

                                          Factor of safety

                                          Fp frac14 Pp

                                          Ppm

                                          frac14 714

                                          394frac14 18

                                          Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                          Figure Q68

                                          Lateral earth pressure 43

                                          (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                          Consider moments (per m) about the tie point A

                                          Force (kN) Arm (m)

                                          (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                          (2)1

                                          2 033 18 452 frac14 601 15

                                          (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                          (4)1

                                          2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                          (5)1

                                          2 134 15 frac14 101 40

                                          (6) 134 30 frac14 402 60

                                          (7)1

                                          2 134 d frac14 67d d3thorn 75

                                          (8) 1

                                          2 30 103 d2 frac141545d2 2d3thorn 75

                                          Moment (kN m)

                                          (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                          XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                          d3 thorn 827d2 466d 1518 frac14 0

                                          By trial

                                          d frac14 544m

                                          The minimum depth of embedment required is 544m

                                          69

                                          For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                          0 frac14 20 98 frac14 102 kN=m3

                                          The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                          44 Lateral earth pressure

                                          uC frac14 147

                                          173 26 98 frac14 216 kN=m2

                                          and the average seepage pressure around the wall is

                                          j frac14 26

                                          173 98 frac14 15 kN=m3

                                          Consider moments about the prop (A) (per m)

                                          Force (kN) Arm (m) Moment (kN m)

                                          (1)1

                                          2 03 17 272 frac14 186 020 37

                                          (2) 03 17 27 53 frac14 730 335 2445

                                          (3)1

                                          2 03 (102thorn 15) 532 frac14 493 423 2085

                                          (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                          (5)1

                                          2 216 26 frac14 281 243 684

                                          (6) 216 27 frac14 583 465 2712

                                          (7)1

                                          2 216 60 frac14 648 800 5184

                                          3055(8)

                                          1

                                          2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                          Factor of safety

                                          Fr frac14 6885

                                          3055frac14 225

                                          Figure Q69

                                          Lateral earth pressure 45

                                          610

                                          For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                          p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                          Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                          Using the recommendations of Twine and Roscoe

                                          p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                          Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                          611

                                          frac14 18 kN=m3 0 frac14 34

                                          H frac14 350m nH frac14 335m mH frac14 185m

                                          Consider a trial value of F frac14 20 Refer to Figure 635

                                          0m frac14 tan1tan 34

                                          20

                                          frac14 186

                                          Then

                                          frac14 45 thorn 0m2frac14 543

                                          W frac14 1

                                          2 18 3502 cot 543 frac14 792 kN=m

                                          Figure Q610

                                          46 Lateral earth pressure

                                          P frac14 1

                                          2 s 3352 frac14 561s kN=m

                                          U frac14 1

                                          2 98 1852 cosec 543 frac14 206 kN=m

                                          Equations 630 and 631 then become

                                          561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                          792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                          ie

                                          561s 0616N 405 frac14 0

                                          792 0857N thorn 563 frac14 0

                                          N frac14 848

                                          0857frac14 989 kN=m

                                          Then

                                          561s 609 405 frac14 0

                                          s frac14 649

                                          561frac14 116 kN=m3

                                          The calculations for trial values of F of 20 15 and 10 are summarized below

                                          F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                          20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                          s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                          Figure Q611

                                          Lateral earth pressure 47

                                          612

                                          For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                          45 thorn 0

                                          2frac14 63

                                          For the retained material between the surface and a depth of 36m

                                          Pa frac14 1

                                          2 030 18 362 frac14 350 kN=m

                                          Weight of reinforced fill between the surface and a depth of 36m is

                                          Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                          eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                          Eccentricity of Rv

                                          e frac14 263 250 frac14 013m

                                          The average vertical stress at a depth of 36m is

                                          z frac14 Rv

                                          L 2efrac14 324

                                          474frac14 68 kN=m2

                                          (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                          Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                          Tensile stress in the element frac14 138 103

                                          65 3frac14 71N=mm2

                                          Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                          Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                          Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                          The weight of ABC is

                                          W frac14 1

                                          2 18 52 265 frac14 124 kN=m

                                          From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                          48 Lateral earth pressure

                                          (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                          Tp frac14 032 68 120 065 frac14 170 kN

                                          Tr frac14 213 420

                                          418frac14 214 kN

                                          Again the tensile failure and slipping limit states are satisfied for this element

                                          Figure Q612

                                          Lateral earth pressure 49

                                          Chapter 7

                                          Consolidation theory

                                          71

                                          Total change in thickness

                                          H frac14 782 602 frac14 180mm

                                          Average thickness frac14 1530thorn 180

                                          2frac14 1620mm

                                          Length of drainage path d frac14 1620

                                          2frac14 810mm

                                          Root time plot (Figure Q71a)

                                          ffiffiffiffiffiffit90p frac14 33

                                          t90 frac14 109min

                                          cv frac14 0848d2

                                          t90frac14 0848 8102

                                          109 1440 365

                                          106frac14 27m2=year

                                          r0 frac14 782 764

                                          782 602frac14 018

                                          180frac14 0100

                                          rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                          10 119

                                          9 180frac14 0735

                                          rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                          Log time plot (Figure Q71b)

                                          t50 frac14 26min

                                          cv frac14 0196d2

                                          t50frac14 0196 8102

                                          26 1440 365

                                          106frac14 26m2=year

                                          r0 frac14 782 763

                                          782 602frac14 019

                                          180frac14 0106

                                          rp frac14 763 623

                                          782 602frac14 140

                                          180frac14 0778

                                          rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                          Figure Q71(a)

                                          Figure Q71(b)

                                          Final void ratio

                                          e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                          e

                                          Hfrac14 1thorn e0

                                          H0frac14 1thorn e1 thorne

                                          H0

                                          ie

                                          e

                                          180frac14 1631thorne

                                          1710

                                          e frac14 2936

                                          1530frac14 0192

                                          Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                          mv frac14 1

                                          1thorn e0 e0 e101 00

                                          frac14 1

                                          1823 0192

                                          0107frac14 098m2=MN

                                          k frac14 cvmvw frac14 265 098 98

                                          60 1440 365 103frac14 81 1010 m=s

                                          72

                                          Using Equation 77 (one-dimensional method)

                                          sc frac14 e0 e11thorn e0 H

                                          Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                          Figure Q72

                                          52 Consolidation theory

                                          Settlement

                                          Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                          1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                          318

                                          Notes 5 92y 460thorn 84

                                          Heave

                                          Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                          1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                          38

                                          73

                                          U frac14 f ethTvTHORN frac14 f cvt

                                          d2

                                          Hence if cv is constant

                                          t1

                                          t2frac14 d

                                          21

                                          d22

                                          where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                          d1 frac14 95mm and d2 frac14 2500mm

                                          for U frac14 050 t2 frac14 t1 d22

                                          d21

                                          frac14 20

                                          60 24 365 25002

                                          952frac14 263 years

                                          for U lt 060 Tv frac14

                                          4U2 (Equation 724(a))

                                          t030 frac14 t050 0302

                                          0502

                                          frac14 263 036 frac14 095 years

                                          Consolidation theory 53

                                          74

                                          The layer is open

                                          d frac14 8

                                          2frac14 4m

                                          Tv frac14 cvtd2frac14 24 3

                                          42frac14 0450

                                          ui frac14 frac14 84 kN=m2

                                          The excess pore water pressure is given by Equation 721

                                          ue frac14Xmfrac141mfrac140

                                          2ui

                                          Msin

                                          Mz

                                          d

                                          expethM2TvTHORN

                                          In this case z frac14 d

                                          sinMz

                                          d

                                          frac14 sinM

                                          where

                                          M frac14

                                          23

                                          25

                                          2

                                          M sin M M2Tv exp (M2Tv)

                                          2thorn1 1110 0329

                                          3

                                          21 9993 457 105

                                          ue frac14 2 84 2

                                          1 0329 ethother terms negligibleTHORN

                                          frac14 352 kN=m2

                                          75

                                          The layer is open

                                          d frac14 6

                                          2frac14 3m

                                          Tv frac14 cvtd2frac14 10 3

                                          32frac14 0333

                                          The layer thickness will be divided into six equal parts ie m frac14 6

                                          54 Consolidation theory

                                          For an open layer

                                          Tv frac14 4n

                                          m2

                                          n frac14 0333 62

                                          4frac14 300

                                          The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                          ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                          i j

                                          0 1 2 3 4 5 6 7 8 9 10 11 12

                                          0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                          The initial and 3-year isochrones are plotted in Figure Q75

                                          Area under initial isochrone frac14 180 units

                                          Area under 3-year isochrone frac14 63 units

                                          The average degree of consolidation is given by Equation 725Thus

                                          U frac14 1 63

                                          180frac14 065

                                          Figure Q75

                                          Consolidation theory 55

                                          76

                                          At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                          0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                          The final consolidation settlement (one-dimensional method) is

                                          sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                          Corrected time t frac14 2 1

                                          2

                                          40

                                          52

                                          frac14 1615 years

                                          Tv frac14 cvtd2frac14 44 1615

                                          42frac14 0444

                                          From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                          77

                                          The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                          Figure Q77

                                          56 Consolidation theory

                                          Point m n Ir (kNm2) sc (mm)

                                          13020frac14 15 20

                                          20frac14 10 0194 (4) 113 124

                                          260

                                          20frac14 30

                                          20

                                          20frac14 10 0204 (2) 59 65

                                          360

                                          20frac14 30

                                          40

                                          20frac14 20 0238 (1) 35 38

                                          430

                                          20frac14 15

                                          40

                                          20frac14 20 0224 (2) 65 72

                                          Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                          78

                                          Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                          (a) Immediate settlement

                                          H

                                          Bfrac14 30

                                          35frac14 086

                                          D

                                          Bfrac14 2

                                          35frac14 006

                                          Figure Q78

                                          Consolidation theory 57

                                          From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                          si frac14 130131qB

                                          Eufrac14 10 032 105 35

                                          40frac14 30mm

                                          (b) Consolidation settlement

                                          Layer z (m) Dz Ic (kNm2) syod (mm)

                                          1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                          3150

                                          Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                          Now

                                          H

                                          Bfrac14 30

                                          35frac14 086 and A frac14 065

                                          from Figure 712 13 frac14 079

                                          sc frac14 13sod frac14 079 315 frac14 250mm

                                          Total settlement

                                          s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                          79

                                          Without sand drains

                                          Uv frac14 025

                                          Tv frac14 0049 ethfrom Figure 718THORN

                                          t frac14 Tvd2

                                          cvfrac14 0049 82

                                          cvWith sand drains

                                          R frac14 0564S frac14 0564 3 frac14 169m

                                          n frac14 Rrfrac14 169

                                          015frac14 113

                                          Tr frac14 cht

                                          4R2frac14 ch

                                          4 1692 0049 82

                                          cvethand ch frac14 cvTHORN

                                          frac14 0275

                                          Ur frac14 073 (from Figure 730)

                                          58 Consolidation theory

                                          Using Equation 740

                                          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                          U frac14 080

                                          710

                                          Without sand drains

                                          Uv frac14 090

                                          Tv frac14 0848

                                          t frac14 Tvd2

                                          cvfrac14 0848 102

                                          96frac14 88 years

                                          With sand drains

                                          R frac14 0564S frac14 0564 4 frac14 226m

                                          n frac14 Rrfrac14 226

                                          015frac14 15

                                          Tr

                                          Tvfrac14 chcv

                                          d2

                                          4R2ethsame tTHORN

                                          Tr

                                          Tvfrac14 140

                                          96 102

                                          4 2262frac14 714 eth1THORN

                                          Using Equation 740

                                          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                          An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                          Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                          040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                          Thus

                                          Uv frac14 0295 and Ur frac14 086

                                          t frac14 88 00683

                                          0848frac14 07 years

                                          Consolidation theory 59

                                          Chapter 8

                                          Bearing capacity

                                          81

                                          (a) The ultimate bearing capacity is given by Equation 83

                                          qf frac14 cNc thorn DNq thorn 1

                                          2BN

                                          For u frac14 0

                                          Nc frac14 514 Nq frac14 1 N frac14 0

                                          qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                          The net ultimate bearing capacity is

                                          qnf frac14 qf D frac14 540 kN=m2

                                          The net foundation pressure is

                                          qn frac14 q D frac14 425

                                          2 eth21 1THORN frac14 192 kN=m2

                                          The factor of safety (Equation 86) is

                                          F frac14 qnfqnfrac14 540

                                          192frac14 28

                                          (b) For 0 frac14 28

                                          Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                          qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                          2 112 2 13

                                          frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                          qnf frac14 574 112 frac14 563 kN=m2

                                          F frac14 563

                                          192frac14 29

                                          (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                          82

                                          For 0 frac14 38

                                          Nq frac14 49 N frac14 67

                                          qnf frac14 DethNq 1THORN thorn 1

                                          2BN ethfrom Equation 83THORN

                                          frac14 eth18 075 48THORN thorn 1

                                          2 18 15 67

                                          frac14 648thorn 905 frac14 1553 kN=m2

                                          qn frac14 500

                                          15 eth18 075THORN frac14 320 kN=m2

                                          F frac14 qnfqnfrac14 1553

                                          320frac14 48

                                          0d frac14 tan1tan 38

                                          125

                                          frac14 32 therefore Nq frac14 23 and N frac14 25

                                          Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                          2 18 15 25

                                          frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                          Design load (action) Vd frac14 500 kN=m

                                          The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                          83

                                          D

                                          Bfrac14 350

                                          225frac14 155

                                          From Figure 85 for a square foundation

                                          Nc frac14 81

                                          Bearing capacity 61

                                          For a rectangular foundation (L frac14 450m B frac14 225m)

                                          Nc frac14 084thorn 016B

                                          L

                                          81 frac14 745

                                          Using Equation 810

                                          qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                          For F frac14 3

                                          qn frac14 1006

                                          3frac14 335 kN=m2

                                          q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                          Design load frac14 405 450 225 frac14 4100 kN

                                          Design undrained strength cud frac14 135

                                          14frac14 96 kN=m2

                                          Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                          frac14 7241 kN

                                          Design load Vd frac14 4100 kN

                                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                                          84

                                          For 0 frac14 40

                                          Nq frac14 64 N frac14 95

                                          qnf frac14 DethNq 1THORN thorn 04BN

                                          (a) Water table 5m below ground level

                                          qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                          qn frac14 400 17 frac14 383 kN=m2

                                          F frac14 2686

                                          383frac14 70

                                          (b) Water table 1m below ground level (ie at foundation level)

                                          0 frac14 20 98 frac14 102 kN=m3

                                          62 Bearing capacity

                                          qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                          F frac14 2040

                                          383frac14 53

                                          (c) Water table at ground level with upward hydraulic gradient 02

                                          eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                          qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                          F frac14 1296

                                          392frac14 33

                                          85

                                          The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                          Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                          Design value of 0 frac14 tan1tan 39

                                          125

                                          frac14 33

                                          For 0 frac14 33 Nq frac14 26 and N frac14 29

                                          Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                                          86

                                          (a) Undrained shear for u frac14 0

                                          Nc frac14 514 Nq frac14 1 N frac14 0

                                          qnf frac14 12cuNc

                                          frac14 12 100 514 frac14 617 kN=m2

                                          qn frac14 qnfFfrac14 617

                                          3frac14 206 kN=m2

                                          q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                          Bearing capacity 63

                                          Drained shear for 0 frac14 32

                                          Nq frac14 23 N frac14 25

                                          0 frac14 21 98 frac14 112 kN=m3

                                          qnf frac14 0DethNq 1THORN thorn 040BN

                                          frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                          frac14 694 kN=m2

                                          q frac14 694

                                          3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                          Design load frac14 42 227 frac14 3632 kN

                                          (b) Design undrained strength cud frac14 100

                                          14frac14 71 kNm2

                                          Design bearing resistance Rd frac14 12cudNe area

                                          frac14 12 71 514 42

                                          frac14 7007 kN

                                          For drained shear 0d frac14 tan1tan 32

                                          125

                                          frac14 26

                                          Nq frac14 12 N frac14 10

                                          Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                          (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                          Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                          1 2 100 0175 0700qn 0182qn

                                          2 6 033 0044 0176qn 0046qn

                                          3 10 020 0017 0068qn 0018qn

                                          0246qn

                                          Diameter of equivalent circle B frac14 45m

                                          H

                                          Bfrac14 12

                                          45frac14 27 and A frac14 042

                                          13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                          64 Bearing capacity

                                          For sc frac14 30mm

                                          qn frac14 30

                                          0147frac14 204 kN=m2

                                          q frac14 204thorn 21 frac14 225 kN=m2

                                          Design load frac14 42 225 frac14 3600 kN

                                          The design load is 3600 kN settlement being the limiting criterion

                                          87

                                          D

                                          Bfrac14 8

                                          4frac14 20

                                          From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                          F frac14 cuNc

                                          Dfrac14 40 71

                                          20 8frac14 18

                                          88

                                          Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                          Design value of 0 frac14 tan1tan 38

                                          125

                                          frac14 32

                                          Figure Q86

                                          Bearing capacity 65

                                          For 0 frac14 32 Nq frac14 23 and N frac14 25

                                          Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                          The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                          Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                          For B frac14 250m qn frac14 3750

                                          2502 17 frac14 583 kN=m2

                                          From Figure 510 m frac14 n frac14 126

                                          6frac14 021

                                          Ir frac14 0019

                                          Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                          Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                          The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                          89

                                          Depth (m) N 0v (kNm2) CN N1

                                          070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                          Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                          (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                          Cw frac14 05thorn 0530

                                          47

                                          frac14 082

                                          66 Bearing capacity

                                          Thus

                                          qa frac14 150 082 frac14 120 kN=m2

                                          (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                          Thus

                                          qa frac14 90 15 frac14 135 kN=m2

                                          (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                          Ic frac14 171

                                          1014frac14 0068

                                          From Equation 819(a) with s frac14 25mm

                                          q frac14 25

                                          3507 0068frac14 150 kN=m2

                                          810

                                          Peak value of strain influence factor occurs at a depth of 27m and is given by

                                          Izp frac14 05thorn 01130

                                          16 27

                                          05

                                          frac14 067

                                          Refer to Figure Q810

                                          E frac14 25qc

                                          Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                          Ez (mm3MN)

                                          1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                          0203

                                          C1 frac14 1 0500qnfrac14 1 05 12 16

                                          130frac14 093

                                          C2 frac14 1 ethsayTHORN

                                          s frac14 C1C2qnX Iz

                                          Ez frac14 093 1 130 0203 frac14 25mm

                                          Bearing capacity 67

                                          811

                                          At pile base level

                                          cu frac14 220 kN=m2

                                          qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                          Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                          00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                          qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                          Then

                                          Qf frac14 Abqb thorn Asqs

                                          frac14

                                          4 32 1980

                                          thorn eth 105 139 86THORN

                                          frac14 13 996thorn 3941 frac14 17 937 kN

                                          0 01 02 03 04 05 06 07

                                          0 2 4 6 8 10 12 14

                                          1

                                          2

                                          3

                                          4

                                          5

                                          6

                                          7

                                          8

                                          (1)

                                          (2)

                                          (3)

                                          (4)

                                          (5)

                                          qc

                                          qc

                                          Iz

                                          Iz

                                          (MNm2)

                                          z (m)

                                          Figure Q810

                                          68 Bearing capacity

                                          Allowable load

                                          ethaTHORN Qf

                                          2frac14 17 937

                                          2frac14 8968 kN

                                          ethbTHORN Abqb

                                          3thorn Asqs frac14 13 996

                                          3thorn 3941 frac14 8606 kN

                                          ie allowable load frac14 8600 kN

                                          Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                          According to the limit state method

                                          Characteristic undrained strength at base level cuk frac14 220

                                          150kN=m2

                                          Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                          150frac14 1320 kN=m2

                                          Characteristic shaft resistance qsk frac14 00150

                                          frac14 86

                                          150frac14 57 kN=m2

                                          Characteristic base and shaft resistances

                                          Rbk frac14

                                          4 32 1320 frac14 9330 kN

                                          Rsk frac14 105 139 86

                                          150frac14 2629 kN

                                          For a bored pile the partial factors are b frac14 160 and s frac14 130

                                          Design bearing resistance Rcd frac14 9330

                                          160thorn 2629

                                          130

                                          frac14 5831thorn 2022

                                          frac14 7850 kN

                                          Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                          812

                                          ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                          qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                          For a single pile

                                          Qf frac14 Abqb thorn Asqs

                                          frac14

                                          4 062 1305

                                          thorn eth 06 15 42THORN

                                          frac14 369thorn 1187 frac14 1556 kN

                                          Bearing capacity 69

                                          Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                          eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                          (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                          (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                          qbkfrac14 9cuk frac14 9 220

                                          150frac14 1320 kN=m2

                                          qskfrac14cuk frac14 040 105

                                          150frac14 28 kN=m2

                                          Rbkfrac14

                                          4 0602 1320 frac14 373 kN

                                          Rskfrac14 060 15 28 frac14 791 kN

                                          Rcdfrac14 373

                                          160thorn 791

                                          130frac14 233thorn 608 frac14 841 kN

                                          Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                          Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                          (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                          q frac14 21 000

                                          1762frac14 68 kN=m2

                                          Immediate settlement

                                          H

                                          Bfrac14 15

                                          176frac14 085

                                          D

                                          Bfrac14 13

                                          176frac14 074

                                          L

                                          Bfrac14 1

                                          Hence from Figure 515

                                          130 frac14 078 and 131 frac14 041

                                          70 Bearing capacity

                                          Thus using Equation 528

                                          si frac14 078 041 68 176

                                          65frac14 6mm

                                          Consolidation settlement

                                          Layer z (m) Area (m2) (kNm2) mvH (mm)

                                          1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                          434 (sod)

                                          Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                          sc frac14 056 434 frac14 24mm

                                          The total settlement is (6thorn 24) frac14 30mm

                                          813

                                          At base level N frac14 26 Then using Equation 830

                                          qb frac14 40NDb

                                          Bfrac14 40 26 2

                                          025frac14 8320 kN=m2

                                          ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                          Figure Q812

                                          Bearing capacity 71

                                          Over the length embedded in sand

                                          N frac14 21 ie18thorn 24

                                          2

                                          Using Equation 831

                                          qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                          For a single pile

                                          Qf frac14 Abqb thorn Asqs

                                          frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                          For the pile group assuming a group efficiency of 12

                                          XQf frac14 12 9 604 frac14 6523 kN

                                          Then the load factor is

                                          F frac14 6523

                                          2000thorn 1000frac14 21

                                          (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                          Characteristic base resistance per unit area qbk frac14 8320

                                          150frac14 5547 kNm2

                                          Characteristic shaft resistance per unit area qsk frac14 42

                                          150frac14 28 kNm2

                                          Characteristic base and shaft resistances for a single pile

                                          Rbk frac14 0252 5547 frac14 347 kN

                                          Rsk frac14 4 025 2 28 frac14 56 kN

                                          For a driven pile the partial factors are b frac14 s frac14 130

                                          Design bearing resistance Rcd frac14 347

                                          130thorn 56

                                          130frac14 310 kN

                                          For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                          Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                          (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                          From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                          72 Bearing capacity

                                          N frac14 24thorn 26thorn 34

                                          3frac14 28

                                          Ic frac14 171

                                          2814frac14 0016 ethEquation 818THORN

                                          s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                          814

                                          Using Equation 841

                                          Tf frac14 DLcu thorn

                                          4ethD2 d2THORNcuNc

                                          frac14 eth 02 5 06 110THORN thorn

                                          4eth022 012THORN110 9

                                          frac14 207thorn 23 frac14 230 kN

                                          Figure Q813

                                          Bearing capacity 73

                                          Chapter 9

                                          Stability of slopes

                                          91

                                          Referring to Figure Q91

                                          W frac14 417 19 frac14 792 kN=m

                                          Q frac14 20 28 frac14 56 kN=m

                                          Arc lengthAB frac14

                                          180 73 90 frac14 115m

                                          Arc length BC frac14

                                          180 28 90 frac14 44m

                                          The factor of safety is given by

                                          F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                          90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                          Depth of tension crack z0 frac14 2cu

                                          frac14 2 20

                                          19frac14 21m

                                          Arc length BD frac14

                                          180 13

                                          1

                                          2 90 frac14 21m

                                          F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                          The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                          Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                          14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                          Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                          The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                          92

                                          u frac14 0

                                          Depth factor D frac14 11

                                          9frac14 122

                                          Using Equation 92 with F frac14 10

                                          Ns frac14 cu

                                          FHfrac14 30

                                          10 19 9frac14 0175

                                          Hence from Figure 93

                                          frac14 50

                                          For F frac14 12

                                          Ns frac14 30

                                          12 19 9frac14 0146

                                          frac14 27

                                          93

                                          Refer to Figure Q93

                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                          1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                          74 m

                                          214 1deg

                                          213 1deg

                                          39 m

                                          WB

                                          D

                                          C

                                          28 m

                                          21 m

                                          A

                                          Q

                                          Soil (1)Soil (2)

                                          73deg

                                          Figure Q91

                                          Stability of slopes 75

                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                          9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                          599 256 328 1372

                                          Figure Q93

                                          76 Stability of slopes

                                          XW cos frac14 b

                                          Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                          W sin frac14 bX

                                          h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                          Arc length La frac14

                                          180 57

                                          1

                                          2 326 frac14 327m

                                          The factor of safety is given by

                                          F frac14 c0La thorn tan0ethW cos ulTHORN

                                          W sin

                                          frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                          frac14 091

                                          According to the limit state method

                                          0d frac14 tan1tan 32

                                          125

                                          frac14 265

                                          c0 frac14 8

                                          160frac14 5 kN=m2

                                          Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                          Design disturbing moment frac14 1075 kN=m

                                          The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                          94

                                          F frac14 1

                                          W sin

                                          Xfc0bthorn ethW ubTHORN tan0g sec

                                          1thorn ethtan tan0=FTHORN

                                          c0 frac14 8 kN=m2

                                          0 frac14 32

                                          c0b frac14 8 2 frac14 16 kN=m

                                          W frac14 bh frac14 21 2 h frac14 42h kN=m

                                          Try F frac14 100

                                          tan0

                                          Ffrac14 0625

                                          Stability of slopes 77

                                          Values of u are as obtained in Figure Q93

                                          SliceNo

                                          h(m)

                                          W frac14 bh(kNm)

                                          W sin(kNm)

                                          ub(kNm)

                                          c0bthorn (W ub) tan0(kNm)

                                          sec

                                          1thorn (tan tan0)FProduct(kNm)

                                          1 05 21 6 2 8 24 1078 262 13 55 31

                                          23 33 30 1042 31

                                          3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                          224 92 72 0931 67

                                          6 50 210 11 40 100 85 0907 777 55 231 14

                                          12 58 112 90 0889 80

                                          8 60 252 1812

                                          80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                          10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                          2154 88 116 0853 99

                                          14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                          1074 1091

                                          F frac14 1091

                                          1074frac14 102 (assumed value 100)

                                          Thus

                                          F frac14 101

                                          95

                                          F frac14 1

                                          W sin

                                          XfWeth1 ruTHORN tan0g sec

                                          1thorn ethtan tan0THORN=F

                                          0 frac14 33

                                          ru frac14 020

                                          W frac14 bh frac14 20 5 h frac14 100h kN=m

                                          eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                          Try F frac14 110

                                          tan 0

                                          Ffrac14 tan 33

                                          110frac14 0590

                                          78 Stability of slopes

                                          Referring to Figure Q95

                                          SliceNo

                                          h(m)

                                          W frac14 bh(kNm)

                                          W sin(kNm)

                                          W(1 ru) tan0(kNm)

                                          sec

                                          1thorn ( tan tan0)FProduct(kNm)

                                          1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                          2120 234 0892 209

                                          4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                          1185 1271

                                          Figure Q95

                                          Stability of slopes 79

                                          F frac14 1271

                                          1185frac14 107

                                          The trial value was 110 therefore take F to be 108

                                          96

                                          (a) Water table at surface the factor of safety is given by Equation 912

                                          F frac14 0

                                          sat

                                          tan0

                                          tan

                                          ptie 15 frac14 92

                                          19

                                          tan 36

                                          tan

                                          tan frac14 0234

                                          frac14 13

                                          Water table well below surface the factor of safety is given by Equation 911

                                          F frac14 tan0

                                          tan

                                          frac14 tan 36

                                          tan 13

                                          frac14 31

                                          (b) 0d frac14 tan1tan 36

                                          125

                                          frac14 30

                                          Depth of potential failure surface frac14 z

                                          Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                          frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                          frac14 504z kN

                                          Design disturbing moment per unit area Sd frac14 sat sin cos

                                          frac14 19 z sin 13 cos 13

                                          frac14 416z kN

                                          Rd gtSd therefore the limit state for overall stability is satisfied

                                          80 Stability of slopes

                                          • Book Cover
                                          • Title
                                          • Contents
                                          • Basic characteristics of soils
                                          • Seepage
                                          • Effective stress
                                          • Shear strength
                                          • Stresses and displacements
                                          • Lateral earth pressure
                                          • Consolidation theory
                                          • Bearing capacity
                                          • Stability of slopes

                                            34

                                            0 frac14 20 98 frac14 102 kN=m3

                                            At 8m depth

                                            0v frac14 eth25 16THORN thorn eth10 20THORN thorn eth45 102THORN frac14 1059 kN=m2

                                            35

                                            0 ethsandTHORN frac14 19 98 frac14 92 kN=m3

                                            0 ethclayTHORN frac14 20 98 frac14 102 kN=m3

                                            Figure Q33

                                            Figure Q34

                                            16 Effective stress

                                            (a) Immediately after WT rise

                                            At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                                            0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                                            At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                                            0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                                            (b) Several years after WT rise

                                            At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                                            At 8m depth

                                            0v frac14 940 kN=m2 (as above)

                                            At 12m depth

                                            0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                                            Figure Q35

                                            Effective stress 17

                                            36

                                            Total weight

                                            ab frac14 210 kN

                                            Effective weight

                                            ac frac14 112 kN

                                            Resultant boundary water force

                                            be frac14 119 kN

                                            Seepage force

                                            ce frac14 34 kN

                                            Resultant body force

                                            ae frac14 99 kN eth73 to horizontalTHORN

                                            (Refer to Figure Q36)

                                            Figure Q36

                                            18 Effective stress

                                            37

                                            Situation (1)(a)

                                            frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                                            u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                            0 frac14 u frac14 694 392 frac14 302 kN=m2

                                            (b)

                                            i frac14 2

                                            4frac14 05

                                            j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                                            Situation (2)(a)

                                            frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                                            u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                            0 frac14 u frac14 498 392 frac14 106 kN=m2

                                            (b)

                                            i frac14 2

                                            4frac14 05

                                            j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                                            38

                                            The flow net is drawn in Figure Q24

                                            Loss in total head between adjacent equipotentials

                                            h frac14 550

                                            Ndfrac14 550

                                            11frac14 050m

                                            Exit hydraulic gradient

                                            ie frac14 h

                                            sfrac14 050

                                            070frac14 071

                                            Effective stress 19

                                            The critical hydraulic gradient is given by Equation 39

                                            ic frac14 0

                                            wfrac14 102

                                            98frac14 104

                                            Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                                            F frac14 iciefrac14 104

                                            071frac14 15

                                            Total head at C

                                            hC frac14 nd

                                            Ndh frac14 24

                                            11 550 frac14 120m

                                            Elevation head at C

                                            zC frac14 250m

                                            Pore water pressure at C

                                            uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                                            Therefore effective vertical stress at C

                                            0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                                            For point D

                                            hD frac14 73

                                            11 550 frac14 365m

                                            zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                                            0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                                            39

                                            The flow net is drawn in Figure Q25

                                            For a soil prism 150 300m adjacent to the piling

                                            hm frac14 26

                                            9 500 frac14 145m

                                            20 Effective stress

                                            Factor of safety against lsquoheavingrsquo (Equation 310)

                                            F frac14 ic

                                            imfrac14 0d

                                            whmfrac14 97 300

                                            98 145frac14 20

                                            With a filter

                                            F frac14 0d thorn wwhm

                                            3 frac14 eth97 300THORN thorn w98 145

                                            w frac14 135 kN=m2

                                            Depth of filterfrac14 13521frac14 065m (if above water level)

                                            Effective stress 21

                                            Chapter 4

                                            Shear strength

                                            41

                                            frac14 295 kN=m2

                                            u frac14 120 kN=m2

                                            0 frac14 u frac14 295 120 frac14 175 kN=m2

                                            f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                                            42

                                            03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                                            100 452 552200 908 1108400 1810 2210800 3624 4424

                                            The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                                            Figure Q42

                                            43

                                            3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                                            200 222 422400 218 618600 220 820

                                            The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                                            44

                                            The modified shear strength parameters are

                                            0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                                            a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                                            The coordinates of the stress point representing failure conditions in the test are

                                            1

                                            2eth1 2THORN frac14 1

                                            2 170 frac14 85 kN=m2

                                            1

                                            2eth1 thorn 3THORN frac14 1

                                            2eth270thorn 100THORN frac14 185 kN=m2

                                            The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                                            uf frac14 36 kN=m2

                                            Figure Q43

                                            Figure Q44

                                            Shear strength 23

                                            45

                                            3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                            150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                            The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                            The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                            46

                                            03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                            200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                            The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                            Figure Q45

                                            24 Shear strength

                                            47

                                            The torque required to produce shear failure is given by

                                            T frac14 dh cud

                                            2thorn 2

                                            Z d=2

                                            0

                                            2r drcur

                                            frac14 cud2h

                                            2thorn 4cu

                                            Z d=2

                                            0

                                            r2dr

                                            frac14 cud2h

                                            2thorn d

                                            3

                                            6

                                            Then

                                            35 frac14 cu52 10

                                            2thorn 53

                                            6

                                            103

                                            cu frac14 76 kN=m3

                                            400

                                            0 400 800 1200 1600

                                            τ (k

                                            Nm

                                            2 )

                                            σprime (kNm2)

                                            34deg

                                            315deg29deg

                                            (a)

                                            (b)

                                            0 400

                                            400

                                            800 1200 1600

                                            Failure envelope

                                            300 500

                                            σprime (kNm2)

                                            τ (k

                                            Nm

                                            2 )

                                            20 (kNm2)

                                            31deg

                                            Figure Q46

                                            Shear strength 25

                                            48

                                            The relevant stress values are calculated as follows

                                            3 frac14 600 kN=m2

                                            1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                            2(1 3) 0 40 79 107 139 159

                                            1

                                            2(01 thorn 03) 400 411 402 389 351 326

                                            1

                                            2(1 thorn 3) 600 640 679 707 739 759

                                            The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                            Af frac14 433 200

                                            319frac14 073

                                            49

                                            B frac14 u33

                                            frac14 144

                                            350 200frac14 096

                                            a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                            0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                            10 283 65 023

                                            Figure Q48

                                            26 Shear strength

                                            The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                            Figure Q49

                                            Shear strength 27

                                            Chapter 5

                                            Stresses and displacements

                                            51

                                            Vertical stress is given by

                                            z frac14 Qz2Ip frac14 5000

                                            52Ip

                                            Values of Ip are obtained from Table 51

                                            r (m) rz Ip z (kNm2)

                                            0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                            10 20 0009 2

                                            The variation of z with radial distance (r) is plotted in Figure Q51

                                            Figure Q51

                                            52

                                            Below the centre load (Figure Q52)

                                            r

                                            zfrac14 0 for the 7500-kN load

                                            Ip frac14 0478

                                            r

                                            zfrac14 5

                                            4frac14 125 for the 10 000- and 9000-kN loads

                                            Ip frac14 0045

                                            Then

                                            z frac14X Q

                                            z2Ip

                                            frac14 7500 0478

                                            42thorn 10 000 0045

                                            42thorn 9000 0045

                                            42

                                            frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                            53

                                            The vertical stress under a corner of a rectangular area is given by

                                            z frac14 qIr

                                            where values of Ir are obtained from Figure 510 In this case

                                            z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                            z

                                            Figure Q52

                                            Stresses and displacements 29

                                            z (m) m n Ir z (kNm2)

                                            0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                            10 010 0005 5

                                            z is plotted against z in Figure Q53

                                            54

                                            (a)

                                            m frac14 125

                                            12frac14 104

                                            n frac14 18

                                            12frac14 150

                                            From Figure 510 Irfrac14 0196

                                            z frac14 2 175 0196 frac14 68 kN=m2

                                            Figure Q53

                                            30 Stresses and displacements

                                            (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                            z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                            55

                                            Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                            Px frac14 2Q

                                            1

                                            m2 thorn 1frac14 2 150

                                            125frac14 76 kN=m

                                            Equation 517 is used to obtain the pressure distribution

                                            px frac14 4Q

                                            h

                                            m2n

                                            ethm2 thorn n2THORN2 frac14150

                                            m2n

                                            ethm2 thorn n2THORN2 ethkN=m2THORN

                                            Figure Q54

                                            Stresses and displacements 31

                                            n m2n

                                            (m2 thorn n2)2

                                            px(kNm2)

                                            0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                            The pressure distribution is plotted in Figure Q55

                                            56

                                            H

                                            Bfrac14 10

                                            2frac14 5

                                            L

                                            Bfrac14 4

                                            2frac14 2

                                            D

                                            Bfrac14 1

                                            2frac14 05

                                            Hence from Figure 515

                                            131 frac14 082

                                            130 frac14 094

                                            Figure Q55

                                            32 Stresses and displacements

                                            The immediate settlement is given by Equation 528

                                            si frac14 130131qB

                                            Eu

                                            frac14 094 082 200 2

                                            45frac14 7mm

                                            Stresses and displacements 33

                                            Chapter 6

                                            Lateral earth pressure

                                            61

                                            For 0 frac14 37 the active pressure coefficient is given by

                                            Ka frac14 1 sin 37

                                            1thorn sin 37frac14 025

                                            The total active thrust (Equation 66a with c0 frac14 0) is

                                            Pa frac14 1

                                            2KaH

                                            2 frac14 1

                                            2 025 17 62 frac14 765 kN=m

                                            If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                            K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                            and the thrust on the wall is

                                            P0 frac14 1

                                            2K0H

                                            2 frac14 1

                                            2 040 17 62 frac14 122 kN=m

                                            62

                                            The active pressure coefficients for the three soil types are as follows

                                            Ka1 frac141 sin 35

                                            1thorn sin 35frac14 0271

                                            Ka2 frac141 sin 27

                                            1thorn sin 27frac14 0375

                                            ffiffiffiffiffiffiffiKa2

                                            p frac14 0613

                                            Ka3 frac141 sin 42

                                            1thorn sin 42frac14 0198

                                            Distribution of active pressure (plotted in Figure Q62)

                                            Depth (m) Soil Active pressure (kNm2)

                                            3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                            12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                            At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                            Total thrust frac14 571 kNm

                                            Point of application is (4893571) m from the top of the wall ie 857m

                                            Force (kN) Arm (m) Moment (kN m)

                                            (1)1

                                            2 0271 16 32 frac14 195 20 390

                                            (2) 0271 16 3 2 frac14 260 40 1040

                                            (3)1

                                            2 0271 92 22 frac14 50 433 217

                                            (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                            (5)1

                                            2 0375 102 32 frac14 172 70 1204

                                            (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                            (7)1

                                            2 0198 112 42 frac14 177 1067 1889

                                            (8)1

                                            2 98 92 frac14 3969 90 35721

                                            5713 48934

                                            Figure Q62

                                            Lateral earth pressure 35

                                            63

                                            (a) For u frac14 0 Ka frac14 Kp frac14 1

                                            Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                            frac14 245

                                            At the lower end of the piling

                                            pa frac14 Kaqthorn Kasatz Kaccu

                                            frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                            frac14 115 kN=m2

                                            pp frac14 Kpsatzthorn Kpccu

                                            frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                            frac14 202 kN=m2

                                            (b) For 0 frac14 26 and frac14 1

                                            20

                                            Ka frac14 035

                                            Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                            pfrac14 145 ethEquation 619THORN

                                            Kp frac14 37

                                            Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                            pfrac14 47 ethEquation 624THORN

                                            At the lower end of the piling

                                            pa frac14 Kaqthorn Ka0z Kacc

                                            0

                                            frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                            frac14 187 kN=m2

                                            pp frac14 Kp0zthorn Kpcc

                                            0

                                            frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                            frac14 198 kN=m2

                                            36 Lateral earth pressure

                                            64

                                            (a) For 0 frac14 38 Ka frac14 024

                                            0 frac14 20 98 frac14 102 kN=m3

                                            The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                            Force (kN) Arm (m) Moment (kN m)

                                            (1) 024 10 66 frac14 159 33 525

                                            (2)1

                                            2 024 17 392 frac14 310 400 1240

                                            (3) 024 17 39 27 frac14 430 135 580

                                            (4)1

                                            2 024 102 272 frac14 89 090 80

                                            (5)1

                                            2 98 272 frac14 357 090 321

                                            Hfrac14 1345 MH frac14 2746

                                            (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                            (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                            XM frac14MV MH frac14 7790 kNm

                                            Lever arm of base resultant

                                            M

                                            Vfrac14 779

                                            488frac14 160

                                            Eccentricity of base resultant

                                            e frac14 200 160 frac14 040m

                                            39 m

                                            27 m

                                            40 m

                                            04 m

                                            04 m

                                            26 m

                                            (7)

                                            (9)

                                            (1)(2)

                                            (3)

                                            (4)

                                            (5)

                                            (8)(6)

                                            (10)

                                            WT

                                            10 kNm2

                                            Hydrostatic

                                            Figure Q64

                                            Lateral earth pressure 37

                                            Base pressures (Equation 627)

                                            p frac14 VB

                                            1 6e

                                            B

                                            frac14 488

                                            4eth1 060THORN

                                            frac14 195 kN=m2 and 49 kN=m2

                                            Factor of safety against sliding (Equation 628)

                                            F frac14 V tan

                                            Hfrac14 488 tan 25

                                            1345frac14 17

                                            (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                            Hfrac14 1633 kN

                                            V frac14 4879 kN

                                            MH frac14 3453 kNm

                                            MV frac14 10536 kNm

                                            The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                            65

                                            For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                            Kp

                                            Ffrac14 385

                                            2

                                            0 frac14 20 98 frac14 102 kN=m3

                                            The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                            Force (kN) Arm (m) Moment (kN m)

                                            (1)1

                                            2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                            (2) 026 17 45 d frac14 199d d2 995d2

                                            (3)1

                                            2 026 102 d2 frac14 133d2 d3 044d3

                                            (4)1

                                            2 385

                                            2 17 152 frac14 368 dthorn 05 368d 184

                                            (5)385

                                            2 17 15 d frac14 491d d2 2455d2

                                            (6)1

                                            2 385

                                            2 102 d2 frac14 982d2 d3 327d3

                                            38 Lateral earth pressure

                                            XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                            d3 thorn 516d2 283d 1724 frac14 0

                                            d frac14 179m

                                            Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                            Over additional 20 embedded depth

                                            pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                            Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                            66

                                            The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                            Ka frac14 sin 69=sin 105

                                            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                            pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                            26664

                                            37775

                                            2

                                            frac14 050

                                            The total active thrust (acting at 25 above the normal) is given by Equation 616

                                            Pa frac14 1

                                            2 050 19 7502 frac14 267 kN=m

                                            Figure Q65

                                            Lateral earth pressure 39

                                            Horizontal component

                                            Ph frac14 267 cos 40 frac14 205 kN=m

                                            Vertical component

                                            Pv frac14 267 sin 40 frac14 172 kN=m

                                            Consider moments about the toe of the wall (Figure Q66) (per m)

                                            Force (kN) Arm (m) Moment (kN m)

                                            (1)1

                                            2 175 650 235 frac14 1337 258 345

                                            (2) 050 650 235 frac14 764 175 134

                                            (3)1

                                            2 070 650 235 frac14 535 127 68

                                            (4) 100 400 235 frac14 940 200 188

                                            (5) 1

                                            2 080 050 235 frac14 47 027 1

                                            Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                            Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                            Lever arm of base resultant

                                            M

                                            Vfrac14 795

                                            525frac14 151m

                                            Eccentricity of base resultant

                                            e frac14 200 151 frac14 049m

                                            Figure Q66

                                            40 Lateral earth pressure

                                            Base pressures (Equation 627)

                                            p frac14 525

                                            41 6 049

                                            4

                                            frac14 228 kN=m2 and 35 kN=m2

                                            The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                            The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                            The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                            67

                                            For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                            Force (kN) Arm (m) Moment (kNm)

                                            (1)1

                                            2 027 17 52 frac14 574 183 1050

                                            (2) 027 17 5 3 frac14 689 500 3445

                                            (3)1

                                            2 027 102 32 frac14 124 550 682

                                            (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                            (5)1

                                            2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                            (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                            (7) 1

                                            2 267

                                            2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                            (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                            2 d frac14 163d d2thorn 650 82d2 1060d

                                            Tie rod force per m frac14 T 0 0

                                            XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                            d3 thorn 77d2 269d 1438 frac14 0

                                            d frac14 467m

                                            Depth of penetration frac14 12d frac14 560m

                                            Lateral earth pressure 41

                                            Algebraic sum of forces for d frac14 467m isX

                                            F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                            T frac14 905 kN=m

                                            Force in each tie rod frac14 25T frac14 226 kN

                                            68

                                            (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                            0 frac14 21 98 frac14 112 kN=m3

                                            The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                            uC frac14 150

                                            165 15 98 frac14 134 kN=m2

                                            The average seepage pressure is

                                            j frac14 15

                                            165 98 frac14 09 kN=m3

                                            Hence

                                            0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                            0 j frac14 112 09 frac14 103 kN=m3

                                            Figure Q67

                                            42 Lateral earth pressure

                                            Consider moments about the anchor point A (per m)

                                            Force (kN) Arm (m) Moment (kN m)

                                            (1) 10 026 150 frac14 390 60 2340

                                            (2)1

                                            2 026 18 452 frac14 474 15 711

                                            (3) 026 18 45 105 frac14 2211 825 18240

                                            (4)1

                                            2 026 121 1052 frac14 1734 100 17340

                                            (5)1

                                            2 134 15 frac14 101 40 404

                                            (6) 134 30 frac14 402 60 2412

                                            (7)1

                                            2 134 60 frac14 402 95 3819

                                            571 4527(8) Ppm

                                            115 115PPm

                                            XM frac14 0

                                            Ppm frac144527

                                            115frac14 394 kN=m

                                            Available passive resistance

                                            Pp frac14 1

                                            2 385 103 62 frac14 714 kN=m

                                            Factor of safety

                                            Fp frac14 Pp

                                            Ppm

                                            frac14 714

                                            394frac14 18

                                            Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                            Figure Q68

                                            Lateral earth pressure 43

                                            (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                            Consider moments (per m) about the tie point A

                                            Force (kN) Arm (m)

                                            (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                            (2)1

                                            2 033 18 452 frac14 601 15

                                            (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                            (4)1

                                            2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                            (5)1

                                            2 134 15 frac14 101 40

                                            (6) 134 30 frac14 402 60

                                            (7)1

                                            2 134 d frac14 67d d3thorn 75

                                            (8) 1

                                            2 30 103 d2 frac141545d2 2d3thorn 75

                                            Moment (kN m)

                                            (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                            XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                            d3 thorn 827d2 466d 1518 frac14 0

                                            By trial

                                            d frac14 544m

                                            The minimum depth of embedment required is 544m

                                            69

                                            For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                            0 frac14 20 98 frac14 102 kN=m3

                                            The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                            44 Lateral earth pressure

                                            uC frac14 147

                                            173 26 98 frac14 216 kN=m2

                                            and the average seepage pressure around the wall is

                                            j frac14 26

                                            173 98 frac14 15 kN=m3

                                            Consider moments about the prop (A) (per m)

                                            Force (kN) Arm (m) Moment (kN m)

                                            (1)1

                                            2 03 17 272 frac14 186 020 37

                                            (2) 03 17 27 53 frac14 730 335 2445

                                            (3)1

                                            2 03 (102thorn 15) 532 frac14 493 423 2085

                                            (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                            (5)1

                                            2 216 26 frac14 281 243 684

                                            (6) 216 27 frac14 583 465 2712

                                            (7)1

                                            2 216 60 frac14 648 800 5184

                                            3055(8)

                                            1

                                            2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                            Factor of safety

                                            Fr frac14 6885

                                            3055frac14 225

                                            Figure Q69

                                            Lateral earth pressure 45

                                            610

                                            For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                            p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                            Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                            Using the recommendations of Twine and Roscoe

                                            p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                            Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                            611

                                            frac14 18 kN=m3 0 frac14 34

                                            H frac14 350m nH frac14 335m mH frac14 185m

                                            Consider a trial value of F frac14 20 Refer to Figure 635

                                            0m frac14 tan1tan 34

                                            20

                                            frac14 186

                                            Then

                                            frac14 45 thorn 0m2frac14 543

                                            W frac14 1

                                            2 18 3502 cot 543 frac14 792 kN=m

                                            Figure Q610

                                            46 Lateral earth pressure

                                            P frac14 1

                                            2 s 3352 frac14 561s kN=m

                                            U frac14 1

                                            2 98 1852 cosec 543 frac14 206 kN=m

                                            Equations 630 and 631 then become

                                            561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                            792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                            ie

                                            561s 0616N 405 frac14 0

                                            792 0857N thorn 563 frac14 0

                                            N frac14 848

                                            0857frac14 989 kN=m

                                            Then

                                            561s 609 405 frac14 0

                                            s frac14 649

                                            561frac14 116 kN=m3

                                            The calculations for trial values of F of 20 15 and 10 are summarized below

                                            F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                            20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                            s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                            Figure Q611

                                            Lateral earth pressure 47

                                            612

                                            For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                            45 thorn 0

                                            2frac14 63

                                            For the retained material between the surface and a depth of 36m

                                            Pa frac14 1

                                            2 030 18 362 frac14 350 kN=m

                                            Weight of reinforced fill between the surface and a depth of 36m is

                                            Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                            eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                            Eccentricity of Rv

                                            e frac14 263 250 frac14 013m

                                            The average vertical stress at a depth of 36m is

                                            z frac14 Rv

                                            L 2efrac14 324

                                            474frac14 68 kN=m2

                                            (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                            Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                            Tensile stress in the element frac14 138 103

                                            65 3frac14 71N=mm2

                                            Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                            Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                            Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                            The weight of ABC is

                                            W frac14 1

                                            2 18 52 265 frac14 124 kN=m

                                            From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                            48 Lateral earth pressure

                                            (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                            Tp frac14 032 68 120 065 frac14 170 kN

                                            Tr frac14 213 420

                                            418frac14 214 kN

                                            Again the tensile failure and slipping limit states are satisfied for this element

                                            Figure Q612

                                            Lateral earth pressure 49

                                            Chapter 7

                                            Consolidation theory

                                            71

                                            Total change in thickness

                                            H frac14 782 602 frac14 180mm

                                            Average thickness frac14 1530thorn 180

                                            2frac14 1620mm

                                            Length of drainage path d frac14 1620

                                            2frac14 810mm

                                            Root time plot (Figure Q71a)

                                            ffiffiffiffiffiffit90p frac14 33

                                            t90 frac14 109min

                                            cv frac14 0848d2

                                            t90frac14 0848 8102

                                            109 1440 365

                                            106frac14 27m2=year

                                            r0 frac14 782 764

                                            782 602frac14 018

                                            180frac14 0100

                                            rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                            10 119

                                            9 180frac14 0735

                                            rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                            Log time plot (Figure Q71b)

                                            t50 frac14 26min

                                            cv frac14 0196d2

                                            t50frac14 0196 8102

                                            26 1440 365

                                            106frac14 26m2=year

                                            r0 frac14 782 763

                                            782 602frac14 019

                                            180frac14 0106

                                            rp frac14 763 623

                                            782 602frac14 140

                                            180frac14 0778

                                            rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                            Figure Q71(a)

                                            Figure Q71(b)

                                            Final void ratio

                                            e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                            e

                                            Hfrac14 1thorn e0

                                            H0frac14 1thorn e1 thorne

                                            H0

                                            ie

                                            e

                                            180frac14 1631thorne

                                            1710

                                            e frac14 2936

                                            1530frac14 0192

                                            Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                            mv frac14 1

                                            1thorn e0 e0 e101 00

                                            frac14 1

                                            1823 0192

                                            0107frac14 098m2=MN

                                            k frac14 cvmvw frac14 265 098 98

                                            60 1440 365 103frac14 81 1010 m=s

                                            72

                                            Using Equation 77 (one-dimensional method)

                                            sc frac14 e0 e11thorn e0 H

                                            Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                            Figure Q72

                                            52 Consolidation theory

                                            Settlement

                                            Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                            1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                            318

                                            Notes 5 92y 460thorn 84

                                            Heave

                                            Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                            1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                            38

                                            73

                                            U frac14 f ethTvTHORN frac14 f cvt

                                            d2

                                            Hence if cv is constant

                                            t1

                                            t2frac14 d

                                            21

                                            d22

                                            where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                            d1 frac14 95mm and d2 frac14 2500mm

                                            for U frac14 050 t2 frac14 t1 d22

                                            d21

                                            frac14 20

                                            60 24 365 25002

                                            952frac14 263 years

                                            for U lt 060 Tv frac14

                                            4U2 (Equation 724(a))

                                            t030 frac14 t050 0302

                                            0502

                                            frac14 263 036 frac14 095 years

                                            Consolidation theory 53

                                            74

                                            The layer is open

                                            d frac14 8

                                            2frac14 4m

                                            Tv frac14 cvtd2frac14 24 3

                                            42frac14 0450

                                            ui frac14 frac14 84 kN=m2

                                            The excess pore water pressure is given by Equation 721

                                            ue frac14Xmfrac141mfrac140

                                            2ui

                                            Msin

                                            Mz

                                            d

                                            expethM2TvTHORN

                                            In this case z frac14 d

                                            sinMz

                                            d

                                            frac14 sinM

                                            where

                                            M frac14

                                            23

                                            25

                                            2

                                            M sin M M2Tv exp (M2Tv)

                                            2thorn1 1110 0329

                                            3

                                            21 9993 457 105

                                            ue frac14 2 84 2

                                            1 0329 ethother terms negligibleTHORN

                                            frac14 352 kN=m2

                                            75

                                            The layer is open

                                            d frac14 6

                                            2frac14 3m

                                            Tv frac14 cvtd2frac14 10 3

                                            32frac14 0333

                                            The layer thickness will be divided into six equal parts ie m frac14 6

                                            54 Consolidation theory

                                            For an open layer

                                            Tv frac14 4n

                                            m2

                                            n frac14 0333 62

                                            4frac14 300

                                            The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                            ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                            i j

                                            0 1 2 3 4 5 6 7 8 9 10 11 12

                                            0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                            The initial and 3-year isochrones are plotted in Figure Q75

                                            Area under initial isochrone frac14 180 units

                                            Area under 3-year isochrone frac14 63 units

                                            The average degree of consolidation is given by Equation 725Thus

                                            U frac14 1 63

                                            180frac14 065

                                            Figure Q75

                                            Consolidation theory 55

                                            76

                                            At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                            0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                            The final consolidation settlement (one-dimensional method) is

                                            sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                            Corrected time t frac14 2 1

                                            2

                                            40

                                            52

                                            frac14 1615 years

                                            Tv frac14 cvtd2frac14 44 1615

                                            42frac14 0444

                                            From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                            77

                                            The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                            Figure Q77

                                            56 Consolidation theory

                                            Point m n Ir (kNm2) sc (mm)

                                            13020frac14 15 20

                                            20frac14 10 0194 (4) 113 124

                                            260

                                            20frac14 30

                                            20

                                            20frac14 10 0204 (2) 59 65

                                            360

                                            20frac14 30

                                            40

                                            20frac14 20 0238 (1) 35 38

                                            430

                                            20frac14 15

                                            40

                                            20frac14 20 0224 (2) 65 72

                                            Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                            78

                                            Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                            (a) Immediate settlement

                                            H

                                            Bfrac14 30

                                            35frac14 086

                                            D

                                            Bfrac14 2

                                            35frac14 006

                                            Figure Q78

                                            Consolidation theory 57

                                            From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                            si frac14 130131qB

                                            Eufrac14 10 032 105 35

                                            40frac14 30mm

                                            (b) Consolidation settlement

                                            Layer z (m) Dz Ic (kNm2) syod (mm)

                                            1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                            3150

                                            Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                            Now

                                            H

                                            Bfrac14 30

                                            35frac14 086 and A frac14 065

                                            from Figure 712 13 frac14 079

                                            sc frac14 13sod frac14 079 315 frac14 250mm

                                            Total settlement

                                            s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                            79

                                            Without sand drains

                                            Uv frac14 025

                                            Tv frac14 0049 ethfrom Figure 718THORN

                                            t frac14 Tvd2

                                            cvfrac14 0049 82

                                            cvWith sand drains

                                            R frac14 0564S frac14 0564 3 frac14 169m

                                            n frac14 Rrfrac14 169

                                            015frac14 113

                                            Tr frac14 cht

                                            4R2frac14 ch

                                            4 1692 0049 82

                                            cvethand ch frac14 cvTHORN

                                            frac14 0275

                                            Ur frac14 073 (from Figure 730)

                                            58 Consolidation theory

                                            Using Equation 740

                                            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                            U frac14 080

                                            710

                                            Without sand drains

                                            Uv frac14 090

                                            Tv frac14 0848

                                            t frac14 Tvd2

                                            cvfrac14 0848 102

                                            96frac14 88 years

                                            With sand drains

                                            R frac14 0564S frac14 0564 4 frac14 226m

                                            n frac14 Rrfrac14 226

                                            015frac14 15

                                            Tr

                                            Tvfrac14 chcv

                                            d2

                                            4R2ethsame tTHORN

                                            Tr

                                            Tvfrac14 140

                                            96 102

                                            4 2262frac14 714 eth1THORN

                                            Using Equation 740

                                            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                            An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                            Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                            040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                            Thus

                                            Uv frac14 0295 and Ur frac14 086

                                            t frac14 88 00683

                                            0848frac14 07 years

                                            Consolidation theory 59

                                            Chapter 8

                                            Bearing capacity

                                            81

                                            (a) The ultimate bearing capacity is given by Equation 83

                                            qf frac14 cNc thorn DNq thorn 1

                                            2BN

                                            For u frac14 0

                                            Nc frac14 514 Nq frac14 1 N frac14 0

                                            qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                            The net ultimate bearing capacity is

                                            qnf frac14 qf D frac14 540 kN=m2

                                            The net foundation pressure is

                                            qn frac14 q D frac14 425

                                            2 eth21 1THORN frac14 192 kN=m2

                                            The factor of safety (Equation 86) is

                                            F frac14 qnfqnfrac14 540

                                            192frac14 28

                                            (b) For 0 frac14 28

                                            Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                            qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                            2 112 2 13

                                            frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                            qnf frac14 574 112 frac14 563 kN=m2

                                            F frac14 563

                                            192frac14 29

                                            (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                            82

                                            For 0 frac14 38

                                            Nq frac14 49 N frac14 67

                                            qnf frac14 DethNq 1THORN thorn 1

                                            2BN ethfrom Equation 83THORN

                                            frac14 eth18 075 48THORN thorn 1

                                            2 18 15 67

                                            frac14 648thorn 905 frac14 1553 kN=m2

                                            qn frac14 500

                                            15 eth18 075THORN frac14 320 kN=m2

                                            F frac14 qnfqnfrac14 1553

                                            320frac14 48

                                            0d frac14 tan1tan 38

                                            125

                                            frac14 32 therefore Nq frac14 23 and N frac14 25

                                            Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                            2 18 15 25

                                            frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                            Design load (action) Vd frac14 500 kN=m

                                            The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                            83

                                            D

                                            Bfrac14 350

                                            225frac14 155

                                            From Figure 85 for a square foundation

                                            Nc frac14 81

                                            Bearing capacity 61

                                            For a rectangular foundation (L frac14 450m B frac14 225m)

                                            Nc frac14 084thorn 016B

                                            L

                                            81 frac14 745

                                            Using Equation 810

                                            qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                            For F frac14 3

                                            qn frac14 1006

                                            3frac14 335 kN=m2

                                            q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                            Design load frac14 405 450 225 frac14 4100 kN

                                            Design undrained strength cud frac14 135

                                            14frac14 96 kN=m2

                                            Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                            frac14 7241 kN

                                            Design load Vd frac14 4100 kN

                                            Rd gt Vd therefore the bearing resistance limit state is satisfied

                                            84

                                            For 0 frac14 40

                                            Nq frac14 64 N frac14 95

                                            qnf frac14 DethNq 1THORN thorn 04BN

                                            (a) Water table 5m below ground level

                                            qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                            qn frac14 400 17 frac14 383 kN=m2

                                            F frac14 2686

                                            383frac14 70

                                            (b) Water table 1m below ground level (ie at foundation level)

                                            0 frac14 20 98 frac14 102 kN=m3

                                            62 Bearing capacity

                                            qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                            F frac14 2040

                                            383frac14 53

                                            (c) Water table at ground level with upward hydraulic gradient 02

                                            eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                            qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                            F frac14 1296

                                            392frac14 33

                                            85

                                            The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                            Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                            Design value of 0 frac14 tan1tan 39

                                            125

                                            frac14 33

                                            For 0 frac14 33 Nq frac14 26 and N frac14 29

                                            Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                            Rd gt Vd therefore the bearing resistance limit state is satisfied

                                            86

                                            (a) Undrained shear for u frac14 0

                                            Nc frac14 514 Nq frac14 1 N frac14 0

                                            qnf frac14 12cuNc

                                            frac14 12 100 514 frac14 617 kN=m2

                                            qn frac14 qnfFfrac14 617

                                            3frac14 206 kN=m2

                                            q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                            Bearing capacity 63

                                            Drained shear for 0 frac14 32

                                            Nq frac14 23 N frac14 25

                                            0 frac14 21 98 frac14 112 kN=m3

                                            qnf frac14 0DethNq 1THORN thorn 040BN

                                            frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                            frac14 694 kN=m2

                                            q frac14 694

                                            3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                            Design load frac14 42 227 frac14 3632 kN

                                            (b) Design undrained strength cud frac14 100

                                            14frac14 71 kNm2

                                            Design bearing resistance Rd frac14 12cudNe area

                                            frac14 12 71 514 42

                                            frac14 7007 kN

                                            For drained shear 0d frac14 tan1tan 32

                                            125

                                            frac14 26

                                            Nq frac14 12 N frac14 10

                                            Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                            (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                            Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                            1 2 100 0175 0700qn 0182qn

                                            2 6 033 0044 0176qn 0046qn

                                            3 10 020 0017 0068qn 0018qn

                                            0246qn

                                            Diameter of equivalent circle B frac14 45m

                                            H

                                            Bfrac14 12

                                            45frac14 27 and A frac14 042

                                            13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                            64 Bearing capacity

                                            For sc frac14 30mm

                                            qn frac14 30

                                            0147frac14 204 kN=m2

                                            q frac14 204thorn 21 frac14 225 kN=m2

                                            Design load frac14 42 225 frac14 3600 kN

                                            The design load is 3600 kN settlement being the limiting criterion

                                            87

                                            D

                                            Bfrac14 8

                                            4frac14 20

                                            From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                            F frac14 cuNc

                                            Dfrac14 40 71

                                            20 8frac14 18

                                            88

                                            Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                            Design value of 0 frac14 tan1tan 38

                                            125

                                            frac14 32

                                            Figure Q86

                                            Bearing capacity 65

                                            For 0 frac14 32 Nq frac14 23 and N frac14 25

                                            Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                            The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                            Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                            For B frac14 250m qn frac14 3750

                                            2502 17 frac14 583 kN=m2

                                            From Figure 510 m frac14 n frac14 126

                                            6frac14 021

                                            Ir frac14 0019

                                            Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                            Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                            The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                            89

                                            Depth (m) N 0v (kNm2) CN N1

                                            070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                            Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                            (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                            Cw frac14 05thorn 0530

                                            47

                                            frac14 082

                                            66 Bearing capacity

                                            Thus

                                            qa frac14 150 082 frac14 120 kN=m2

                                            (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                            Thus

                                            qa frac14 90 15 frac14 135 kN=m2

                                            (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                            Ic frac14 171

                                            1014frac14 0068

                                            From Equation 819(a) with s frac14 25mm

                                            q frac14 25

                                            3507 0068frac14 150 kN=m2

                                            810

                                            Peak value of strain influence factor occurs at a depth of 27m and is given by

                                            Izp frac14 05thorn 01130

                                            16 27

                                            05

                                            frac14 067

                                            Refer to Figure Q810

                                            E frac14 25qc

                                            Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                            Ez (mm3MN)

                                            1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                            0203

                                            C1 frac14 1 0500qnfrac14 1 05 12 16

                                            130frac14 093

                                            C2 frac14 1 ethsayTHORN

                                            s frac14 C1C2qnX Iz

                                            Ez frac14 093 1 130 0203 frac14 25mm

                                            Bearing capacity 67

                                            811

                                            At pile base level

                                            cu frac14 220 kN=m2

                                            qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                            Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                            00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                            qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                            Then

                                            Qf frac14 Abqb thorn Asqs

                                            frac14

                                            4 32 1980

                                            thorn eth 105 139 86THORN

                                            frac14 13 996thorn 3941 frac14 17 937 kN

                                            0 01 02 03 04 05 06 07

                                            0 2 4 6 8 10 12 14

                                            1

                                            2

                                            3

                                            4

                                            5

                                            6

                                            7

                                            8

                                            (1)

                                            (2)

                                            (3)

                                            (4)

                                            (5)

                                            qc

                                            qc

                                            Iz

                                            Iz

                                            (MNm2)

                                            z (m)

                                            Figure Q810

                                            68 Bearing capacity

                                            Allowable load

                                            ethaTHORN Qf

                                            2frac14 17 937

                                            2frac14 8968 kN

                                            ethbTHORN Abqb

                                            3thorn Asqs frac14 13 996

                                            3thorn 3941 frac14 8606 kN

                                            ie allowable load frac14 8600 kN

                                            Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                            According to the limit state method

                                            Characteristic undrained strength at base level cuk frac14 220

                                            150kN=m2

                                            Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                            150frac14 1320 kN=m2

                                            Characteristic shaft resistance qsk frac14 00150

                                            frac14 86

                                            150frac14 57 kN=m2

                                            Characteristic base and shaft resistances

                                            Rbk frac14

                                            4 32 1320 frac14 9330 kN

                                            Rsk frac14 105 139 86

                                            150frac14 2629 kN

                                            For a bored pile the partial factors are b frac14 160 and s frac14 130

                                            Design bearing resistance Rcd frac14 9330

                                            160thorn 2629

                                            130

                                            frac14 5831thorn 2022

                                            frac14 7850 kN

                                            Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                            812

                                            ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                            qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                            For a single pile

                                            Qf frac14 Abqb thorn Asqs

                                            frac14

                                            4 062 1305

                                            thorn eth 06 15 42THORN

                                            frac14 369thorn 1187 frac14 1556 kN

                                            Bearing capacity 69

                                            Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                            eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                            (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                            (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                            qbkfrac14 9cuk frac14 9 220

                                            150frac14 1320 kN=m2

                                            qskfrac14cuk frac14 040 105

                                            150frac14 28 kN=m2

                                            Rbkfrac14

                                            4 0602 1320 frac14 373 kN

                                            Rskfrac14 060 15 28 frac14 791 kN

                                            Rcdfrac14 373

                                            160thorn 791

                                            130frac14 233thorn 608 frac14 841 kN

                                            Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                            Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                            (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                            q frac14 21 000

                                            1762frac14 68 kN=m2

                                            Immediate settlement

                                            H

                                            Bfrac14 15

                                            176frac14 085

                                            D

                                            Bfrac14 13

                                            176frac14 074

                                            L

                                            Bfrac14 1

                                            Hence from Figure 515

                                            130 frac14 078 and 131 frac14 041

                                            70 Bearing capacity

                                            Thus using Equation 528

                                            si frac14 078 041 68 176

                                            65frac14 6mm

                                            Consolidation settlement

                                            Layer z (m) Area (m2) (kNm2) mvH (mm)

                                            1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                            434 (sod)

                                            Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                            sc frac14 056 434 frac14 24mm

                                            The total settlement is (6thorn 24) frac14 30mm

                                            813

                                            At base level N frac14 26 Then using Equation 830

                                            qb frac14 40NDb

                                            Bfrac14 40 26 2

                                            025frac14 8320 kN=m2

                                            ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                            Figure Q812

                                            Bearing capacity 71

                                            Over the length embedded in sand

                                            N frac14 21 ie18thorn 24

                                            2

                                            Using Equation 831

                                            qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                            For a single pile

                                            Qf frac14 Abqb thorn Asqs

                                            frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                            For the pile group assuming a group efficiency of 12

                                            XQf frac14 12 9 604 frac14 6523 kN

                                            Then the load factor is

                                            F frac14 6523

                                            2000thorn 1000frac14 21

                                            (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                            Characteristic base resistance per unit area qbk frac14 8320

                                            150frac14 5547 kNm2

                                            Characteristic shaft resistance per unit area qsk frac14 42

                                            150frac14 28 kNm2

                                            Characteristic base and shaft resistances for a single pile

                                            Rbk frac14 0252 5547 frac14 347 kN

                                            Rsk frac14 4 025 2 28 frac14 56 kN

                                            For a driven pile the partial factors are b frac14 s frac14 130

                                            Design bearing resistance Rcd frac14 347

                                            130thorn 56

                                            130frac14 310 kN

                                            For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                            Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                            (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                            From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                            72 Bearing capacity

                                            N frac14 24thorn 26thorn 34

                                            3frac14 28

                                            Ic frac14 171

                                            2814frac14 0016 ethEquation 818THORN

                                            s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                            814

                                            Using Equation 841

                                            Tf frac14 DLcu thorn

                                            4ethD2 d2THORNcuNc

                                            frac14 eth 02 5 06 110THORN thorn

                                            4eth022 012THORN110 9

                                            frac14 207thorn 23 frac14 230 kN

                                            Figure Q813

                                            Bearing capacity 73

                                            Chapter 9

                                            Stability of slopes

                                            91

                                            Referring to Figure Q91

                                            W frac14 417 19 frac14 792 kN=m

                                            Q frac14 20 28 frac14 56 kN=m

                                            Arc lengthAB frac14

                                            180 73 90 frac14 115m

                                            Arc length BC frac14

                                            180 28 90 frac14 44m

                                            The factor of safety is given by

                                            F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                            90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                            Depth of tension crack z0 frac14 2cu

                                            frac14 2 20

                                            19frac14 21m

                                            Arc length BD frac14

                                            180 13

                                            1

                                            2 90 frac14 21m

                                            F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                            The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                            Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                            14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                            Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                            The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                            92

                                            u frac14 0

                                            Depth factor D frac14 11

                                            9frac14 122

                                            Using Equation 92 with F frac14 10

                                            Ns frac14 cu

                                            FHfrac14 30

                                            10 19 9frac14 0175

                                            Hence from Figure 93

                                            frac14 50

                                            For F frac14 12

                                            Ns frac14 30

                                            12 19 9frac14 0146

                                            frac14 27

                                            93

                                            Refer to Figure Q93

                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                            1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                            74 m

                                            214 1deg

                                            213 1deg

                                            39 m

                                            WB

                                            D

                                            C

                                            28 m

                                            21 m

                                            A

                                            Q

                                            Soil (1)Soil (2)

                                            73deg

                                            Figure Q91

                                            Stability of slopes 75

                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                            9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                            599 256 328 1372

                                            Figure Q93

                                            76 Stability of slopes

                                            XW cos frac14 b

                                            Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                            W sin frac14 bX

                                            h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                            Arc length La frac14

                                            180 57

                                            1

                                            2 326 frac14 327m

                                            The factor of safety is given by

                                            F frac14 c0La thorn tan0ethW cos ulTHORN

                                            W sin

                                            frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                            frac14 091

                                            According to the limit state method

                                            0d frac14 tan1tan 32

                                            125

                                            frac14 265

                                            c0 frac14 8

                                            160frac14 5 kN=m2

                                            Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                            Design disturbing moment frac14 1075 kN=m

                                            The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                            94

                                            F frac14 1

                                            W sin

                                            Xfc0bthorn ethW ubTHORN tan0g sec

                                            1thorn ethtan tan0=FTHORN

                                            c0 frac14 8 kN=m2

                                            0 frac14 32

                                            c0b frac14 8 2 frac14 16 kN=m

                                            W frac14 bh frac14 21 2 h frac14 42h kN=m

                                            Try F frac14 100

                                            tan0

                                            Ffrac14 0625

                                            Stability of slopes 77

                                            Values of u are as obtained in Figure Q93

                                            SliceNo

                                            h(m)

                                            W frac14 bh(kNm)

                                            W sin(kNm)

                                            ub(kNm)

                                            c0bthorn (W ub) tan0(kNm)

                                            sec

                                            1thorn (tan tan0)FProduct(kNm)

                                            1 05 21 6 2 8 24 1078 262 13 55 31

                                            23 33 30 1042 31

                                            3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                            224 92 72 0931 67

                                            6 50 210 11 40 100 85 0907 777 55 231 14

                                            12 58 112 90 0889 80

                                            8 60 252 1812

                                            80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                            10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                            2154 88 116 0853 99

                                            14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                            1074 1091

                                            F frac14 1091

                                            1074frac14 102 (assumed value 100)

                                            Thus

                                            F frac14 101

                                            95

                                            F frac14 1

                                            W sin

                                            XfWeth1 ruTHORN tan0g sec

                                            1thorn ethtan tan0THORN=F

                                            0 frac14 33

                                            ru frac14 020

                                            W frac14 bh frac14 20 5 h frac14 100h kN=m

                                            eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                            Try F frac14 110

                                            tan 0

                                            Ffrac14 tan 33

                                            110frac14 0590

                                            78 Stability of slopes

                                            Referring to Figure Q95

                                            SliceNo

                                            h(m)

                                            W frac14 bh(kNm)

                                            W sin(kNm)

                                            W(1 ru) tan0(kNm)

                                            sec

                                            1thorn ( tan tan0)FProduct(kNm)

                                            1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                            2120 234 0892 209

                                            4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                            1185 1271

                                            Figure Q95

                                            Stability of slopes 79

                                            F frac14 1271

                                            1185frac14 107

                                            The trial value was 110 therefore take F to be 108

                                            96

                                            (a) Water table at surface the factor of safety is given by Equation 912

                                            F frac14 0

                                            sat

                                            tan0

                                            tan

                                            ptie 15 frac14 92

                                            19

                                            tan 36

                                            tan

                                            tan frac14 0234

                                            frac14 13

                                            Water table well below surface the factor of safety is given by Equation 911

                                            F frac14 tan0

                                            tan

                                            frac14 tan 36

                                            tan 13

                                            frac14 31

                                            (b) 0d frac14 tan1tan 36

                                            125

                                            frac14 30

                                            Depth of potential failure surface frac14 z

                                            Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                            frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                            frac14 504z kN

                                            Design disturbing moment per unit area Sd frac14 sat sin cos

                                            frac14 19 z sin 13 cos 13

                                            frac14 416z kN

                                            Rd gtSd therefore the limit state for overall stability is satisfied

                                            80 Stability of slopes

                                            • Book Cover
                                            • Title
                                            • Contents
                                            • Basic characteristics of soils
                                            • Seepage
                                            • Effective stress
                                            • Shear strength
                                            • Stresses and displacements
                                            • Lateral earth pressure
                                            • Consolidation theory
                                            • Bearing capacity
                                            • Stability of slopes

                                              (a) Immediately after WT rise

                                              At 8m depth pore water pressure is governed by the new WT level because thepermeability of the sand is high

                                              0v frac14 eth3 16THORN thorn eth5 92THORN frac14 940 kN=m2

                                              At 12m depth pore water pressure is governed by the old WT level because thepermeability of the clay is very low (However there will be an increase in total stressof 9 kNm2 due to the increase in unit weight from 16 to 19 kNm2 between 3 and 6mdepth this is accompanied by an immediate increase of 9 kNm2 in pore waterpressure)

                                              0v frac14 eth6 16THORN thorn eth3 92THORN thorn eth3 102THORN frac14 1542 kN=m2

                                              (b) Several years after WT rise

                                              At both depths pore water pressure is governed by the newWT level it being assumedthat swelling of the clay is complete

                                              At 8m depth

                                              0v frac14 940 kN=m2 (as above)

                                              At 12m depth

                                              0v frac14 eth3 16THORN thorn eth6 92THORN thorn eth3 102THORN frac14 1338 kN=m2

                                              Figure Q35

                                              Effective stress 17

                                              36

                                              Total weight

                                              ab frac14 210 kN

                                              Effective weight

                                              ac frac14 112 kN

                                              Resultant boundary water force

                                              be frac14 119 kN

                                              Seepage force

                                              ce frac14 34 kN

                                              Resultant body force

                                              ae frac14 99 kN eth73 to horizontalTHORN

                                              (Refer to Figure Q36)

                                              Figure Q36

                                              18 Effective stress

                                              37

                                              Situation (1)(a)

                                              frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                                              u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                              0 frac14 u frac14 694 392 frac14 302 kN=m2

                                              (b)

                                              i frac14 2

                                              4frac14 05

                                              j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                                              Situation (2)(a)

                                              frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                                              u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                              0 frac14 u frac14 498 392 frac14 106 kN=m2

                                              (b)

                                              i frac14 2

                                              4frac14 05

                                              j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                                              38

                                              The flow net is drawn in Figure Q24

                                              Loss in total head between adjacent equipotentials

                                              h frac14 550

                                              Ndfrac14 550

                                              11frac14 050m

                                              Exit hydraulic gradient

                                              ie frac14 h

                                              sfrac14 050

                                              070frac14 071

                                              Effective stress 19

                                              The critical hydraulic gradient is given by Equation 39

                                              ic frac14 0

                                              wfrac14 102

                                              98frac14 104

                                              Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                                              F frac14 iciefrac14 104

                                              071frac14 15

                                              Total head at C

                                              hC frac14 nd

                                              Ndh frac14 24

                                              11 550 frac14 120m

                                              Elevation head at C

                                              zC frac14 250m

                                              Pore water pressure at C

                                              uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                                              Therefore effective vertical stress at C

                                              0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                                              For point D

                                              hD frac14 73

                                              11 550 frac14 365m

                                              zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                                              0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                                              39

                                              The flow net is drawn in Figure Q25

                                              For a soil prism 150 300m adjacent to the piling

                                              hm frac14 26

                                              9 500 frac14 145m

                                              20 Effective stress

                                              Factor of safety against lsquoheavingrsquo (Equation 310)

                                              F frac14 ic

                                              imfrac14 0d

                                              whmfrac14 97 300

                                              98 145frac14 20

                                              With a filter

                                              F frac14 0d thorn wwhm

                                              3 frac14 eth97 300THORN thorn w98 145

                                              w frac14 135 kN=m2

                                              Depth of filterfrac14 13521frac14 065m (if above water level)

                                              Effective stress 21

                                              Chapter 4

                                              Shear strength

                                              41

                                              frac14 295 kN=m2

                                              u frac14 120 kN=m2

                                              0 frac14 u frac14 295 120 frac14 175 kN=m2

                                              f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                                              42

                                              03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                                              100 452 552200 908 1108400 1810 2210800 3624 4424

                                              The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                                              Figure Q42

                                              43

                                              3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                                              200 222 422400 218 618600 220 820

                                              The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                                              44

                                              The modified shear strength parameters are

                                              0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                                              a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                                              The coordinates of the stress point representing failure conditions in the test are

                                              1

                                              2eth1 2THORN frac14 1

                                              2 170 frac14 85 kN=m2

                                              1

                                              2eth1 thorn 3THORN frac14 1

                                              2eth270thorn 100THORN frac14 185 kN=m2

                                              The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                                              uf frac14 36 kN=m2

                                              Figure Q43

                                              Figure Q44

                                              Shear strength 23

                                              45

                                              3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                              150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                              The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                              The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                              46

                                              03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                              200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                              The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                              Figure Q45

                                              24 Shear strength

                                              47

                                              The torque required to produce shear failure is given by

                                              T frac14 dh cud

                                              2thorn 2

                                              Z d=2

                                              0

                                              2r drcur

                                              frac14 cud2h

                                              2thorn 4cu

                                              Z d=2

                                              0

                                              r2dr

                                              frac14 cud2h

                                              2thorn d

                                              3

                                              6

                                              Then

                                              35 frac14 cu52 10

                                              2thorn 53

                                              6

                                              103

                                              cu frac14 76 kN=m3

                                              400

                                              0 400 800 1200 1600

                                              τ (k

                                              Nm

                                              2 )

                                              σprime (kNm2)

                                              34deg

                                              315deg29deg

                                              (a)

                                              (b)

                                              0 400

                                              400

                                              800 1200 1600

                                              Failure envelope

                                              300 500

                                              σprime (kNm2)

                                              τ (k

                                              Nm

                                              2 )

                                              20 (kNm2)

                                              31deg

                                              Figure Q46

                                              Shear strength 25

                                              48

                                              The relevant stress values are calculated as follows

                                              3 frac14 600 kN=m2

                                              1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                              2(1 3) 0 40 79 107 139 159

                                              1

                                              2(01 thorn 03) 400 411 402 389 351 326

                                              1

                                              2(1 thorn 3) 600 640 679 707 739 759

                                              The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                              Af frac14 433 200

                                              319frac14 073

                                              49

                                              B frac14 u33

                                              frac14 144

                                              350 200frac14 096

                                              a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                              0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                              10 283 65 023

                                              Figure Q48

                                              26 Shear strength

                                              The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                              Figure Q49

                                              Shear strength 27

                                              Chapter 5

                                              Stresses and displacements

                                              51

                                              Vertical stress is given by

                                              z frac14 Qz2Ip frac14 5000

                                              52Ip

                                              Values of Ip are obtained from Table 51

                                              r (m) rz Ip z (kNm2)

                                              0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                              10 20 0009 2

                                              The variation of z with radial distance (r) is plotted in Figure Q51

                                              Figure Q51

                                              52

                                              Below the centre load (Figure Q52)

                                              r

                                              zfrac14 0 for the 7500-kN load

                                              Ip frac14 0478

                                              r

                                              zfrac14 5

                                              4frac14 125 for the 10 000- and 9000-kN loads

                                              Ip frac14 0045

                                              Then

                                              z frac14X Q

                                              z2Ip

                                              frac14 7500 0478

                                              42thorn 10 000 0045

                                              42thorn 9000 0045

                                              42

                                              frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                              53

                                              The vertical stress under a corner of a rectangular area is given by

                                              z frac14 qIr

                                              where values of Ir are obtained from Figure 510 In this case

                                              z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                              z

                                              Figure Q52

                                              Stresses and displacements 29

                                              z (m) m n Ir z (kNm2)

                                              0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                              10 010 0005 5

                                              z is plotted against z in Figure Q53

                                              54

                                              (a)

                                              m frac14 125

                                              12frac14 104

                                              n frac14 18

                                              12frac14 150

                                              From Figure 510 Irfrac14 0196

                                              z frac14 2 175 0196 frac14 68 kN=m2

                                              Figure Q53

                                              30 Stresses and displacements

                                              (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                              z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                              55

                                              Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                              Px frac14 2Q

                                              1

                                              m2 thorn 1frac14 2 150

                                              125frac14 76 kN=m

                                              Equation 517 is used to obtain the pressure distribution

                                              px frac14 4Q

                                              h

                                              m2n

                                              ethm2 thorn n2THORN2 frac14150

                                              m2n

                                              ethm2 thorn n2THORN2 ethkN=m2THORN

                                              Figure Q54

                                              Stresses and displacements 31

                                              n m2n

                                              (m2 thorn n2)2

                                              px(kNm2)

                                              0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                              The pressure distribution is plotted in Figure Q55

                                              56

                                              H

                                              Bfrac14 10

                                              2frac14 5

                                              L

                                              Bfrac14 4

                                              2frac14 2

                                              D

                                              Bfrac14 1

                                              2frac14 05

                                              Hence from Figure 515

                                              131 frac14 082

                                              130 frac14 094

                                              Figure Q55

                                              32 Stresses and displacements

                                              The immediate settlement is given by Equation 528

                                              si frac14 130131qB

                                              Eu

                                              frac14 094 082 200 2

                                              45frac14 7mm

                                              Stresses and displacements 33

                                              Chapter 6

                                              Lateral earth pressure

                                              61

                                              For 0 frac14 37 the active pressure coefficient is given by

                                              Ka frac14 1 sin 37

                                              1thorn sin 37frac14 025

                                              The total active thrust (Equation 66a with c0 frac14 0) is

                                              Pa frac14 1

                                              2KaH

                                              2 frac14 1

                                              2 025 17 62 frac14 765 kN=m

                                              If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                              K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                              and the thrust on the wall is

                                              P0 frac14 1

                                              2K0H

                                              2 frac14 1

                                              2 040 17 62 frac14 122 kN=m

                                              62

                                              The active pressure coefficients for the three soil types are as follows

                                              Ka1 frac141 sin 35

                                              1thorn sin 35frac14 0271

                                              Ka2 frac141 sin 27

                                              1thorn sin 27frac14 0375

                                              ffiffiffiffiffiffiffiKa2

                                              p frac14 0613

                                              Ka3 frac141 sin 42

                                              1thorn sin 42frac14 0198

                                              Distribution of active pressure (plotted in Figure Q62)

                                              Depth (m) Soil Active pressure (kNm2)

                                              3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                              12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                              At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                              Total thrust frac14 571 kNm

                                              Point of application is (4893571) m from the top of the wall ie 857m

                                              Force (kN) Arm (m) Moment (kN m)

                                              (1)1

                                              2 0271 16 32 frac14 195 20 390

                                              (2) 0271 16 3 2 frac14 260 40 1040

                                              (3)1

                                              2 0271 92 22 frac14 50 433 217

                                              (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                              (5)1

                                              2 0375 102 32 frac14 172 70 1204

                                              (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                              (7)1

                                              2 0198 112 42 frac14 177 1067 1889

                                              (8)1

                                              2 98 92 frac14 3969 90 35721

                                              5713 48934

                                              Figure Q62

                                              Lateral earth pressure 35

                                              63

                                              (a) For u frac14 0 Ka frac14 Kp frac14 1

                                              Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                              frac14 245

                                              At the lower end of the piling

                                              pa frac14 Kaqthorn Kasatz Kaccu

                                              frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                              frac14 115 kN=m2

                                              pp frac14 Kpsatzthorn Kpccu

                                              frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                              frac14 202 kN=m2

                                              (b) For 0 frac14 26 and frac14 1

                                              20

                                              Ka frac14 035

                                              Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                              pfrac14 145 ethEquation 619THORN

                                              Kp frac14 37

                                              Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                              pfrac14 47 ethEquation 624THORN

                                              At the lower end of the piling

                                              pa frac14 Kaqthorn Ka0z Kacc

                                              0

                                              frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                              frac14 187 kN=m2

                                              pp frac14 Kp0zthorn Kpcc

                                              0

                                              frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                              frac14 198 kN=m2

                                              36 Lateral earth pressure

                                              64

                                              (a) For 0 frac14 38 Ka frac14 024

                                              0 frac14 20 98 frac14 102 kN=m3

                                              The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                              Force (kN) Arm (m) Moment (kN m)

                                              (1) 024 10 66 frac14 159 33 525

                                              (2)1

                                              2 024 17 392 frac14 310 400 1240

                                              (3) 024 17 39 27 frac14 430 135 580

                                              (4)1

                                              2 024 102 272 frac14 89 090 80

                                              (5)1

                                              2 98 272 frac14 357 090 321

                                              Hfrac14 1345 MH frac14 2746

                                              (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                              (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                              XM frac14MV MH frac14 7790 kNm

                                              Lever arm of base resultant

                                              M

                                              Vfrac14 779

                                              488frac14 160

                                              Eccentricity of base resultant

                                              e frac14 200 160 frac14 040m

                                              39 m

                                              27 m

                                              40 m

                                              04 m

                                              04 m

                                              26 m

                                              (7)

                                              (9)

                                              (1)(2)

                                              (3)

                                              (4)

                                              (5)

                                              (8)(6)

                                              (10)

                                              WT

                                              10 kNm2

                                              Hydrostatic

                                              Figure Q64

                                              Lateral earth pressure 37

                                              Base pressures (Equation 627)

                                              p frac14 VB

                                              1 6e

                                              B

                                              frac14 488

                                              4eth1 060THORN

                                              frac14 195 kN=m2 and 49 kN=m2

                                              Factor of safety against sliding (Equation 628)

                                              F frac14 V tan

                                              Hfrac14 488 tan 25

                                              1345frac14 17

                                              (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                              Hfrac14 1633 kN

                                              V frac14 4879 kN

                                              MH frac14 3453 kNm

                                              MV frac14 10536 kNm

                                              The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                              65

                                              For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                              Kp

                                              Ffrac14 385

                                              2

                                              0 frac14 20 98 frac14 102 kN=m3

                                              The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                              Force (kN) Arm (m) Moment (kN m)

                                              (1)1

                                              2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                              (2) 026 17 45 d frac14 199d d2 995d2

                                              (3)1

                                              2 026 102 d2 frac14 133d2 d3 044d3

                                              (4)1

                                              2 385

                                              2 17 152 frac14 368 dthorn 05 368d 184

                                              (5)385

                                              2 17 15 d frac14 491d d2 2455d2

                                              (6)1

                                              2 385

                                              2 102 d2 frac14 982d2 d3 327d3

                                              38 Lateral earth pressure

                                              XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                              d3 thorn 516d2 283d 1724 frac14 0

                                              d frac14 179m

                                              Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                              Over additional 20 embedded depth

                                              pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                              Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                              66

                                              The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                              Ka frac14 sin 69=sin 105

                                              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                              pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                              26664

                                              37775

                                              2

                                              frac14 050

                                              The total active thrust (acting at 25 above the normal) is given by Equation 616

                                              Pa frac14 1

                                              2 050 19 7502 frac14 267 kN=m

                                              Figure Q65

                                              Lateral earth pressure 39

                                              Horizontal component

                                              Ph frac14 267 cos 40 frac14 205 kN=m

                                              Vertical component

                                              Pv frac14 267 sin 40 frac14 172 kN=m

                                              Consider moments about the toe of the wall (Figure Q66) (per m)

                                              Force (kN) Arm (m) Moment (kN m)

                                              (1)1

                                              2 175 650 235 frac14 1337 258 345

                                              (2) 050 650 235 frac14 764 175 134

                                              (3)1

                                              2 070 650 235 frac14 535 127 68

                                              (4) 100 400 235 frac14 940 200 188

                                              (5) 1

                                              2 080 050 235 frac14 47 027 1

                                              Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                              Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                              Lever arm of base resultant

                                              M

                                              Vfrac14 795

                                              525frac14 151m

                                              Eccentricity of base resultant

                                              e frac14 200 151 frac14 049m

                                              Figure Q66

                                              40 Lateral earth pressure

                                              Base pressures (Equation 627)

                                              p frac14 525

                                              41 6 049

                                              4

                                              frac14 228 kN=m2 and 35 kN=m2

                                              The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                              The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                              The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                              67

                                              For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                              Force (kN) Arm (m) Moment (kNm)

                                              (1)1

                                              2 027 17 52 frac14 574 183 1050

                                              (2) 027 17 5 3 frac14 689 500 3445

                                              (3)1

                                              2 027 102 32 frac14 124 550 682

                                              (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                              (5)1

                                              2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                              (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                              (7) 1

                                              2 267

                                              2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                              (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                              2 d frac14 163d d2thorn 650 82d2 1060d

                                              Tie rod force per m frac14 T 0 0

                                              XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                              d3 thorn 77d2 269d 1438 frac14 0

                                              d frac14 467m

                                              Depth of penetration frac14 12d frac14 560m

                                              Lateral earth pressure 41

                                              Algebraic sum of forces for d frac14 467m isX

                                              F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                              T frac14 905 kN=m

                                              Force in each tie rod frac14 25T frac14 226 kN

                                              68

                                              (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                              0 frac14 21 98 frac14 112 kN=m3

                                              The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                              uC frac14 150

                                              165 15 98 frac14 134 kN=m2

                                              The average seepage pressure is

                                              j frac14 15

                                              165 98 frac14 09 kN=m3

                                              Hence

                                              0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                              0 j frac14 112 09 frac14 103 kN=m3

                                              Figure Q67

                                              42 Lateral earth pressure

                                              Consider moments about the anchor point A (per m)

                                              Force (kN) Arm (m) Moment (kN m)

                                              (1) 10 026 150 frac14 390 60 2340

                                              (2)1

                                              2 026 18 452 frac14 474 15 711

                                              (3) 026 18 45 105 frac14 2211 825 18240

                                              (4)1

                                              2 026 121 1052 frac14 1734 100 17340

                                              (5)1

                                              2 134 15 frac14 101 40 404

                                              (6) 134 30 frac14 402 60 2412

                                              (7)1

                                              2 134 60 frac14 402 95 3819

                                              571 4527(8) Ppm

                                              115 115PPm

                                              XM frac14 0

                                              Ppm frac144527

                                              115frac14 394 kN=m

                                              Available passive resistance

                                              Pp frac14 1

                                              2 385 103 62 frac14 714 kN=m

                                              Factor of safety

                                              Fp frac14 Pp

                                              Ppm

                                              frac14 714

                                              394frac14 18

                                              Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                              Figure Q68

                                              Lateral earth pressure 43

                                              (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                              Consider moments (per m) about the tie point A

                                              Force (kN) Arm (m)

                                              (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                              (2)1

                                              2 033 18 452 frac14 601 15

                                              (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                              (4)1

                                              2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                              (5)1

                                              2 134 15 frac14 101 40

                                              (6) 134 30 frac14 402 60

                                              (7)1

                                              2 134 d frac14 67d d3thorn 75

                                              (8) 1

                                              2 30 103 d2 frac141545d2 2d3thorn 75

                                              Moment (kN m)

                                              (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                              XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                              d3 thorn 827d2 466d 1518 frac14 0

                                              By trial

                                              d frac14 544m

                                              The minimum depth of embedment required is 544m

                                              69

                                              For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                              0 frac14 20 98 frac14 102 kN=m3

                                              The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                              44 Lateral earth pressure

                                              uC frac14 147

                                              173 26 98 frac14 216 kN=m2

                                              and the average seepage pressure around the wall is

                                              j frac14 26

                                              173 98 frac14 15 kN=m3

                                              Consider moments about the prop (A) (per m)

                                              Force (kN) Arm (m) Moment (kN m)

                                              (1)1

                                              2 03 17 272 frac14 186 020 37

                                              (2) 03 17 27 53 frac14 730 335 2445

                                              (3)1

                                              2 03 (102thorn 15) 532 frac14 493 423 2085

                                              (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                              (5)1

                                              2 216 26 frac14 281 243 684

                                              (6) 216 27 frac14 583 465 2712

                                              (7)1

                                              2 216 60 frac14 648 800 5184

                                              3055(8)

                                              1

                                              2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                              Factor of safety

                                              Fr frac14 6885

                                              3055frac14 225

                                              Figure Q69

                                              Lateral earth pressure 45

                                              610

                                              For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                              p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                              Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                              Using the recommendations of Twine and Roscoe

                                              p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                              Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                              611

                                              frac14 18 kN=m3 0 frac14 34

                                              H frac14 350m nH frac14 335m mH frac14 185m

                                              Consider a trial value of F frac14 20 Refer to Figure 635

                                              0m frac14 tan1tan 34

                                              20

                                              frac14 186

                                              Then

                                              frac14 45 thorn 0m2frac14 543

                                              W frac14 1

                                              2 18 3502 cot 543 frac14 792 kN=m

                                              Figure Q610

                                              46 Lateral earth pressure

                                              P frac14 1

                                              2 s 3352 frac14 561s kN=m

                                              U frac14 1

                                              2 98 1852 cosec 543 frac14 206 kN=m

                                              Equations 630 and 631 then become

                                              561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                              792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                              ie

                                              561s 0616N 405 frac14 0

                                              792 0857N thorn 563 frac14 0

                                              N frac14 848

                                              0857frac14 989 kN=m

                                              Then

                                              561s 609 405 frac14 0

                                              s frac14 649

                                              561frac14 116 kN=m3

                                              The calculations for trial values of F of 20 15 and 10 are summarized below

                                              F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                              20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                              s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                              Figure Q611

                                              Lateral earth pressure 47

                                              612

                                              For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                              45 thorn 0

                                              2frac14 63

                                              For the retained material between the surface and a depth of 36m

                                              Pa frac14 1

                                              2 030 18 362 frac14 350 kN=m

                                              Weight of reinforced fill between the surface and a depth of 36m is

                                              Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                              eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                              Eccentricity of Rv

                                              e frac14 263 250 frac14 013m

                                              The average vertical stress at a depth of 36m is

                                              z frac14 Rv

                                              L 2efrac14 324

                                              474frac14 68 kN=m2

                                              (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                              Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                              Tensile stress in the element frac14 138 103

                                              65 3frac14 71N=mm2

                                              Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                              Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                              Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                              The weight of ABC is

                                              W frac14 1

                                              2 18 52 265 frac14 124 kN=m

                                              From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                              48 Lateral earth pressure

                                              (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                              Tp frac14 032 68 120 065 frac14 170 kN

                                              Tr frac14 213 420

                                              418frac14 214 kN

                                              Again the tensile failure and slipping limit states are satisfied for this element

                                              Figure Q612

                                              Lateral earth pressure 49

                                              Chapter 7

                                              Consolidation theory

                                              71

                                              Total change in thickness

                                              H frac14 782 602 frac14 180mm

                                              Average thickness frac14 1530thorn 180

                                              2frac14 1620mm

                                              Length of drainage path d frac14 1620

                                              2frac14 810mm

                                              Root time plot (Figure Q71a)

                                              ffiffiffiffiffiffit90p frac14 33

                                              t90 frac14 109min

                                              cv frac14 0848d2

                                              t90frac14 0848 8102

                                              109 1440 365

                                              106frac14 27m2=year

                                              r0 frac14 782 764

                                              782 602frac14 018

                                              180frac14 0100

                                              rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                              10 119

                                              9 180frac14 0735

                                              rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                              Log time plot (Figure Q71b)

                                              t50 frac14 26min

                                              cv frac14 0196d2

                                              t50frac14 0196 8102

                                              26 1440 365

                                              106frac14 26m2=year

                                              r0 frac14 782 763

                                              782 602frac14 019

                                              180frac14 0106

                                              rp frac14 763 623

                                              782 602frac14 140

                                              180frac14 0778

                                              rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                              Figure Q71(a)

                                              Figure Q71(b)

                                              Final void ratio

                                              e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                              e

                                              Hfrac14 1thorn e0

                                              H0frac14 1thorn e1 thorne

                                              H0

                                              ie

                                              e

                                              180frac14 1631thorne

                                              1710

                                              e frac14 2936

                                              1530frac14 0192

                                              Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                              mv frac14 1

                                              1thorn e0 e0 e101 00

                                              frac14 1

                                              1823 0192

                                              0107frac14 098m2=MN

                                              k frac14 cvmvw frac14 265 098 98

                                              60 1440 365 103frac14 81 1010 m=s

                                              72

                                              Using Equation 77 (one-dimensional method)

                                              sc frac14 e0 e11thorn e0 H

                                              Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                              Figure Q72

                                              52 Consolidation theory

                                              Settlement

                                              Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                              1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                              318

                                              Notes 5 92y 460thorn 84

                                              Heave

                                              Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                              1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                              38

                                              73

                                              U frac14 f ethTvTHORN frac14 f cvt

                                              d2

                                              Hence if cv is constant

                                              t1

                                              t2frac14 d

                                              21

                                              d22

                                              where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                              d1 frac14 95mm and d2 frac14 2500mm

                                              for U frac14 050 t2 frac14 t1 d22

                                              d21

                                              frac14 20

                                              60 24 365 25002

                                              952frac14 263 years

                                              for U lt 060 Tv frac14

                                              4U2 (Equation 724(a))

                                              t030 frac14 t050 0302

                                              0502

                                              frac14 263 036 frac14 095 years

                                              Consolidation theory 53

                                              74

                                              The layer is open

                                              d frac14 8

                                              2frac14 4m

                                              Tv frac14 cvtd2frac14 24 3

                                              42frac14 0450

                                              ui frac14 frac14 84 kN=m2

                                              The excess pore water pressure is given by Equation 721

                                              ue frac14Xmfrac141mfrac140

                                              2ui

                                              Msin

                                              Mz

                                              d

                                              expethM2TvTHORN

                                              In this case z frac14 d

                                              sinMz

                                              d

                                              frac14 sinM

                                              where

                                              M frac14

                                              23

                                              25

                                              2

                                              M sin M M2Tv exp (M2Tv)

                                              2thorn1 1110 0329

                                              3

                                              21 9993 457 105

                                              ue frac14 2 84 2

                                              1 0329 ethother terms negligibleTHORN

                                              frac14 352 kN=m2

                                              75

                                              The layer is open

                                              d frac14 6

                                              2frac14 3m

                                              Tv frac14 cvtd2frac14 10 3

                                              32frac14 0333

                                              The layer thickness will be divided into six equal parts ie m frac14 6

                                              54 Consolidation theory

                                              For an open layer

                                              Tv frac14 4n

                                              m2

                                              n frac14 0333 62

                                              4frac14 300

                                              The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                              ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                              i j

                                              0 1 2 3 4 5 6 7 8 9 10 11 12

                                              0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                              The initial and 3-year isochrones are plotted in Figure Q75

                                              Area under initial isochrone frac14 180 units

                                              Area under 3-year isochrone frac14 63 units

                                              The average degree of consolidation is given by Equation 725Thus

                                              U frac14 1 63

                                              180frac14 065

                                              Figure Q75

                                              Consolidation theory 55

                                              76

                                              At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                              0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                              The final consolidation settlement (one-dimensional method) is

                                              sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                              Corrected time t frac14 2 1

                                              2

                                              40

                                              52

                                              frac14 1615 years

                                              Tv frac14 cvtd2frac14 44 1615

                                              42frac14 0444

                                              From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                              77

                                              The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                              Figure Q77

                                              56 Consolidation theory

                                              Point m n Ir (kNm2) sc (mm)

                                              13020frac14 15 20

                                              20frac14 10 0194 (4) 113 124

                                              260

                                              20frac14 30

                                              20

                                              20frac14 10 0204 (2) 59 65

                                              360

                                              20frac14 30

                                              40

                                              20frac14 20 0238 (1) 35 38

                                              430

                                              20frac14 15

                                              40

                                              20frac14 20 0224 (2) 65 72

                                              Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                              78

                                              Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                              (a) Immediate settlement

                                              H

                                              Bfrac14 30

                                              35frac14 086

                                              D

                                              Bfrac14 2

                                              35frac14 006

                                              Figure Q78

                                              Consolidation theory 57

                                              From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                              si frac14 130131qB

                                              Eufrac14 10 032 105 35

                                              40frac14 30mm

                                              (b) Consolidation settlement

                                              Layer z (m) Dz Ic (kNm2) syod (mm)

                                              1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                              3150

                                              Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                              Now

                                              H

                                              Bfrac14 30

                                              35frac14 086 and A frac14 065

                                              from Figure 712 13 frac14 079

                                              sc frac14 13sod frac14 079 315 frac14 250mm

                                              Total settlement

                                              s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                              79

                                              Without sand drains

                                              Uv frac14 025

                                              Tv frac14 0049 ethfrom Figure 718THORN

                                              t frac14 Tvd2

                                              cvfrac14 0049 82

                                              cvWith sand drains

                                              R frac14 0564S frac14 0564 3 frac14 169m

                                              n frac14 Rrfrac14 169

                                              015frac14 113

                                              Tr frac14 cht

                                              4R2frac14 ch

                                              4 1692 0049 82

                                              cvethand ch frac14 cvTHORN

                                              frac14 0275

                                              Ur frac14 073 (from Figure 730)

                                              58 Consolidation theory

                                              Using Equation 740

                                              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                              U frac14 080

                                              710

                                              Without sand drains

                                              Uv frac14 090

                                              Tv frac14 0848

                                              t frac14 Tvd2

                                              cvfrac14 0848 102

                                              96frac14 88 years

                                              With sand drains

                                              R frac14 0564S frac14 0564 4 frac14 226m

                                              n frac14 Rrfrac14 226

                                              015frac14 15

                                              Tr

                                              Tvfrac14 chcv

                                              d2

                                              4R2ethsame tTHORN

                                              Tr

                                              Tvfrac14 140

                                              96 102

                                              4 2262frac14 714 eth1THORN

                                              Using Equation 740

                                              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                              An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                              Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                              040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                              Thus

                                              Uv frac14 0295 and Ur frac14 086

                                              t frac14 88 00683

                                              0848frac14 07 years

                                              Consolidation theory 59

                                              Chapter 8

                                              Bearing capacity

                                              81

                                              (a) The ultimate bearing capacity is given by Equation 83

                                              qf frac14 cNc thorn DNq thorn 1

                                              2BN

                                              For u frac14 0

                                              Nc frac14 514 Nq frac14 1 N frac14 0

                                              qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                              The net ultimate bearing capacity is

                                              qnf frac14 qf D frac14 540 kN=m2

                                              The net foundation pressure is

                                              qn frac14 q D frac14 425

                                              2 eth21 1THORN frac14 192 kN=m2

                                              The factor of safety (Equation 86) is

                                              F frac14 qnfqnfrac14 540

                                              192frac14 28

                                              (b) For 0 frac14 28

                                              Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                              qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                              2 112 2 13

                                              frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                              qnf frac14 574 112 frac14 563 kN=m2

                                              F frac14 563

                                              192frac14 29

                                              (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                              82

                                              For 0 frac14 38

                                              Nq frac14 49 N frac14 67

                                              qnf frac14 DethNq 1THORN thorn 1

                                              2BN ethfrom Equation 83THORN

                                              frac14 eth18 075 48THORN thorn 1

                                              2 18 15 67

                                              frac14 648thorn 905 frac14 1553 kN=m2

                                              qn frac14 500

                                              15 eth18 075THORN frac14 320 kN=m2

                                              F frac14 qnfqnfrac14 1553

                                              320frac14 48

                                              0d frac14 tan1tan 38

                                              125

                                              frac14 32 therefore Nq frac14 23 and N frac14 25

                                              Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                              2 18 15 25

                                              frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                              Design load (action) Vd frac14 500 kN=m

                                              The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                              83

                                              D

                                              Bfrac14 350

                                              225frac14 155

                                              From Figure 85 for a square foundation

                                              Nc frac14 81

                                              Bearing capacity 61

                                              For a rectangular foundation (L frac14 450m B frac14 225m)

                                              Nc frac14 084thorn 016B

                                              L

                                              81 frac14 745

                                              Using Equation 810

                                              qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                              For F frac14 3

                                              qn frac14 1006

                                              3frac14 335 kN=m2

                                              q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                              Design load frac14 405 450 225 frac14 4100 kN

                                              Design undrained strength cud frac14 135

                                              14frac14 96 kN=m2

                                              Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                              frac14 7241 kN

                                              Design load Vd frac14 4100 kN

                                              Rd gt Vd therefore the bearing resistance limit state is satisfied

                                              84

                                              For 0 frac14 40

                                              Nq frac14 64 N frac14 95

                                              qnf frac14 DethNq 1THORN thorn 04BN

                                              (a) Water table 5m below ground level

                                              qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                              qn frac14 400 17 frac14 383 kN=m2

                                              F frac14 2686

                                              383frac14 70

                                              (b) Water table 1m below ground level (ie at foundation level)

                                              0 frac14 20 98 frac14 102 kN=m3

                                              62 Bearing capacity

                                              qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                              F frac14 2040

                                              383frac14 53

                                              (c) Water table at ground level with upward hydraulic gradient 02

                                              eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                              qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                              F frac14 1296

                                              392frac14 33

                                              85

                                              The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                              Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                              Design value of 0 frac14 tan1tan 39

                                              125

                                              frac14 33

                                              For 0 frac14 33 Nq frac14 26 and N frac14 29

                                              Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                              Rd gt Vd therefore the bearing resistance limit state is satisfied

                                              86

                                              (a) Undrained shear for u frac14 0

                                              Nc frac14 514 Nq frac14 1 N frac14 0

                                              qnf frac14 12cuNc

                                              frac14 12 100 514 frac14 617 kN=m2

                                              qn frac14 qnfFfrac14 617

                                              3frac14 206 kN=m2

                                              q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                              Bearing capacity 63

                                              Drained shear for 0 frac14 32

                                              Nq frac14 23 N frac14 25

                                              0 frac14 21 98 frac14 112 kN=m3

                                              qnf frac14 0DethNq 1THORN thorn 040BN

                                              frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                              frac14 694 kN=m2

                                              q frac14 694

                                              3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                              Design load frac14 42 227 frac14 3632 kN

                                              (b) Design undrained strength cud frac14 100

                                              14frac14 71 kNm2

                                              Design bearing resistance Rd frac14 12cudNe area

                                              frac14 12 71 514 42

                                              frac14 7007 kN

                                              For drained shear 0d frac14 tan1tan 32

                                              125

                                              frac14 26

                                              Nq frac14 12 N frac14 10

                                              Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                              (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                              Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                              1 2 100 0175 0700qn 0182qn

                                              2 6 033 0044 0176qn 0046qn

                                              3 10 020 0017 0068qn 0018qn

                                              0246qn

                                              Diameter of equivalent circle B frac14 45m

                                              H

                                              Bfrac14 12

                                              45frac14 27 and A frac14 042

                                              13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                              64 Bearing capacity

                                              For sc frac14 30mm

                                              qn frac14 30

                                              0147frac14 204 kN=m2

                                              q frac14 204thorn 21 frac14 225 kN=m2

                                              Design load frac14 42 225 frac14 3600 kN

                                              The design load is 3600 kN settlement being the limiting criterion

                                              87

                                              D

                                              Bfrac14 8

                                              4frac14 20

                                              From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                              F frac14 cuNc

                                              Dfrac14 40 71

                                              20 8frac14 18

                                              88

                                              Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                              Design value of 0 frac14 tan1tan 38

                                              125

                                              frac14 32

                                              Figure Q86

                                              Bearing capacity 65

                                              For 0 frac14 32 Nq frac14 23 and N frac14 25

                                              Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                              The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                              Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                              For B frac14 250m qn frac14 3750

                                              2502 17 frac14 583 kN=m2

                                              From Figure 510 m frac14 n frac14 126

                                              6frac14 021

                                              Ir frac14 0019

                                              Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                              Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                              The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                              89

                                              Depth (m) N 0v (kNm2) CN N1

                                              070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                              Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                              (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                              Cw frac14 05thorn 0530

                                              47

                                              frac14 082

                                              66 Bearing capacity

                                              Thus

                                              qa frac14 150 082 frac14 120 kN=m2

                                              (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                              Thus

                                              qa frac14 90 15 frac14 135 kN=m2

                                              (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                              Ic frac14 171

                                              1014frac14 0068

                                              From Equation 819(a) with s frac14 25mm

                                              q frac14 25

                                              3507 0068frac14 150 kN=m2

                                              810

                                              Peak value of strain influence factor occurs at a depth of 27m and is given by

                                              Izp frac14 05thorn 01130

                                              16 27

                                              05

                                              frac14 067

                                              Refer to Figure Q810

                                              E frac14 25qc

                                              Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                              Ez (mm3MN)

                                              1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                              0203

                                              C1 frac14 1 0500qnfrac14 1 05 12 16

                                              130frac14 093

                                              C2 frac14 1 ethsayTHORN

                                              s frac14 C1C2qnX Iz

                                              Ez frac14 093 1 130 0203 frac14 25mm

                                              Bearing capacity 67

                                              811

                                              At pile base level

                                              cu frac14 220 kN=m2

                                              qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                              Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                              00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                              qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                              Then

                                              Qf frac14 Abqb thorn Asqs

                                              frac14

                                              4 32 1980

                                              thorn eth 105 139 86THORN

                                              frac14 13 996thorn 3941 frac14 17 937 kN

                                              0 01 02 03 04 05 06 07

                                              0 2 4 6 8 10 12 14

                                              1

                                              2

                                              3

                                              4

                                              5

                                              6

                                              7

                                              8

                                              (1)

                                              (2)

                                              (3)

                                              (4)

                                              (5)

                                              qc

                                              qc

                                              Iz

                                              Iz

                                              (MNm2)

                                              z (m)

                                              Figure Q810

                                              68 Bearing capacity

                                              Allowable load

                                              ethaTHORN Qf

                                              2frac14 17 937

                                              2frac14 8968 kN

                                              ethbTHORN Abqb

                                              3thorn Asqs frac14 13 996

                                              3thorn 3941 frac14 8606 kN

                                              ie allowable load frac14 8600 kN

                                              Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                              According to the limit state method

                                              Characteristic undrained strength at base level cuk frac14 220

                                              150kN=m2

                                              Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                              150frac14 1320 kN=m2

                                              Characteristic shaft resistance qsk frac14 00150

                                              frac14 86

                                              150frac14 57 kN=m2

                                              Characteristic base and shaft resistances

                                              Rbk frac14

                                              4 32 1320 frac14 9330 kN

                                              Rsk frac14 105 139 86

                                              150frac14 2629 kN

                                              For a bored pile the partial factors are b frac14 160 and s frac14 130

                                              Design bearing resistance Rcd frac14 9330

                                              160thorn 2629

                                              130

                                              frac14 5831thorn 2022

                                              frac14 7850 kN

                                              Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                              812

                                              ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                              qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                              For a single pile

                                              Qf frac14 Abqb thorn Asqs

                                              frac14

                                              4 062 1305

                                              thorn eth 06 15 42THORN

                                              frac14 369thorn 1187 frac14 1556 kN

                                              Bearing capacity 69

                                              Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                              eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                              (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                              (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                              qbkfrac14 9cuk frac14 9 220

                                              150frac14 1320 kN=m2

                                              qskfrac14cuk frac14 040 105

                                              150frac14 28 kN=m2

                                              Rbkfrac14

                                              4 0602 1320 frac14 373 kN

                                              Rskfrac14 060 15 28 frac14 791 kN

                                              Rcdfrac14 373

                                              160thorn 791

                                              130frac14 233thorn 608 frac14 841 kN

                                              Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                              Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                              (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                              q frac14 21 000

                                              1762frac14 68 kN=m2

                                              Immediate settlement

                                              H

                                              Bfrac14 15

                                              176frac14 085

                                              D

                                              Bfrac14 13

                                              176frac14 074

                                              L

                                              Bfrac14 1

                                              Hence from Figure 515

                                              130 frac14 078 and 131 frac14 041

                                              70 Bearing capacity

                                              Thus using Equation 528

                                              si frac14 078 041 68 176

                                              65frac14 6mm

                                              Consolidation settlement

                                              Layer z (m) Area (m2) (kNm2) mvH (mm)

                                              1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                              434 (sod)

                                              Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                              sc frac14 056 434 frac14 24mm

                                              The total settlement is (6thorn 24) frac14 30mm

                                              813

                                              At base level N frac14 26 Then using Equation 830

                                              qb frac14 40NDb

                                              Bfrac14 40 26 2

                                              025frac14 8320 kN=m2

                                              ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                              Figure Q812

                                              Bearing capacity 71

                                              Over the length embedded in sand

                                              N frac14 21 ie18thorn 24

                                              2

                                              Using Equation 831

                                              qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                              For a single pile

                                              Qf frac14 Abqb thorn Asqs

                                              frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                              For the pile group assuming a group efficiency of 12

                                              XQf frac14 12 9 604 frac14 6523 kN

                                              Then the load factor is

                                              F frac14 6523

                                              2000thorn 1000frac14 21

                                              (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                              Characteristic base resistance per unit area qbk frac14 8320

                                              150frac14 5547 kNm2

                                              Characteristic shaft resistance per unit area qsk frac14 42

                                              150frac14 28 kNm2

                                              Characteristic base and shaft resistances for a single pile

                                              Rbk frac14 0252 5547 frac14 347 kN

                                              Rsk frac14 4 025 2 28 frac14 56 kN

                                              For a driven pile the partial factors are b frac14 s frac14 130

                                              Design bearing resistance Rcd frac14 347

                                              130thorn 56

                                              130frac14 310 kN

                                              For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                              Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                              (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                              From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                              72 Bearing capacity

                                              N frac14 24thorn 26thorn 34

                                              3frac14 28

                                              Ic frac14 171

                                              2814frac14 0016 ethEquation 818THORN

                                              s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                              814

                                              Using Equation 841

                                              Tf frac14 DLcu thorn

                                              4ethD2 d2THORNcuNc

                                              frac14 eth 02 5 06 110THORN thorn

                                              4eth022 012THORN110 9

                                              frac14 207thorn 23 frac14 230 kN

                                              Figure Q813

                                              Bearing capacity 73

                                              Chapter 9

                                              Stability of slopes

                                              91

                                              Referring to Figure Q91

                                              W frac14 417 19 frac14 792 kN=m

                                              Q frac14 20 28 frac14 56 kN=m

                                              Arc lengthAB frac14

                                              180 73 90 frac14 115m

                                              Arc length BC frac14

                                              180 28 90 frac14 44m

                                              The factor of safety is given by

                                              F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                              90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                              Depth of tension crack z0 frac14 2cu

                                              frac14 2 20

                                              19frac14 21m

                                              Arc length BD frac14

                                              180 13

                                              1

                                              2 90 frac14 21m

                                              F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                              The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                              Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                              14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                              Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                              The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                              92

                                              u frac14 0

                                              Depth factor D frac14 11

                                              9frac14 122

                                              Using Equation 92 with F frac14 10

                                              Ns frac14 cu

                                              FHfrac14 30

                                              10 19 9frac14 0175

                                              Hence from Figure 93

                                              frac14 50

                                              For F frac14 12

                                              Ns frac14 30

                                              12 19 9frac14 0146

                                              frac14 27

                                              93

                                              Refer to Figure Q93

                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                              1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                              74 m

                                              214 1deg

                                              213 1deg

                                              39 m

                                              WB

                                              D

                                              C

                                              28 m

                                              21 m

                                              A

                                              Q

                                              Soil (1)Soil (2)

                                              73deg

                                              Figure Q91

                                              Stability of slopes 75

                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                              9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                              599 256 328 1372

                                              Figure Q93

                                              76 Stability of slopes

                                              XW cos frac14 b

                                              Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                              W sin frac14 bX

                                              h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                              Arc length La frac14

                                              180 57

                                              1

                                              2 326 frac14 327m

                                              The factor of safety is given by

                                              F frac14 c0La thorn tan0ethW cos ulTHORN

                                              W sin

                                              frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                              frac14 091

                                              According to the limit state method

                                              0d frac14 tan1tan 32

                                              125

                                              frac14 265

                                              c0 frac14 8

                                              160frac14 5 kN=m2

                                              Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                              Design disturbing moment frac14 1075 kN=m

                                              The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                              94

                                              F frac14 1

                                              W sin

                                              Xfc0bthorn ethW ubTHORN tan0g sec

                                              1thorn ethtan tan0=FTHORN

                                              c0 frac14 8 kN=m2

                                              0 frac14 32

                                              c0b frac14 8 2 frac14 16 kN=m

                                              W frac14 bh frac14 21 2 h frac14 42h kN=m

                                              Try F frac14 100

                                              tan0

                                              Ffrac14 0625

                                              Stability of slopes 77

                                              Values of u are as obtained in Figure Q93

                                              SliceNo

                                              h(m)

                                              W frac14 bh(kNm)

                                              W sin(kNm)

                                              ub(kNm)

                                              c0bthorn (W ub) tan0(kNm)

                                              sec

                                              1thorn (tan tan0)FProduct(kNm)

                                              1 05 21 6 2 8 24 1078 262 13 55 31

                                              23 33 30 1042 31

                                              3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                              224 92 72 0931 67

                                              6 50 210 11 40 100 85 0907 777 55 231 14

                                              12 58 112 90 0889 80

                                              8 60 252 1812

                                              80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                              10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                              2154 88 116 0853 99

                                              14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                              1074 1091

                                              F frac14 1091

                                              1074frac14 102 (assumed value 100)

                                              Thus

                                              F frac14 101

                                              95

                                              F frac14 1

                                              W sin

                                              XfWeth1 ruTHORN tan0g sec

                                              1thorn ethtan tan0THORN=F

                                              0 frac14 33

                                              ru frac14 020

                                              W frac14 bh frac14 20 5 h frac14 100h kN=m

                                              eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                              Try F frac14 110

                                              tan 0

                                              Ffrac14 tan 33

                                              110frac14 0590

                                              78 Stability of slopes

                                              Referring to Figure Q95

                                              SliceNo

                                              h(m)

                                              W frac14 bh(kNm)

                                              W sin(kNm)

                                              W(1 ru) tan0(kNm)

                                              sec

                                              1thorn ( tan tan0)FProduct(kNm)

                                              1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                              2120 234 0892 209

                                              4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                              1185 1271

                                              Figure Q95

                                              Stability of slopes 79

                                              F frac14 1271

                                              1185frac14 107

                                              The trial value was 110 therefore take F to be 108

                                              96

                                              (a) Water table at surface the factor of safety is given by Equation 912

                                              F frac14 0

                                              sat

                                              tan0

                                              tan

                                              ptie 15 frac14 92

                                              19

                                              tan 36

                                              tan

                                              tan frac14 0234

                                              frac14 13

                                              Water table well below surface the factor of safety is given by Equation 911

                                              F frac14 tan0

                                              tan

                                              frac14 tan 36

                                              tan 13

                                              frac14 31

                                              (b) 0d frac14 tan1tan 36

                                              125

                                              frac14 30

                                              Depth of potential failure surface frac14 z

                                              Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                              frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                              frac14 504z kN

                                              Design disturbing moment per unit area Sd frac14 sat sin cos

                                              frac14 19 z sin 13 cos 13

                                              frac14 416z kN

                                              Rd gtSd therefore the limit state for overall stability is satisfied

                                              80 Stability of slopes

                                              • Book Cover
                                              • Title
                                              • Contents
                                              • Basic characteristics of soils
                                              • Seepage
                                              • Effective stress
                                              • Shear strength
                                              • Stresses and displacements
                                              • Lateral earth pressure
                                              • Consolidation theory
                                              • Bearing capacity
                                              • Stability of slopes

                                                36

                                                Total weight

                                                ab frac14 210 kN

                                                Effective weight

                                                ac frac14 112 kN

                                                Resultant boundary water force

                                                be frac14 119 kN

                                                Seepage force

                                                ce frac14 34 kN

                                                Resultant body force

                                                ae frac14 99 kN eth73 to horizontalTHORN

                                                (Refer to Figure Q36)

                                                Figure Q36

                                                18 Effective stress

                                                37

                                                Situation (1)(a)

                                                frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                                                u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                                0 frac14 u frac14 694 392 frac14 302 kN=m2

                                                (b)

                                                i frac14 2

                                                4frac14 05

                                                j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                                                Situation (2)(a)

                                                frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                                                u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                                0 frac14 u frac14 498 392 frac14 106 kN=m2

                                                (b)

                                                i frac14 2

                                                4frac14 05

                                                j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                                                38

                                                The flow net is drawn in Figure Q24

                                                Loss in total head between adjacent equipotentials

                                                h frac14 550

                                                Ndfrac14 550

                                                11frac14 050m

                                                Exit hydraulic gradient

                                                ie frac14 h

                                                sfrac14 050

                                                070frac14 071

                                                Effective stress 19

                                                The critical hydraulic gradient is given by Equation 39

                                                ic frac14 0

                                                wfrac14 102

                                                98frac14 104

                                                Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                                                F frac14 iciefrac14 104

                                                071frac14 15

                                                Total head at C

                                                hC frac14 nd

                                                Ndh frac14 24

                                                11 550 frac14 120m

                                                Elevation head at C

                                                zC frac14 250m

                                                Pore water pressure at C

                                                uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                                                Therefore effective vertical stress at C

                                                0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                                                For point D

                                                hD frac14 73

                                                11 550 frac14 365m

                                                zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                                                0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                                                39

                                                The flow net is drawn in Figure Q25

                                                For a soil prism 150 300m adjacent to the piling

                                                hm frac14 26

                                                9 500 frac14 145m

                                                20 Effective stress

                                                Factor of safety against lsquoheavingrsquo (Equation 310)

                                                F frac14 ic

                                                imfrac14 0d

                                                whmfrac14 97 300

                                                98 145frac14 20

                                                With a filter

                                                F frac14 0d thorn wwhm

                                                3 frac14 eth97 300THORN thorn w98 145

                                                w frac14 135 kN=m2

                                                Depth of filterfrac14 13521frac14 065m (if above water level)

                                                Effective stress 21

                                                Chapter 4

                                                Shear strength

                                                41

                                                frac14 295 kN=m2

                                                u frac14 120 kN=m2

                                                0 frac14 u frac14 295 120 frac14 175 kN=m2

                                                f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                                                42

                                                03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                                                100 452 552200 908 1108400 1810 2210800 3624 4424

                                                The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                                                Figure Q42

                                                43

                                                3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                                                200 222 422400 218 618600 220 820

                                                The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                                                44

                                                The modified shear strength parameters are

                                                0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                                                a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                                                The coordinates of the stress point representing failure conditions in the test are

                                                1

                                                2eth1 2THORN frac14 1

                                                2 170 frac14 85 kN=m2

                                                1

                                                2eth1 thorn 3THORN frac14 1

                                                2eth270thorn 100THORN frac14 185 kN=m2

                                                The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                                                uf frac14 36 kN=m2

                                                Figure Q43

                                                Figure Q44

                                                Shear strength 23

                                                45

                                                3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                                150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                                The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                                The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                                46

                                                03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                                200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                                The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                                Figure Q45

                                                24 Shear strength

                                                47

                                                The torque required to produce shear failure is given by

                                                T frac14 dh cud

                                                2thorn 2

                                                Z d=2

                                                0

                                                2r drcur

                                                frac14 cud2h

                                                2thorn 4cu

                                                Z d=2

                                                0

                                                r2dr

                                                frac14 cud2h

                                                2thorn d

                                                3

                                                6

                                                Then

                                                35 frac14 cu52 10

                                                2thorn 53

                                                6

                                                103

                                                cu frac14 76 kN=m3

                                                400

                                                0 400 800 1200 1600

                                                τ (k

                                                Nm

                                                2 )

                                                σprime (kNm2)

                                                34deg

                                                315deg29deg

                                                (a)

                                                (b)

                                                0 400

                                                400

                                                800 1200 1600

                                                Failure envelope

                                                300 500

                                                σprime (kNm2)

                                                τ (k

                                                Nm

                                                2 )

                                                20 (kNm2)

                                                31deg

                                                Figure Q46

                                                Shear strength 25

                                                48

                                                The relevant stress values are calculated as follows

                                                3 frac14 600 kN=m2

                                                1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                                2(1 3) 0 40 79 107 139 159

                                                1

                                                2(01 thorn 03) 400 411 402 389 351 326

                                                1

                                                2(1 thorn 3) 600 640 679 707 739 759

                                                The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                                Af frac14 433 200

                                                319frac14 073

                                                49

                                                B frac14 u33

                                                frac14 144

                                                350 200frac14 096

                                                a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                                0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                                10 283 65 023

                                                Figure Q48

                                                26 Shear strength

                                                The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                                Figure Q49

                                                Shear strength 27

                                                Chapter 5

                                                Stresses and displacements

                                                51

                                                Vertical stress is given by

                                                z frac14 Qz2Ip frac14 5000

                                                52Ip

                                                Values of Ip are obtained from Table 51

                                                r (m) rz Ip z (kNm2)

                                                0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                                10 20 0009 2

                                                The variation of z with radial distance (r) is plotted in Figure Q51

                                                Figure Q51

                                                52

                                                Below the centre load (Figure Q52)

                                                r

                                                zfrac14 0 for the 7500-kN load

                                                Ip frac14 0478

                                                r

                                                zfrac14 5

                                                4frac14 125 for the 10 000- and 9000-kN loads

                                                Ip frac14 0045

                                                Then

                                                z frac14X Q

                                                z2Ip

                                                frac14 7500 0478

                                                42thorn 10 000 0045

                                                42thorn 9000 0045

                                                42

                                                frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                                53

                                                The vertical stress under a corner of a rectangular area is given by

                                                z frac14 qIr

                                                where values of Ir are obtained from Figure 510 In this case

                                                z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                                z

                                                Figure Q52

                                                Stresses and displacements 29

                                                z (m) m n Ir z (kNm2)

                                                0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                                10 010 0005 5

                                                z is plotted against z in Figure Q53

                                                54

                                                (a)

                                                m frac14 125

                                                12frac14 104

                                                n frac14 18

                                                12frac14 150

                                                From Figure 510 Irfrac14 0196

                                                z frac14 2 175 0196 frac14 68 kN=m2

                                                Figure Q53

                                                30 Stresses and displacements

                                                (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                                z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                                55

                                                Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                                Px frac14 2Q

                                                1

                                                m2 thorn 1frac14 2 150

                                                125frac14 76 kN=m

                                                Equation 517 is used to obtain the pressure distribution

                                                px frac14 4Q

                                                h

                                                m2n

                                                ethm2 thorn n2THORN2 frac14150

                                                m2n

                                                ethm2 thorn n2THORN2 ethkN=m2THORN

                                                Figure Q54

                                                Stresses and displacements 31

                                                n m2n

                                                (m2 thorn n2)2

                                                px(kNm2)

                                                0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                The pressure distribution is plotted in Figure Q55

                                                56

                                                H

                                                Bfrac14 10

                                                2frac14 5

                                                L

                                                Bfrac14 4

                                                2frac14 2

                                                D

                                                Bfrac14 1

                                                2frac14 05

                                                Hence from Figure 515

                                                131 frac14 082

                                                130 frac14 094

                                                Figure Q55

                                                32 Stresses and displacements

                                                The immediate settlement is given by Equation 528

                                                si frac14 130131qB

                                                Eu

                                                frac14 094 082 200 2

                                                45frac14 7mm

                                                Stresses and displacements 33

                                                Chapter 6

                                                Lateral earth pressure

                                                61

                                                For 0 frac14 37 the active pressure coefficient is given by

                                                Ka frac14 1 sin 37

                                                1thorn sin 37frac14 025

                                                The total active thrust (Equation 66a with c0 frac14 0) is

                                                Pa frac14 1

                                                2KaH

                                                2 frac14 1

                                                2 025 17 62 frac14 765 kN=m

                                                If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                and the thrust on the wall is

                                                P0 frac14 1

                                                2K0H

                                                2 frac14 1

                                                2 040 17 62 frac14 122 kN=m

                                                62

                                                The active pressure coefficients for the three soil types are as follows

                                                Ka1 frac141 sin 35

                                                1thorn sin 35frac14 0271

                                                Ka2 frac141 sin 27

                                                1thorn sin 27frac14 0375

                                                ffiffiffiffiffiffiffiKa2

                                                p frac14 0613

                                                Ka3 frac141 sin 42

                                                1thorn sin 42frac14 0198

                                                Distribution of active pressure (plotted in Figure Q62)

                                                Depth (m) Soil Active pressure (kNm2)

                                                3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                Total thrust frac14 571 kNm

                                                Point of application is (4893571) m from the top of the wall ie 857m

                                                Force (kN) Arm (m) Moment (kN m)

                                                (1)1

                                                2 0271 16 32 frac14 195 20 390

                                                (2) 0271 16 3 2 frac14 260 40 1040

                                                (3)1

                                                2 0271 92 22 frac14 50 433 217

                                                (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                (5)1

                                                2 0375 102 32 frac14 172 70 1204

                                                (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                (7)1

                                                2 0198 112 42 frac14 177 1067 1889

                                                (8)1

                                                2 98 92 frac14 3969 90 35721

                                                5713 48934

                                                Figure Q62

                                                Lateral earth pressure 35

                                                63

                                                (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                frac14 245

                                                At the lower end of the piling

                                                pa frac14 Kaqthorn Kasatz Kaccu

                                                frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                frac14 115 kN=m2

                                                pp frac14 Kpsatzthorn Kpccu

                                                frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                frac14 202 kN=m2

                                                (b) For 0 frac14 26 and frac14 1

                                                20

                                                Ka frac14 035

                                                Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                pfrac14 145 ethEquation 619THORN

                                                Kp frac14 37

                                                Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                pfrac14 47 ethEquation 624THORN

                                                At the lower end of the piling

                                                pa frac14 Kaqthorn Ka0z Kacc

                                                0

                                                frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                frac14 187 kN=m2

                                                pp frac14 Kp0zthorn Kpcc

                                                0

                                                frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                frac14 198 kN=m2

                                                36 Lateral earth pressure

                                                64

                                                (a) For 0 frac14 38 Ka frac14 024

                                                0 frac14 20 98 frac14 102 kN=m3

                                                The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                Force (kN) Arm (m) Moment (kN m)

                                                (1) 024 10 66 frac14 159 33 525

                                                (2)1

                                                2 024 17 392 frac14 310 400 1240

                                                (3) 024 17 39 27 frac14 430 135 580

                                                (4)1

                                                2 024 102 272 frac14 89 090 80

                                                (5)1

                                                2 98 272 frac14 357 090 321

                                                Hfrac14 1345 MH frac14 2746

                                                (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                XM frac14MV MH frac14 7790 kNm

                                                Lever arm of base resultant

                                                M

                                                Vfrac14 779

                                                488frac14 160

                                                Eccentricity of base resultant

                                                e frac14 200 160 frac14 040m

                                                39 m

                                                27 m

                                                40 m

                                                04 m

                                                04 m

                                                26 m

                                                (7)

                                                (9)

                                                (1)(2)

                                                (3)

                                                (4)

                                                (5)

                                                (8)(6)

                                                (10)

                                                WT

                                                10 kNm2

                                                Hydrostatic

                                                Figure Q64

                                                Lateral earth pressure 37

                                                Base pressures (Equation 627)

                                                p frac14 VB

                                                1 6e

                                                B

                                                frac14 488

                                                4eth1 060THORN

                                                frac14 195 kN=m2 and 49 kN=m2

                                                Factor of safety against sliding (Equation 628)

                                                F frac14 V tan

                                                Hfrac14 488 tan 25

                                                1345frac14 17

                                                (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                Hfrac14 1633 kN

                                                V frac14 4879 kN

                                                MH frac14 3453 kNm

                                                MV frac14 10536 kNm

                                                The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                65

                                                For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                Kp

                                                Ffrac14 385

                                                2

                                                0 frac14 20 98 frac14 102 kN=m3

                                                The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                Force (kN) Arm (m) Moment (kN m)

                                                (1)1

                                                2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                (2) 026 17 45 d frac14 199d d2 995d2

                                                (3)1

                                                2 026 102 d2 frac14 133d2 d3 044d3

                                                (4)1

                                                2 385

                                                2 17 152 frac14 368 dthorn 05 368d 184

                                                (5)385

                                                2 17 15 d frac14 491d d2 2455d2

                                                (6)1

                                                2 385

                                                2 102 d2 frac14 982d2 d3 327d3

                                                38 Lateral earth pressure

                                                XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                d3 thorn 516d2 283d 1724 frac14 0

                                                d frac14 179m

                                                Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                Over additional 20 embedded depth

                                                pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                66

                                                The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                Ka frac14 sin 69=sin 105

                                                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                26664

                                                37775

                                                2

                                                frac14 050

                                                The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                Pa frac14 1

                                                2 050 19 7502 frac14 267 kN=m

                                                Figure Q65

                                                Lateral earth pressure 39

                                                Horizontal component

                                                Ph frac14 267 cos 40 frac14 205 kN=m

                                                Vertical component

                                                Pv frac14 267 sin 40 frac14 172 kN=m

                                                Consider moments about the toe of the wall (Figure Q66) (per m)

                                                Force (kN) Arm (m) Moment (kN m)

                                                (1)1

                                                2 175 650 235 frac14 1337 258 345

                                                (2) 050 650 235 frac14 764 175 134

                                                (3)1

                                                2 070 650 235 frac14 535 127 68

                                                (4) 100 400 235 frac14 940 200 188

                                                (5) 1

                                                2 080 050 235 frac14 47 027 1

                                                Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                Lever arm of base resultant

                                                M

                                                Vfrac14 795

                                                525frac14 151m

                                                Eccentricity of base resultant

                                                e frac14 200 151 frac14 049m

                                                Figure Q66

                                                40 Lateral earth pressure

                                                Base pressures (Equation 627)

                                                p frac14 525

                                                41 6 049

                                                4

                                                frac14 228 kN=m2 and 35 kN=m2

                                                The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                67

                                                For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                Force (kN) Arm (m) Moment (kNm)

                                                (1)1

                                                2 027 17 52 frac14 574 183 1050

                                                (2) 027 17 5 3 frac14 689 500 3445

                                                (3)1

                                                2 027 102 32 frac14 124 550 682

                                                (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                (5)1

                                                2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                (7) 1

                                                2 267

                                                2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                2 d frac14 163d d2thorn 650 82d2 1060d

                                                Tie rod force per m frac14 T 0 0

                                                XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                d3 thorn 77d2 269d 1438 frac14 0

                                                d frac14 467m

                                                Depth of penetration frac14 12d frac14 560m

                                                Lateral earth pressure 41

                                                Algebraic sum of forces for d frac14 467m isX

                                                F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                T frac14 905 kN=m

                                                Force in each tie rod frac14 25T frac14 226 kN

                                                68

                                                (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                0 frac14 21 98 frac14 112 kN=m3

                                                The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                uC frac14 150

                                                165 15 98 frac14 134 kN=m2

                                                The average seepage pressure is

                                                j frac14 15

                                                165 98 frac14 09 kN=m3

                                                Hence

                                                0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                0 j frac14 112 09 frac14 103 kN=m3

                                                Figure Q67

                                                42 Lateral earth pressure

                                                Consider moments about the anchor point A (per m)

                                                Force (kN) Arm (m) Moment (kN m)

                                                (1) 10 026 150 frac14 390 60 2340

                                                (2)1

                                                2 026 18 452 frac14 474 15 711

                                                (3) 026 18 45 105 frac14 2211 825 18240

                                                (4)1

                                                2 026 121 1052 frac14 1734 100 17340

                                                (5)1

                                                2 134 15 frac14 101 40 404

                                                (6) 134 30 frac14 402 60 2412

                                                (7)1

                                                2 134 60 frac14 402 95 3819

                                                571 4527(8) Ppm

                                                115 115PPm

                                                XM frac14 0

                                                Ppm frac144527

                                                115frac14 394 kN=m

                                                Available passive resistance

                                                Pp frac14 1

                                                2 385 103 62 frac14 714 kN=m

                                                Factor of safety

                                                Fp frac14 Pp

                                                Ppm

                                                frac14 714

                                                394frac14 18

                                                Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                Figure Q68

                                                Lateral earth pressure 43

                                                (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                Consider moments (per m) about the tie point A

                                                Force (kN) Arm (m)

                                                (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                (2)1

                                                2 033 18 452 frac14 601 15

                                                (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                (4)1

                                                2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                (5)1

                                                2 134 15 frac14 101 40

                                                (6) 134 30 frac14 402 60

                                                (7)1

                                                2 134 d frac14 67d d3thorn 75

                                                (8) 1

                                                2 30 103 d2 frac141545d2 2d3thorn 75

                                                Moment (kN m)

                                                (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                d3 thorn 827d2 466d 1518 frac14 0

                                                By trial

                                                d frac14 544m

                                                The minimum depth of embedment required is 544m

                                                69

                                                For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                0 frac14 20 98 frac14 102 kN=m3

                                                The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                44 Lateral earth pressure

                                                uC frac14 147

                                                173 26 98 frac14 216 kN=m2

                                                and the average seepage pressure around the wall is

                                                j frac14 26

                                                173 98 frac14 15 kN=m3

                                                Consider moments about the prop (A) (per m)

                                                Force (kN) Arm (m) Moment (kN m)

                                                (1)1

                                                2 03 17 272 frac14 186 020 37

                                                (2) 03 17 27 53 frac14 730 335 2445

                                                (3)1

                                                2 03 (102thorn 15) 532 frac14 493 423 2085

                                                (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                (5)1

                                                2 216 26 frac14 281 243 684

                                                (6) 216 27 frac14 583 465 2712

                                                (7)1

                                                2 216 60 frac14 648 800 5184

                                                3055(8)

                                                1

                                                2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                Factor of safety

                                                Fr frac14 6885

                                                3055frac14 225

                                                Figure Q69

                                                Lateral earth pressure 45

                                                610

                                                For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                Using the recommendations of Twine and Roscoe

                                                p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                611

                                                frac14 18 kN=m3 0 frac14 34

                                                H frac14 350m nH frac14 335m mH frac14 185m

                                                Consider a trial value of F frac14 20 Refer to Figure 635

                                                0m frac14 tan1tan 34

                                                20

                                                frac14 186

                                                Then

                                                frac14 45 thorn 0m2frac14 543

                                                W frac14 1

                                                2 18 3502 cot 543 frac14 792 kN=m

                                                Figure Q610

                                                46 Lateral earth pressure

                                                P frac14 1

                                                2 s 3352 frac14 561s kN=m

                                                U frac14 1

                                                2 98 1852 cosec 543 frac14 206 kN=m

                                                Equations 630 and 631 then become

                                                561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                ie

                                                561s 0616N 405 frac14 0

                                                792 0857N thorn 563 frac14 0

                                                N frac14 848

                                                0857frac14 989 kN=m

                                                Then

                                                561s 609 405 frac14 0

                                                s frac14 649

                                                561frac14 116 kN=m3

                                                The calculations for trial values of F of 20 15 and 10 are summarized below

                                                F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                Figure Q611

                                                Lateral earth pressure 47

                                                612

                                                For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                45 thorn 0

                                                2frac14 63

                                                For the retained material between the surface and a depth of 36m

                                                Pa frac14 1

                                                2 030 18 362 frac14 350 kN=m

                                                Weight of reinforced fill between the surface and a depth of 36m is

                                                Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                Eccentricity of Rv

                                                e frac14 263 250 frac14 013m

                                                The average vertical stress at a depth of 36m is

                                                z frac14 Rv

                                                L 2efrac14 324

                                                474frac14 68 kN=m2

                                                (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                Tensile stress in the element frac14 138 103

                                                65 3frac14 71N=mm2

                                                Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                The weight of ABC is

                                                W frac14 1

                                                2 18 52 265 frac14 124 kN=m

                                                From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                48 Lateral earth pressure

                                                (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                Tp frac14 032 68 120 065 frac14 170 kN

                                                Tr frac14 213 420

                                                418frac14 214 kN

                                                Again the tensile failure and slipping limit states are satisfied for this element

                                                Figure Q612

                                                Lateral earth pressure 49

                                                Chapter 7

                                                Consolidation theory

                                                71

                                                Total change in thickness

                                                H frac14 782 602 frac14 180mm

                                                Average thickness frac14 1530thorn 180

                                                2frac14 1620mm

                                                Length of drainage path d frac14 1620

                                                2frac14 810mm

                                                Root time plot (Figure Q71a)

                                                ffiffiffiffiffiffit90p frac14 33

                                                t90 frac14 109min

                                                cv frac14 0848d2

                                                t90frac14 0848 8102

                                                109 1440 365

                                                106frac14 27m2=year

                                                r0 frac14 782 764

                                                782 602frac14 018

                                                180frac14 0100

                                                rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                10 119

                                                9 180frac14 0735

                                                rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                Log time plot (Figure Q71b)

                                                t50 frac14 26min

                                                cv frac14 0196d2

                                                t50frac14 0196 8102

                                                26 1440 365

                                                106frac14 26m2=year

                                                r0 frac14 782 763

                                                782 602frac14 019

                                                180frac14 0106

                                                rp frac14 763 623

                                                782 602frac14 140

                                                180frac14 0778

                                                rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                Figure Q71(a)

                                                Figure Q71(b)

                                                Final void ratio

                                                e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                e

                                                Hfrac14 1thorn e0

                                                H0frac14 1thorn e1 thorne

                                                H0

                                                ie

                                                e

                                                180frac14 1631thorne

                                                1710

                                                e frac14 2936

                                                1530frac14 0192

                                                Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                mv frac14 1

                                                1thorn e0 e0 e101 00

                                                frac14 1

                                                1823 0192

                                                0107frac14 098m2=MN

                                                k frac14 cvmvw frac14 265 098 98

                                                60 1440 365 103frac14 81 1010 m=s

                                                72

                                                Using Equation 77 (one-dimensional method)

                                                sc frac14 e0 e11thorn e0 H

                                                Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                Figure Q72

                                                52 Consolidation theory

                                                Settlement

                                                Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                318

                                                Notes 5 92y 460thorn 84

                                                Heave

                                                Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                38

                                                73

                                                U frac14 f ethTvTHORN frac14 f cvt

                                                d2

                                                Hence if cv is constant

                                                t1

                                                t2frac14 d

                                                21

                                                d22

                                                where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                d1 frac14 95mm and d2 frac14 2500mm

                                                for U frac14 050 t2 frac14 t1 d22

                                                d21

                                                frac14 20

                                                60 24 365 25002

                                                952frac14 263 years

                                                for U lt 060 Tv frac14

                                                4U2 (Equation 724(a))

                                                t030 frac14 t050 0302

                                                0502

                                                frac14 263 036 frac14 095 years

                                                Consolidation theory 53

                                                74

                                                The layer is open

                                                d frac14 8

                                                2frac14 4m

                                                Tv frac14 cvtd2frac14 24 3

                                                42frac14 0450

                                                ui frac14 frac14 84 kN=m2

                                                The excess pore water pressure is given by Equation 721

                                                ue frac14Xmfrac141mfrac140

                                                2ui

                                                Msin

                                                Mz

                                                d

                                                expethM2TvTHORN

                                                In this case z frac14 d

                                                sinMz

                                                d

                                                frac14 sinM

                                                where

                                                M frac14

                                                23

                                                25

                                                2

                                                M sin M M2Tv exp (M2Tv)

                                                2thorn1 1110 0329

                                                3

                                                21 9993 457 105

                                                ue frac14 2 84 2

                                                1 0329 ethother terms negligibleTHORN

                                                frac14 352 kN=m2

                                                75

                                                The layer is open

                                                d frac14 6

                                                2frac14 3m

                                                Tv frac14 cvtd2frac14 10 3

                                                32frac14 0333

                                                The layer thickness will be divided into six equal parts ie m frac14 6

                                                54 Consolidation theory

                                                For an open layer

                                                Tv frac14 4n

                                                m2

                                                n frac14 0333 62

                                                4frac14 300

                                                The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                i j

                                                0 1 2 3 4 5 6 7 8 9 10 11 12

                                                0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                The initial and 3-year isochrones are plotted in Figure Q75

                                                Area under initial isochrone frac14 180 units

                                                Area under 3-year isochrone frac14 63 units

                                                The average degree of consolidation is given by Equation 725Thus

                                                U frac14 1 63

                                                180frac14 065

                                                Figure Q75

                                                Consolidation theory 55

                                                76

                                                At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                The final consolidation settlement (one-dimensional method) is

                                                sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                Corrected time t frac14 2 1

                                                2

                                                40

                                                52

                                                frac14 1615 years

                                                Tv frac14 cvtd2frac14 44 1615

                                                42frac14 0444

                                                From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                77

                                                The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                Figure Q77

                                                56 Consolidation theory

                                                Point m n Ir (kNm2) sc (mm)

                                                13020frac14 15 20

                                                20frac14 10 0194 (4) 113 124

                                                260

                                                20frac14 30

                                                20

                                                20frac14 10 0204 (2) 59 65

                                                360

                                                20frac14 30

                                                40

                                                20frac14 20 0238 (1) 35 38

                                                430

                                                20frac14 15

                                                40

                                                20frac14 20 0224 (2) 65 72

                                                Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                78

                                                Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                (a) Immediate settlement

                                                H

                                                Bfrac14 30

                                                35frac14 086

                                                D

                                                Bfrac14 2

                                                35frac14 006

                                                Figure Q78

                                                Consolidation theory 57

                                                From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                si frac14 130131qB

                                                Eufrac14 10 032 105 35

                                                40frac14 30mm

                                                (b) Consolidation settlement

                                                Layer z (m) Dz Ic (kNm2) syod (mm)

                                                1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                3150

                                                Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                Now

                                                H

                                                Bfrac14 30

                                                35frac14 086 and A frac14 065

                                                from Figure 712 13 frac14 079

                                                sc frac14 13sod frac14 079 315 frac14 250mm

                                                Total settlement

                                                s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                79

                                                Without sand drains

                                                Uv frac14 025

                                                Tv frac14 0049 ethfrom Figure 718THORN

                                                t frac14 Tvd2

                                                cvfrac14 0049 82

                                                cvWith sand drains

                                                R frac14 0564S frac14 0564 3 frac14 169m

                                                n frac14 Rrfrac14 169

                                                015frac14 113

                                                Tr frac14 cht

                                                4R2frac14 ch

                                                4 1692 0049 82

                                                cvethand ch frac14 cvTHORN

                                                frac14 0275

                                                Ur frac14 073 (from Figure 730)

                                                58 Consolidation theory

                                                Using Equation 740

                                                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                U frac14 080

                                                710

                                                Without sand drains

                                                Uv frac14 090

                                                Tv frac14 0848

                                                t frac14 Tvd2

                                                cvfrac14 0848 102

                                                96frac14 88 years

                                                With sand drains

                                                R frac14 0564S frac14 0564 4 frac14 226m

                                                n frac14 Rrfrac14 226

                                                015frac14 15

                                                Tr

                                                Tvfrac14 chcv

                                                d2

                                                4R2ethsame tTHORN

                                                Tr

                                                Tvfrac14 140

                                                96 102

                                                4 2262frac14 714 eth1THORN

                                                Using Equation 740

                                                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                Thus

                                                Uv frac14 0295 and Ur frac14 086

                                                t frac14 88 00683

                                                0848frac14 07 years

                                                Consolidation theory 59

                                                Chapter 8

                                                Bearing capacity

                                                81

                                                (a) The ultimate bearing capacity is given by Equation 83

                                                qf frac14 cNc thorn DNq thorn 1

                                                2BN

                                                For u frac14 0

                                                Nc frac14 514 Nq frac14 1 N frac14 0

                                                qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                The net ultimate bearing capacity is

                                                qnf frac14 qf D frac14 540 kN=m2

                                                The net foundation pressure is

                                                qn frac14 q D frac14 425

                                                2 eth21 1THORN frac14 192 kN=m2

                                                The factor of safety (Equation 86) is

                                                F frac14 qnfqnfrac14 540

                                                192frac14 28

                                                (b) For 0 frac14 28

                                                Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                2 112 2 13

                                                frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                qnf frac14 574 112 frac14 563 kN=m2

                                                F frac14 563

                                                192frac14 29

                                                (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                82

                                                For 0 frac14 38

                                                Nq frac14 49 N frac14 67

                                                qnf frac14 DethNq 1THORN thorn 1

                                                2BN ethfrom Equation 83THORN

                                                frac14 eth18 075 48THORN thorn 1

                                                2 18 15 67

                                                frac14 648thorn 905 frac14 1553 kN=m2

                                                qn frac14 500

                                                15 eth18 075THORN frac14 320 kN=m2

                                                F frac14 qnfqnfrac14 1553

                                                320frac14 48

                                                0d frac14 tan1tan 38

                                                125

                                                frac14 32 therefore Nq frac14 23 and N frac14 25

                                                Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                2 18 15 25

                                                frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                Design load (action) Vd frac14 500 kN=m

                                                The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                83

                                                D

                                                Bfrac14 350

                                                225frac14 155

                                                From Figure 85 for a square foundation

                                                Nc frac14 81

                                                Bearing capacity 61

                                                For a rectangular foundation (L frac14 450m B frac14 225m)

                                                Nc frac14 084thorn 016B

                                                L

                                                81 frac14 745

                                                Using Equation 810

                                                qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                For F frac14 3

                                                qn frac14 1006

                                                3frac14 335 kN=m2

                                                q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                Design load frac14 405 450 225 frac14 4100 kN

                                                Design undrained strength cud frac14 135

                                                14frac14 96 kN=m2

                                                Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                frac14 7241 kN

                                                Design load Vd frac14 4100 kN

                                                Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                84

                                                For 0 frac14 40

                                                Nq frac14 64 N frac14 95

                                                qnf frac14 DethNq 1THORN thorn 04BN

                                                (a) Water table 5m below ground level

                                                qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                qn frac14 400 17 frac14 383 kN=m2

                                                F frac14 2686

                                                383frac14 70

                                                (b) Water table 1m below ground level (ie at foundation level)

                                                0 frac14 20 98 frac14 102 kN=m3

                                                62 Bearing capacity

                                                qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                F frac14 2040

                                                383frac14 53

                                                (c) Water table at ground level with upward hydraulic gradient 02

                                                eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                F frac14 1296

                                                392frac14 33

                                                85

                                                The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                Design value of 0 frac14 tan1tan 39

                                                125

                                                frac14 33

                                                For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                86

                                                (a) Undrained shear for u frac14 0

                                                Nc frac14 514 Nq frac14 1 N frac14 0

                                                qnf frac14 12cuNc

                                                frac14 12 100 514 frac14 617 kN=m2

                                                qn frac14 qnfFfrac14 617

                                                3frac14 206 kN=m2

                                                q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                Bearing capacity 63

                                                Drained shear for 0 frac14 32

                                                Nq frac14 23 N frac14 25

                                                0 frac14 21 98 frac14 112 kN=m3

                                                qnf frac14 0DethNq 1THORN thorn 040BN

                                                frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                frac14 694 kN=m2

                                                q frac14 694

                                                3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                Design load frac14 42 227 frac14 3632 kN

                                                (b) Design undrained strength cud frac14 100

                                                14frac14 71 kNm2

                                                Design bearing resistance Rd frac14 12cudNe area

                                                frac14 12 71 514 42

                                                frac14 7007 kN

                                                For drained shear 0d frac14 tan1tan 32

                                                125

                                                frac14 26

                                                Nq frac14 12 N frac14 10

                                                Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                1 2 100 0175 0700qn 0182qn

                                                2 6 033 0044 0176qn 0046qn

                                                3 10 020 0017 0068qn 0018qn

                                                0246qn

                                                Diameter of equivalent circle B frac14 45m

                                                H

                                                Bfrac14 12

                                                45frac14 27 and A frac14 042

                                                13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                64 Bearing capacity

                                                For sc frac14 30mm

                                                qn frac14 30

                                                0147frac14 204 kN=m2

                                                q frac14 204thorn 21 frac14 225 kN=m2

                                                Design load frac14 42 225 frac14 3600 kN

                                                The design load is 3600 kN settlement being the limiting criterion

                                                87

                                                D

                                                Bfrac14 8

                                                4frac14 20

                                                From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                F frac14 cuNc

                                                Dfrac14 40 71

                                                20 8frac14 18

                                                88

                                                Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                Design value of 0 frac14 tan1tan 38

                                                125

                                                frac14 32

                                                Figure Q86

                                                Bearing capacity 65

                                                For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                For B frac14 250m qn frac14 3750

                                                2502 17 frac14 583 kN=m2

                                                From Figure 510 m frac14 n frac14 126

                                                6frac14 021

                                                Ir frac14 0019

                                                Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                89

                                                Depth (m) N 0v (kNm2) CN N1

                                                070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                Cw frac14 05thorn 0530

                                                47

                                                frac14 082

                                                66 Bearing capacity

                                                Thus

                                                qa frac14 150 082 frac14 120 kN=m2

                                                (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                Thus

                                                qa frac14 90 15 frac14 135 kN=m2

                                                (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                Ic frac14 171

                                                1014frac14 0068

                                                From Equation 819(a) with s frac14 25mm

                                                q frac14 25

                                                3507 0068frac14 150 kN=m2

                                                810

                                                Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                Izp frac14 05thorn 01130

                                                16 27

                                                05

                                                frac14 067

                                                Refer to Figure Q810

                                                E frac14 25qc

                                                Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                Ez (mm3MN)

                                                1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                0203

                                                C1 frac14 1 0500qnfrac14 1 05 12 16

                                                130frac14 093

                                                C2 frac14 1 ethsayTHORN

                                                s frac14 C1C2qnX Iz

                                                Ez frac14 093 1 130 0203 frac14 25mm

                                                Bearing capacity 67

                                                811

                                                At pile base level

                                                cu frac14 220 kN=m2

                                                qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                Then

                                                Qf frac14 Abqb thorn Asqs

                                                frac14

                                                4 32 1980

                                                thorn eth 105 139 86THORN

                                                frac14 13 996thorn 3941 frac14 17 937 kN

                                                0 01 02 03 04 05 06 07

                                                0 2 4 6 8 10 12 14

                                                1

                                                2

                                                3

                                                4

                                                5

                                                6

                                                7

                                                8

                                                (1)

                                                (2)

                                                (3)

                                                (4)

                                                (5)

                                                qc

                                                qc

                                                Iz

                                                Iz

                                                (MNm2)

                                                z (m)

                                                Figure Q810

                                                68 Bearing capacity

                                                Allowable load

                                                ethaTHORN Qf

                                                2frac14 17 937

                                                2frac14 8968 kN

                                                ethbTHORN Abqb

                                                3thorn Asqs frac14 13 996

                                                3thorn 3941 frac14 8606 kN

                                                ie allowable load frac14 8600 kN

                                                Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                According to the limit state method

                                                Characteristic undrained strength at base level cuk frac14 220

                                                150kN=m2

                                                Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                150frac14 1320 kN=m2

                                                Characteristic shaft resistance qsk frac14 00150

                                                frac14 86

                                                150frac14 57 kN=m2

                                                Characteristic base and shaft resistances

                                                Rbk frac14

                                                4 32 1320 frac14 9330 kN

                                                Rsk frac14 105 139 86

                                                150frac14 2629 kN

                                                For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                Design bearing resistance Rcd frac14 9330

                                                160thorn 2629

                                                130

                                                frac14 5831thorn 2022

                                                frac14 7850 kN

                                                Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                812

                                                ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                For a single pile

                                                Qf frac14 Abqb thorn Asqs

                                                frac14

                                                4 062 1305

                                                thorn eth 06 15 42THORN

                                                frac14 369thorn 1187 frac14 1556 kN

                                                Bearing capacity 69

                                                Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                qbkfrac14 9cuk frac14 9 220

                                                150frac14 1320 kN=m2

                                                qskfrac14cuk frac14 040 105

                                                150frac14 28 kN=m2

                                                Rbkfrac14

                                                4 0602 1320 frac14 373 kN

                                                Rskfrac14 060 15 28 frac14 791 kN

                                                Rcdfrac14 373

                                                160thorn 791

                                                130frac14 233thorn 608 frac14 841 kN

                                                Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                q frac14 21 000

                                                1762frac14 68 kN=m2

                                                Immediate settlement

                                                H

                                                Bfrac14 15

                                                176frac14 085

                                                D

                                                Bfrac14 13

                                                176frac14 074

                                                L

                                                Bfrac14 1

                                                Hence from Figure 515

                                                130 frac14 078 and 131 frac14 041

                                                70 Bearing capacity

                                                Thus using Equation 528

                                                si frac14 078 041 68 176

                                                65frac14 6mm

                                                Consolidation settlement

                                                Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                434 (sod)

                                                Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                sc frac14 056 434 frac14 24mm

                                                The total settlement is (6thorn 24) frac14 30mm

                                                813

                                                At base level N frac14 26 Then using Equation 830

                                                qb frac14 40NDb

                                                Bfrac14 40 26 2

                                                025frac14 8320 kN=m2

                                                ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                Figure Q812

                                                Bearing capacity 71

                                                Over the length embedded in sand

                                                N frac14 21 ie18thorn 24

                                                2

                                                Using Equation 831

                                                qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                For a single pile

                                                Qf frac14 Abqb thorn Asqs

                                                frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                For the pile group assuming a group efficiency of 12

                                                XQf frac14 12 9 604 frac14 6523 kN

                                                Then the load factor is

                                                F frac14 6523

                                                2000thorn 1000frac14 21

                                                (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                Characteristic base resistance per unit area qbk frac14 8320

                                                150frac14 5547 kNm2

                                                Characteristic shaft resistance per unit area qsk frac14 42

                                                150frac14 28 kNm2

                                                Characteristic base and shaft resistances for a single pile

                                                Rbk frac14 0252 5547 frac14 347 kN

                                                Rsk frac14 4 025 2 28 frac14 56 kN

                                                For a driven pile the partial factors are b frac14 s frac14 130

                                                Design bearing resistance Rcd frac14 347

                                                130thorn 56

                                                130frac14 310 kN

                                                For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                72 Bearing capacity

                                                N frac14 24thorn 26thorn 34

                                                3frac14 28

                                                Ic frac14 171

                                                2814frac14 0016 ethEquation 818THORN

                                                s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                814

                                                Using Equation 841

                                                Tf frac14 DLcu thorn

                                                4ethD2 d2THORNcuNc

                                                frac14 eth 02 5 06 110THORN thorn

                                                4eth022 012THORN110 9

                                                frac14 207thorn 23 frac14 230 kN

                                                Figure Q813

                                                Bearing capacity 73

                                                Chapter 9

                                                Stability of slopes

                                                91

                                                Referring to Figure Q91

                                                W frac14 417 19 frac14 792 kN=m

                                                Q frac14 20 28 frac14 56 kN=m

                                                Arc lengthAB frac14

                                                180 73 90 frac14 115m

                                                Arc length BC frac14

                                                180 28 90 frac14 44m

                                                The factor of safety is given by

                                                F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                Depth of tension crack z0 frac14 2cu

                                                frac14 2 20

                                                19frac14 21m

                                                Arc length BD frac14

                                                180 13

                                                1

                                                2 90 frac14 21m

                                                F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                92

                                                u frac14 0

                                                Depth factor D frac14 11

                                                9frac14 122

                                                Using Equation 92 with F frac14 10

                                                Ns frac14 cu

                                                FHfrac14 30

                                                10 19 9frac14 0175

                                                Hence from Figure 93

                                                frac14 50

                                                For F frac14 12

                                                Ns frac14 30

                                                12 19 9frac14 0146

                                                frac14 27

                                                93

                                                Refer to Figure Q93

                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                74 m

                                                214 1deg

                                                213 1deg

                                                39 m

                                                WB

                                                D

                                                C

                                                28 m

                                                21 m

                                                A

                                                Q

                                                Soil (1)Soil (2)

                                                73deg

                                                Figure Q91

                                                Stability of slopes 75

                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                599 256 328 1372

                                                Figure Q93

                                                76 Stability of slopes

                                                XW cos frac14 b

                                                Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                W sin frac14 bX

                                                h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                Arc length La frac14

                                                180 57

                                                1

                                                2 326 frac14 327m

                                                The factor of safety is given by

                                                F frac14 c0La thorn tan0ethW cos ulTHORN

                                                W sin

                                                frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                frac14 091

                                                According to the limit state method

                                                0d frac14 tan1tan 32

                                                125

                                                frac14 265

                                                c0 frac14 8

                                                160frac14 5 kN=m2

                                                Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                Design disturbing moment frac14 1075 kN=m

                                                The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                94

                                                F frac14 1

                                                W sin

                                                Xfc0bthorn ethW ubTHORN tan0g sec

                                                1thorn ethtan tan0=FTHORN

                                                c0 frac14 8 kN=m2

                                                0 frac14 32

                                                c0b frac14 8 2 frac14 16 kN=m

                                                W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                Try F frac14 100

                                                tan0

                                                Ffrac14 0625

                                                Stability of slopes 77

                                                Values of u are as obtained in Figure Q93

                                                SliceNo

                                                h(m)

                                                W frac14 bh(kNm)

                                                W sin(kNm)

                                                ub(kNm)

                                                c0bthorn (W ub) tan0(kNm)

                                                sec

                                                1thorn (tan tan0)FProduct(kNm)

                                                1 05 21 6 2 8 24 1078 262 13 55 31

                                                23 33 30 1042 31

                                                3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                224 92 72 0931 67

                                                6 50 210 11 40 100 85 0907 777 55 231 14

                                                12 58 112 90 0889 80

                                                8 60 252 1812

                                                80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                2154 88 116 0853 99

                                                14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                1074 1091

                                                F frac14 1091

                                                1074frac14 102 (assumed value 100)

                                                Thus

                                                F frac14 101

                                                95

                                                F frac14 1

                                                W sin

                                                XfWeth1 ruTHORN tan0g sec

                                                1thorn ethtan tan0THORN=F

                                                0 frac14 33

                                                ru frac14 020

                                                W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                Try F frac14 110

                                                tan 0

                                                Ffrac14 tan 33

                                                110frac14 0590

                                                78 Stability of slopes

                                                Referring to Figure Q95

                                                SliceNo

                                                h(m)

                                                W frac14 bh(kNm)

                                                W sin(kNm)

                                                W(1 ru) tan0(kNm)

                                                sec

                                                1thorn ( tan tan0)FProduct(kNm)

                                                1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                2120 234 0892 209

                                                4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                1185 1271

                                                Figure Q95

                                                Stability of slopes 79

                                                F frac14 1271

                                                1185frac14 107

                                                The trial value was 110 therefore take F to be 108

                                                96

                                                (a) Water table at surface the factor of safety is given by Equation 912

                                                F frac14 0

                                                sat

                                                tan0

                                                tan

                                                ptie 15 frac14 92

                                                19

                                                tan 36

                                                tan

                                                tan frac14 0234

                                                frac14 13

                                                Water table well below surface the factor of safety is given by Equation 911

                                                F frac14 tan0

                                                tan

                                                frac14 tan 36

                                                tan 13

                                                frac14 31

                                                (b) 0d frac14 tan1tan 36

                                                125

                                                frac14 30

                                                Depth of potential failure surface frac14 z

                                                Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                frac14 504z kN

                                                Design disturbing moment per unit area Sd frac14 sat sin cos

                                                frac14 19 z sin 13 cos 13

                                                frac14 416z kN

                                                Rd gtSd therefore the limit state for overall stability is satisfied

                                                80 Stability of slopes

                                                • Book Cover
                                                • Title
                                                • Contents
                                                • Basic characteristics of soils
                                                • Seepage
                                                • Effective stress
                                                • Shear strength
                                                • Stresses and displacements
                                                • Lateral earth pressure
                                                • Consolidation theory
                                                • Bearing capacity
                                                • Stability of slopes

                                                  37

                                                  Situation (1)(a)

                                                  frac14 3w thorn 2sat frac14 eth3 98THORN thorn eth2 20THORN frac14 694 kN=m2

                                                  u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                                  0 frac14 u frac14 694 392 frac14 302 kN=m2

                                                  (b)

                                                  i frac14 2

                                                  4frac14 05

                                                  j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 thorn jTHORN frac14 2eth102thorn 49THORN frac14 302 kN=m2

                                                  Situation (2)(a)

                                                  frac14 1w thorn 2sat frac14 eth1 98THORN thorn eth2 20THORN frac14 498 kN=m2

                                                  u frac14 wethh zTHORN frac14 98f1 eth3THORNg frac14 392 kN=m2

                                                  0 frac14 u frac14 498 392 frac14 106 kN=m2

                                                  (b)

                                                  i frac14 2

                                                  4frac14 05

                                                  j frac14 iw frac14 05 98 frac14 49 kN=m3 0 frac14 2eth0 jTHORN frac14 2eth102 49THORN frac14 106 kN=m2

                                                  38

                                                  The flow net is drawn in Figure Q24

                                                  Loss in total head between adjacent equipotentials

                                                  h frac14 550

                                                  Ndfrac14 550

                                                  11frac14 050m

                                                  Exit hydraulic gradient

                                                  ie frac14 h

                                                  sfrac14 050

                                                  070frac14 071

                                                  Effective stress 19

                                                  The critical hydraulic gradient is given by Equation 39

                                                  ic frac14 0

                                                  wfrac14 102

                                                  98frac14 104

                                                  Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                                                  F frac14 iciefrac14 104

                                                  071frac14 15

                                                  Total head at C

                                                  hC frac14 nd

                                                  Ndh frac14 24

                                                  11 550 frac14 120m

                                                  Elevation head at C

                                                  zC frac14 250m

                                                  Pore water pressure at C

                                                  uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                                                  Therefore effective vertical stress at C

                                                  0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                                                  For point D

                                                  hD frac14 73

                                                  11 550 frac14 365m

                                                  zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                                                  0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                                                  39

                                                  The flow net is drawn in Figure Q25

                                                  For a soil prism 150 300m adjacent to the piling

                                                  hm frac14 26

                                                  9 500 frac14 145m

                                                  20 Effective stress

                                                  Factor of safety against lsquoheavingrsquo (Equation 310)

                                                  F frac14 ic

                                                  imfrac14 0d

                                                  whmfrac14 97 300

                                                  98 145frac14 20

                                                  With a filter

                                                  F frac14 0d thorn wwhm

                                                  3 frac14 eth97 300THORN thorn w98 145

                                                  w frac14 135 kN=m2

                                                  Depth of filterfrac14 13521frac14 065m (if above water level)

                                                  Effective stress 21

                                                  Chapter 4

                                                  Shear strength

                                                  41

                                                  frac14 295 kN=m2

                                                  u frac14 120 kN=m2

                                                  0 frac14 u frac14 295 120 frac14 175 kN=m2

                                                  f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                                                  42

                                                  03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                                                  100 452 552200 908 1108400 1810 2210800 3624 4424

                                                  The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                                                  Figure Q42

                                                  43

                                                  3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                                                  200 222 422400 218 618600 220 820

                                                  The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                                                  44

                                                  The modified shear strength parameters are

                                                  0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                                                  a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                                                  The coordinates of the stress point representing failure conditions in the test are

                                                  1

                                                  2eth1 2THORN frac14 1

                                                  2 170 frac14 85 kN=m2

                                                  1

                                                  2eth1 thorn 3THORN frac14 1

                                                  2eth270thorn 100THORN frac14 185 kN=m2

                                                  The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                                                  uf frac14 36 kN=m2

                                                  Figure Q43

                                                  Figure Q44

                                                  Shear strength 23

                                                  45

                                                  3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                                  150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                                  The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                                  The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                                  46

                                                  03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                                  200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                                  The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                                  Figure Q45

                                                  24 Shear strength

                                                  47

                                                  The torque required to produce shear failure is given by

                                                  T frac14 dh cud

                                                  2thorn 2

                                                  Z d=2

                                                  0

                                                  2r drcur

                                                  frac14 cud2h

                                                  2thorn 4cu

                                                  Z d=2

                                                  0

                                                  r2dr

                                                  frac14 cud2h

                                                  2thorn d

                                                  3

                                                  6

                                                  Then

                                                  35 frac14 cu52 10

                                                  2thorn 53

                                                  6

                                                  103

                                                  cu frac14 76 kN=m3

                                                  400

                                                  0 400 800 1200 1600

                                                  τ (k

                                                  Nm

                                                  2 )

                                                  σprime (kNm2)

                                                  34deg

                                                  315deg29deg

                                                  (a)

                                                  (b)

                                                  0 400

                                                  400

                                                  800 1200 1600

                                                  Failure envelope

                                                  300 500

                                                  σprime (kNm2)

                                                  τ (k

                                                  Nm

                                                  2 )

                                                  20 (kNm2)

                                                  31deg

                                                  Figure Q46

                                                  Shear strength 25

                                                  48

                                                  The relevant stress values are calculated as follows

                                                  3 frac14 600 kN=m2

                                                  1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                                  2(1 3) 0 40 79 107 139 159

                                                  1

                                                  2(01 thorn 03) 400 411 402 389 351 326

                                                  1

                                                  2(1 thorn 3) 600 640 679 707 739 759

                                                  The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                                  Af frac14 433 200

                                                  319frac14 073

                                                  49

                                                  B frac14 u33

                                                  frac14 144

                                                  350 200frac14 096

                                                  a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                                  0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                                  10 283 65 023

                                                  Figure Q48

                                                  26 Shear strength

                                                  The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                                  Figure Q49

                                                  Shear strength 27

                                                  Chapter 5

                                                  Stresses and displacements

                                                  51

                                                  Vertical stress is given by

                                                  z frac14 Qz2Ip frac14 5000

                                                  52Ip

                                                  Values of Ip are obtained from Table 51

                                                  r (m) rz Ip z (kNm2)

                                                  0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                                  10 20 0009 2

                                                  The variation of z with radial distance (r) is plotted in Figure Q51

                                                  Figure Q51

                                                  52

                                                  Below the centre load (Figure Q52)

                                                  r

                                                  zfrac14 0 for the 7500-kN load

                                                  Ip frac14 0478

                                                  r

                                                  zfrac14 5

                                                  4frac14 125 for the 10 000- and 9000-kN loads

                                                  Ip frac14 0045

                                                  Then

                                                  z frac14X Q

                                                  z2Ip

                                                  frac14 7500 0478

                                                  42thorn 10 000 0045

                                                  42thorn 9000 0045

                                                  42

                                                  frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                                  53

                                                  The vertical stress under a corner of a rectangular area is given by

                                                  z frac14 qIr

                                                  where values of Ir are obtained from Figure 510 In this case

                                                  z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                                  z

                                                  Figure Q52

                                                  Stresses and displacements 29

                                                  z (m) m n Ir z (kNm2)

                                                  0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                                  10 010 0005 5

                                                  z is plotted against z in Figure Q53

                                                  54

                                                  (a)

                                                  m frac14 125

                                                  12frac14 104

                                                  n frac14 18

                                                  12frac14 150

                                                  From Figure 510 Irfrac14 0196

                                                  z frac14 2 175 0196 frac14 68 kN=m2

                                                  Figure Q53

                                                  30 Stresses and displacements

                                                  (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                                  z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                                  55

                                                  Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                                  Px frac14 2Q

                                                  1

                                                  m2 thorn 1frac14 2 150

                                                  125frac14 76 kN=m

                                                  Equation 517 is used to obtain the pressure distribution

                                                  px frac14 4Q

                                                  h

                                                  m2n

                                                  ethm2 thorn n2THORN2 frac14150

                                                  m2n

                                                  ethm2 thorn n2THORN2 ethkN=m2THORN

                                                  Figure Q54

                                                  Stresses and displacements 31

                                                  n m2n

                                                  (m2 thorn n2)2

                                                  px(kNm2)

                                                  0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                  The pressure distribution is plotted in Figure Q55

                                                  56

                                                  H

                                                  Bfrac14 10

                                                  2frac14 5

                                                  L

                                                  Bfrac14 4

                                                  2frac14 2

                                                  D

                                                  Bfrac14 1

                                                  2frac14 05

                                                  Hence from Figure 515

                                                  131 frac14 082

                                                  130 frac14 094

                                                  Figure Q55

                                                  32 Stresses and displacements

                                                  The immediate settlement is given by Equation 528

                                                  si frac14 130131qB

                                                  Eu

                                                  frac14 094 082 200 2

                                                  45frac14 7mm

                                                  Stresses and displacements 33

                                                  Chapter 6

                                                  Lateral earth pressure

                                                  61

                                                  For 0 frac14 37 the active pressure coefficient is given by

                                                  Ka frac14 1 sin 37

                                                  1thorn sin 37frac14 025

                                                  The total active thrust (Equation 66a with c0 frac14 0) is

                                                  Pa frac14 1

                                                  2KaH

                                                  2 frac14 1

                                                  2 025 17 62 frac14 765 kN=m

                                                  If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                  K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                  and the thrust on the wall is

                                                  P0 frac14 1

                                                  2K0H

                                                  2 frac14 1

                                                  2 040 17 62 frac14 122 kN=m

                                                  62

                                                  The active pressure coefficients for the three soil types are as follows

                                                  Ka1 frac141 sin 35

                                                  1thorn sin 35frac14 0271

                                                  Ka2 frac141 sin 27

                                                  1thorn sin 27frac14 0375

                                                  ffiffiffiffiffiffiffiKa2

                                                  p frac14 0613

                                                  Ka3 frac141 sin 42

                                                  1thorn sin 42frac14 0198

                                                  Distribution of active pressure (plotted in Figure Q62)

                                                  Depth (m) Soil Active pressure (kNm2)

                                                  3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                  12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                  At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                  Total thrust frac14 571 kNm

                                                  Point of application is (4893571) m from the top of the wall ie 857m

                                                  Force (kN) Arm (m) Moment (kN m)

                                                  (1)1

                                                  2 0271 16 32 frac14 195 20 390

                                                  (2) 0271 16 3 2 frac14 260 40 1040

                                                  (3)1

                                                  2 0271 92 22 frac14 50 433 217

                                                  (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                  (5)1

                                                  2 0375 102 32 frac14 172 70 1204

                                                  (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                  (7)1

                                                  2 0198 112 42 frac14 177 1067 1889

                                                  (8)1

                                                  2 98 92 frac14 3969 90 35721

                                                  5713 48934

                                                  Figure Q62

                                                  Lateral earth pressure 35

                                                  63

                                                  (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                  Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                  frac14 245

                                                  At the lower end of the piling

                                                  pa frac14 Kaqthorn Kasatz Kaccu

                                                  frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                  frac14 115 kN=m2

                                                  pp frac14 Kpsatzthorn Kpccu

                                                  frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                  frac14 202 kN=m2

                                                  (b) For 0 frac14 26 and frac14 1

                                                  20

                                                  Ka frac14 035

                                                  Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                  pfrac14 145 ethEquation 619THORN

                                                  Kp frac14 37

                                                  Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                  pfrac14 47 ethEquation 624THORN

                                                  At the lower end of the piling

                                                  pa frac14 Kaqthorn Ka0z Kacc

                                                  0

                                                  frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                  frac14 187 kN=m2

                                                  pp frac14 Kp0zthorn Kpcc

                                                  0

                                                  frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                  frac14 198 kN=m2

                                                  36 Lateral earth pressure

                                                  64

                                                  (a) For 0 frac14 38 Ka frac14 024

                                                  0 frac14 20 98 frac14 102 kN=m3

                                                  The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                  Force (kN) Arm (m) Moment (kN m)

                                                  (1) 024 10 66 frac14 159 33 525

                                                  (2)1

                                                  2 024 17 392 frac14 310 400 1240

                                                  (3) 024 17 39 27 frac14 430 135 580

                                                  (4)1

                                                  2 024 102 272 frac14 89 090 80

                                                  (5)1

                                                  2 98 272 frac14 357 090 321

                                                  Hfrac14 1345 MH frac14 2746

                                                  (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                  (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                  XM frac14MV MH frac14 7790 kNm

                                                  Lever arm of base resultant

                                                  M

                                                  Vfrac14 779

                                                  488frac14 160

                                                  Eccentricity of base resultant

                                                  e frac14 200 160 frac14 040m

                                                  39 m

                                                  27 m

                                                  40 m

                                                  04 m

                                                  04 m

                                                  26 m

                                                  (7)

                                                  (9)

                                                  (1)(2)

                                                  (3)

                                                  (4)

                                                  (5)

                                                  (8)(6)

                                                  (10)

                                                  WT

                                                  10 kNm2

                                                  Hydrostatic

                                                  Figure Q64

                                                  Lateral earth pressure 37

                                                  Base pressures (Equation 627)

                                                  p frac14 VB

                                                  1 6e

                                                  B

                                                  frac14 488

                                                  4eth1 060THORN

                                                  frac14 195 kN=m2 and 49 kN=m2

                                                  Factor of safety against sliding (Equation 628)

                                                  F frac14 V tan

                                                  Hfrac14 488 tan 25

                                                  1345frac14 17

                                                  (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                  Hfrac14 1633 kN

                                                  V frac14 4879 kN

                                                  MH frac14 3453 kNm

                                                  MV frac14 10536 kNm

                                                  The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                  65

                                                  For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                  Kp

                                                  Ffrac14 385

                                                  2

                                                  0 frac14 20 98 frac14 102 kN=m3

                                                  The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                  Force (kN) Arm (m) Moment (kN m)

                                                  (1)1

                                                  2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                  (2) 026 17 45 d frac14 199d d2 995d2

                                                  (3)1

                                                  2 026 102 d2 frac14 133d2 d3 044d3

                                                  (4)1

                                                  2 385

                                                  2 17 152 frac14 368 dthorn 05 368d 184

                                                  (5)385

                                                  2 17 15 d frac14 491d d2 2455d2

                                                  (6)1

                                                  2 385

                                                  2 102 d2 frac14 982d2 d3 327d3

                                                  38 Lateral earth pressure

                                                  XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                  d3 thorn 516d2 283d 1724 frac14 0

                                                  d frac14 179m

                                                  Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                  Over additional 20 embedded depth

                                                  pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                  Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                  66

                                                  The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                  Ka frac14 sin 69=sin 105

                                                  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                  pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                  26664

                                                  37775

                                                  2

                                                  frac14 050

                                                  The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                  Pa frac14 1

                                                  2 050 19 7502 frac14 267 kN=m

                                                  Figure Q65

                                                  Lateral earth pressure 39

                                                  Horizontal component

                                                  Ph frac14 267 cos 40 frac14 205 kN=m

                                                  Vertical component

                                                  Pv frac14 267 sin 40 frac14 172 kN=m

                                                  Consider moments about the toe of the wall (Figure Q66) (per m)

                                                  Force (kN) Arm (m) Moment (kN m)

                                                  (1)1

                                                  2 175 650 235 frac14 1337 258 345

                                                  (2) 050 650 235 frac14 764 175 134

                                                  (3)1

                                                  2 070 650 235 frac14 535 127 68

                                                  (4) 100 400 235 frac14 940 200 188

                                                  (5) 1

                                                  2 080 050 235 frac14 47 027 1

                                                  Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                  Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                  Lever arm of base resultant

                                                  M

                                                  Vfrac14 795

                                                  525frac14 151m

                                                  Eccentricity of base resultant

                                                  e frac14 200 151 frac14 049m

                                                  Figure Q66

                                                  40 Lateral earth pressure

                                                  Base pressures (Equation 627)

                                                  p frac14 525

                                                  41 6 049

                                                  4

                                                  frac14 228 kN=m2 and 35 kN=m2

                                                  The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                  The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                  The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                  67

                                                  For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                  Force (kN) Arm (m) Moment (kNm)

                                                  (1)1

                                                  2 027 17 52 frac14 574 183 1050

                                                  (2) 027 17 5 3 frac14 689 500 3445

                                                  (3)1

                                                  2 027 102 32 frac14 124 550 682

                                                  (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                  (5)1

                                                  2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                  (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                  (7) 1

                                                  2 267

                                                  2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                  (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                  2 d frac14 163d d2thorn 650 82d2 1060d

                                                  Tie rod force per m frac14 T 0 0

                                                  XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                  d3 thorn 77d2 269d 1438 frac14 0

                                                  d frac14 467m

                                                  Depth of penetration frac14 12d frac14 560m

                                                  Lateral earth pressure 41

                                                  Algebraic sum of forces for d frac14 467m isX

                                                  F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                  T frac14 905 kN=m

                                                  Force in each tie rod frac14 25T frac14 226 kN

                                                  68

                                                  (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                  0 frac14 21 98 frac14 112 kN=m3

                                                  The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                  uC frac14 150

                                                  165 15 98 frac14 134 kN=m2

                                                  The average seepage pressure is

                                                  j frac14 15

                                                  165 98 frac14 09 kN=m3

                                                  Hence

                                                  0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                  0 j frac14 112 09 frac14 103 kN=m3

                                                  Figure Q67

                                                  42 Lateral earth pressure

                                                  Consider moments about the anchor point A (per m)

                                                  Force (kN) Arm (m) Moment (kN m)

                                                  (1) 10 026 150 frac14 390 60 2340

                                                  (2)1

                                                  2 026 18 452 frac14 474 15 711

                                                  (3) 026 18 45 105 frac14 2211 825 18240

                                                  (4)1

                                                  2 026 121 1052 frac14 1734 100 17340

                                                  (5)1

                                                  2 134 15 frac14 101 40 404

                                                  (6) 134 30 frac14 402 60 2412

                                                  (7)1

                                                  2 134 60 frac14 402 95 3819

                                                  571 4527(8) Ppm

                                                  115 115PPm

                                                  XM frac14 0

                                                  Ppm frac144527

                                                  115frac14 394 kN=m

                                                  Available passive resistance

                                                  Pp frac14 1

                                                  2 385 103 62 frac14 714 kN=m

                                                  Factor of safety

                                                  Fp frac14 Pp

                                                  Ppm

                                                  frac14 714

                                                  394frac14 18

                                                  Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                  Figure Q68

                                                  Lateral earth pressure 43

                                                  (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                  Consider moments (per m) about the tie point A

                                                  Force (kN) Arm (m)

                                                  (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                  (2)1

                                                  2 033 18 452 frac14 601 15

                                                  (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                  (4)1

                                                  2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                  (5)1

                                                  2 134 15 frac14 101 40

                                                  (6) 134 30 frac14 402 60

                                                  (7)1

                                                  2 134 d frac14 67d d3thorn 75

                                                  (8) 1

                                                  2 30 103 d2 frac141545d2 2d3thorn 75

                                                  Moment (kN m)

                                                  (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                  XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                  d3 thorn 827d2 466d 1518 frac14 0

                                                  By trial

                                                  d frac14 544m

                                                  The minimum depth of embedment required is 544m

                                                  69

                                                  For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                  0 frac14 20 98 frac14 102 kN=m3

                                                  The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                  44 Lateral earth pressure

                                                  uC frac14 147

                                                  173 26 98 frac14 216 kN=m2

                                                  and the average seepage pressure around the wall is

                                                  j frac14 26

                                                  173 98 frac14 15 kN=m3

                                                  Consider moments about the prop (A) (per m)

                                                  Force (kN) Arm (m) Moment (kN m)

                                                  (1)1

                                                  2 03 17 272 frac14 186 020 37

                                                  (2) 03 17 27 53 frac14 730 335 2445

                                                  (3)1

                                                  2 03 (102thorn 15) 532 frac14 493 423 2085

                                                  (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                  (5)1

                                                  2 216 26 frac14 281 243 684

                                                  (6) 216 27 frac14 583 465 2712

                                                  (7)1

                                                  2 216 60 frac14 648 800 5184

                                                  3055(8)

                                                  1

                                                  2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                  Factor of safety

                                                  Fr frac14 6885

                                                  3055frac14 225

                                                  Figure Q69

                                                  Lateral earth pressure 45

                                                  610

                                                  For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                  p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                  Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                  Using the recommendations of Twine and Roscoe

                                                  p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                  Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                  611

                                                  frac14 18 kN=m3 0 frac14 34

                                                  H frac14 350m nH frac14 335m mH frac14 185m

                                                  Consider a trial value of F frac14 20 Refer to Figure 635

                                                  0m frac14 tan1tan 34

                                                  20

                                                  frac14 186

                                                  Then

                                                  frac14 45 thorn 0m2frac14 543

                                                  W frac14 1

                                                  2 18 3502 cot 543 frac14 792 kN=m

                                                  Figure Q610

                                                  46 Lateral earth pressure

                                                  P frac14 1

                                                  2 s 3352 frac14 561s kN=m

                                                  U frac14 1

                                                  2 98 1852 cosec 543 frac14 206 kN=m

                                                  Equations 630 and 631 then become

                                                  561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                  792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                  ie

                                                  561s 0616N 405 frac14 0

                                                  792 0857N thorn 563 frac14 0

                                                  N frac14 848

                                                  0857frac14 989 kN=m

                                                  Then

                                                  561s 609 405 frac14 0

                                                  s frac14 649

                                                  561frac14 116 kN=m3

                                                  The calculations for trial values of F of 20 15 and 10 are summarized below

                                                  F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                  20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                  s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                  Figure Q611

                                                  Lateral earth pressure 47

                                                  612

                                                  For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                  45 thorn 0

                                                  2frac14 63

                                                  For the retained material between the surface and a depth of 36m

                                                  Pa frac14 1

                                                  2 030 18 362 frac14 350 kN=m

                                                  Weight of reinforced fill between the surface and a depth of 36m is

                                                  Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                  eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                  Eccentricity of Rv

                                                  e frac14 263 250 frac14 013m

                                                  The average vertical stress at a depth of 36m is

                                                  z frac14 Rv

                                                  L 2efrac14 324

                                                  474frac14 68 kN=m2

                                                  (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                  Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                  Tensile stress in the element frac14 138 103

                                                  65 3frac14 71N=mm2

                                                  Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                  Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                  Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                  The weight of ABC is

                                                  W frac14 1

                                                  2 18 52 265 frac14 124 kN=m

                                                  From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                  48 Lateral earth pressure

                                                  (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                  Tp frac14 032 68 120 065 frac14 170 kN

                                                  Tr frac14 213 420

                                                  418frac14 214 kN

                                                  Again the tensile failure and slipping limit states are satisfied for this element

                                                  Figure Q612

                                                  Lateral earth pressure 49

                                                  Chapter 7

                                                  Consolidation theory

                                                  71

                                                  Total change in thickness

                                                  H frac14 782 602 frac14 180mm

                                                  Average thickness frac14 1530thorn 180

                                                  2frac14 1620mm

                                                  Length of drainage path d frac14 1620

                                                  2frac14 810mm

                                                  Root time plot (Figure Q71a)

                                                  ffiffiffiffiffiffit90p frac14 33

                                                  t90 frac14 109min

                                                  cv frac14 0848d2

                                                  t90frac14 0848 8102

                                                  109 1440 365

                                                  106frac14 27m2=year

                                                  r0 frac14 782 764

                                                  782 602frac14 018

                                                  180frac14 0100

                                                  rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                  10 119

                                                  9 180frac14 0735

                                                  rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                  Log time plot (Figure Q71b)

                                                  t50 frac14 26min

                                                  cv frac14 0196d2

                                                  t50frac14 0196 8102

                                                  26 1440 365

                                                  106frac14 26m2=year

                                                  r0 frac14 782 763

                                                  782 602frac14 019

                                                  180frac14 0106

                                                  rp frac14 763 623

                                                  782 602frac14 140

                                                  180frac14 0778

                                                  rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                  Figure Q71(a)

                                                  Figure Q71(b)

                                                  Final void ratio

                                                  e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                  e

                                                  Hfrac14 1thorn e0

                                                  H0frac14 1thorn e1 thorne

                                                  H0

                                                  ie

                                                  e

                                                  180frac14 1631thorne

                                                  1710

                                                  e frac14 2936

                                                  1530frac14 0192

                                                  Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                  mv frac14 1

                                                  1thorn e0 e0 e101 00

                                                  frac14 1

                                                  1823 0192

                                                  0107frac14 098m2=MN

                                                  k frac14 cvmvw frac14 265 098 98

                                                  60 1440 365 103frac14 81 1010 m=s

                                                  72

                                                  Using Equation 77 (one-dimensional method)

                                                  sc frac14 e0 e11thorn e0 H

                                                  Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                  Figure Q72

                                                  52 Consolidation theory

                                                  Settlement

                                                  Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                  1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                  318

                                                  Notes 5 92y 460thorn 84

                                                  Heave

                                                  Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                  1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                  38

                                                  73

                                                  U frac14 f ethTvTHORN frac14 f cvt

                                                  d2

                                                  Hence if cv is constant

                                                  t1

                                                  t2frac14 d

                                                  21

                                                  d22

                                                  where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                  d1 frac14 95mm and d2 frac14 2500mm

                                                  for U frac14 050 t2 frac14 t1 d22

                                                  d21

                                                  frac14 20

                                                  60 24 365 25002

                                                  952frac14 263 years

                                                  for U lt 060 Tv frac14

                                                  4U2 (Equation 724(a))

                                                  t030 frac14 t050 0302

                                                  0502

                                                  frac14 263 036 frac14 095 years

                                                  Consolidation theory 53

                                                  74

                                                  The layer is open

                                                  d frac14 8

                                                  2frac14 4m

                                                  Tv frac14 cvtd2frac14 24 3

                                                  42frac14 0450

                                                  ui frac14 frac14 84 kN=m2

                                                  The excess pore water pressure is given by Equation 721

                                                  ue frac14Xmfrac141mfrac140

                                                  2ui

                                                  Msin

                                                  Mz

                                                  d

                                                  expethM2TvTHORN

                                                  In this case z frac14 d

                                                  sinMz

                                                  d

                                                  frac14 sinM

                                                  where

                                                  M frac14

                                                  23

                                                  25

                                                  2

                                                  M sin M M2Tv exp (M2Tv)

                                                  2thorn1 1110 0329

                                                  3

                                                  21 9993 457 105

                                                  ue frac14 2 84 2

                                                  1 0329 ethother terms negligibleTHORN

                                                  frac14 352 kN=m2

                                                  75

                                                  The layer is open

                                                  d frac14 6

                                                  2frac14 3m

                                                  Tv frac14 cvtd2frac14 10 3

                                                  32frac14 0333

                                                  The layer thickness will be divided into six equal parts ie m frac14 6

                                                  54 Consolidation theory

                                                  For an open layer

                                                  Tv frac14 4n

                                                  m2

                                                  n frac14 0333 62

                                                  4frac14 300

                                                  The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                  ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                  i j

                                                  0 1 2 3 4 5 6 7 8 9 10 11 12

                                                  0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                  The initial and 3-year isochrones are plotted in Figure Q75

                                                  Area under initial isochrone frac14 180 units

                                                  Area under 3-year isochrone frac14 63 units

                                                  The average degree of consolidation is given by Equation 725Thus

                                                  U frac14 1 63

                                                  180frac14 065

                                                  Figure Q75

                                                  Consolidation theory 55

                                                  76

                                                  At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                  0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                  The final consolidation settlement (one-dimensional method) is

                                                  sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                  Corrected time t frac14 2 1

                                                  2

                                                  40

                                                  52

                                                  frac14 1615 years

                                                  Tv frac14 cvtd2frac14 44 1615

                                                  42frac14 0444

                                                  From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                  77

                                                  The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                  Figure Q77

                                                  56 Consolidation theory

                                                  Point m n Ir (kNm2) sc (mm)

                                                  13020frac14 15 20

                                                  20frac14 10 0194 (4) 113 124

                                                  260

                                                  20frac14 30

                                                  20

                                                  20frac14 10 0204 (2) 59 65

                                                  360

                                                  20frac14 30

                                                  40

                                                  20frac14 20 0238 (1) 35 38

                                                  430

                                                  20frac14 15

                                                  40

                                                  20frac14 20 0224 (2) 65 72

                                                  Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                  78

                                                  Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                  (a) Immediate settlement

                                                  H

                                                  Bfrac14 30

                                                  35frac14 086

                                                  D

                                                  Bfrac14 2

                                                  35frac14 006

                                                  Figure Q78

                                                  Consolidation theory 57

                                                  From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                  si frac14 130131qB

                                                  Eufrac14 10 032 105 35

                                                  40frac14 30mm

                                                  (b) Consolidation settlement

                                                  Layer z (m) Dz Ic (kNm2) syod (mm)

                                                  1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                  3150

                                                  Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                  Now

                                                  H

                                                  Bfrac14 30

                                                  35frac14 086 and A frac14 065

                                                  from Figure 712 13 frac14 079

                                                  sc frac14 13sod frac14 079 315 frac14 250mm

                                                  Total settlement

                                                  s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                  79

                                                  Without sand drains

                                                  Uv frac14 025

                                                  Tv frac14 0049 ethfrom Figure 718THORN

                                                  t frac14 Tvd2

                                                  cvfrac14 0049 82

                                                  cvWith sand drains

                                                  R frac14 0564S frac14 0564 3 frac14 169m

                                                  n frac14 Rrfrac14 169

                                                  015frac14 113

                                                  Tr frac14 cht

                                                  4R2frac14 ch

                                                  4 1692 0049 82

                                                  cvethand ch frac14 cvTHORN

                                                  frac14 0275

                                                  Ur frac14 073 (from Figure 730)

                                                  58 Consolidation theory

                                                  Using Equation 740

                                                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                  U frac14 080

                                                  710

                                                  Without sand drains

                                                  Uv frac14 090

                                                  Tv frac14 0848

                                                  t frac14 Tvd2

                                                  cvfrac14 0848 102

                                                  96frac14 88 years

                                                  With sand drains

                                                  R frac14 0564S frac14 0564 4 frac14 226m

                                                  n frac14 Rrfrac14 226

                                                  015frac14 15

                                                  Tr

                                                  Tvfrac14 chcv

                                                  d2

                                                  4R2ethsame tTHORN

                                                  Tr

                                                  Tvfrac14 140

                                                  96 102

                                                  4 2262frac14 714 eth1THORN

                                                  Using Equation 740

                                                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                  An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                  Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                  040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                  Thus

                                                  Uv frac14 0295 and Ur frac14 086

                                                  t frac14 88 00683

                                                  0848frac14 07 years

                                                  Consolidation theory 59

                                                  Chapter 8

                                                  Bearing capacity

                                                  81

                                                  (a) The ultimate bearing capacity is given by Equation 83

                                                  qf frac14 cNc thorn DNq thorn 1

                                                  2BN

                                                  For u frac14 0

                                                  Nc frac14 514 Nq frac14 1 N frac14 0

                                                  qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                  The net ultimate bearing capacity is

                                                  qnf frac14 qf D frac14 540 kN=m2

                                                  The net foundation pressure is

                                                  qn frac14 q D frac14 425

                                                  2 eth21 1THORN frac14 192 kN=m2

                                                  The factor of safety (Equation 86) is

                                                  F frac14 qnfqnfrac14 540

                                                  192frac14 28

                                                  (b) For 0 frac14 28

                                                  Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                  qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                  2 112 2 13

                                                  frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                  qnf frac14 574 112 frac14 563 kN=m2

                                                  F frac14 563

                                                  192frac14 29

                                                  (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                  82

                                                  For 0 frac14 38

                                                  Nq frac14 49 N frac14 67

                                                  qnf frac14 DethNq 1THORN thorn 1

                                                  2BN ethfrom Equation 83THORN

                                                  frac14 eth18 075 48THORN thorn 1

                                                  2 18 15 67

                                                  frac14 648thorn 905 frac14 1553 kN=m2

                                                  qn frac14 500

                                                  15 eth18 075THORN frac14 320 kN=m2

                                                  F frac14 qnfqnfrac14 1553

                                                  320frac14 48

                                                  0d frac14 tan1tan 38

                                                  125

                                                  frac14 32 therefore Nq frac14 23 and N frac14 25

                                                  Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                  2 18 15 25

                                                  frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                  Design load (action) Vd frac14 500 kN=m

                                                  The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                  83

                                                  D

                                                  Bfrac14 350

                                                  225frac14 155

                                                  From Figure 85 for a square foundation

                                                  Nc frac14 81

                                                  Bearing capacity 61

                                                  For a rectangular foundation (L frac14 450m B frac14 225m)

                                                  Nc frac14 084thorn 016B

                                                  L

                                                  81 frac14 745

                                                  Using Equation 810

                                                  qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                  For F frac14 3

                                                  qn frac14 1006

                                                  3frac14 335 kN=m2

                                                  q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                  Design load frac14 405 450 225 frac14 4100 kN

                                                  Design undrained strength cud frac14 135

                                                  14frac14 96 kN=m2

                                                  Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                  frac14 7241 kN

                                                  Design load Vd frac14 4100 kN

                                                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                  84

                                                  For 0 frac14 40

                                                  Nq frac14 64 N frac14 95

                                                  qnf frac14 DethNq 1THORN thorn 04BN

                                                  (a) Water table 5m below ground level

                                                  qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                  qn frac14 400 17 frac14 383 kN=m2

                                                  F frac14 2686

                                                  383frac14 70

                                                  (b) Water table 1m below ground level (ie at foundation level)

                                                  0 frac14 20 98 frac14 102 kN=m3

                                                  62 Bearing capacity

                                                  qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                  F frac14 2040

                                                  383frac14 53

                                                  (c) Water table at ground level with upward hydraulic gradient 02

                                                  eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                  qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                  F frac14 1296

                                                  392frac14 33

                                                  85

                                                  The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                  Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                  Design value of 0 frac14 tan1tan 39

                                                  125

                                                  frac14 33

                                                  For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                  Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                  86

                                                  (a) Undrained shear for u frac14 0

                                                  Nc frac14 514 Nq frac14 1 N frac14 0

                                                  qnf frac14 12cuNc

                                                  frac14 12 100 514 frac14 617 kN=m2

                                                  qn frac14 qnfFfrac14 617

                                                  3frac14 206 kN=m2

                                                  q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                  Bearing capacity 63

                                                  Drained shear for 0 frac14 32

                                                  Nq frac14 23 N frac14 25

                                                  0 frac14 21 98 frac14 112 kN=m3

                                                  qnf frac14 0DethNq 1THORN thorn 040BN

                                                  frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                  frac14 694 kN=m2

                                                  q frac14 694

                                                  3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                  Design load frac14 42 227 frac14 3632 kN

                                                  (b) Design undrained strength cud frac14 100

                                                  14frac14 71 kNm2

                                                  Design bearing resistance Rd frac14 12cudNe area

                                                  frac14 12 71 514 42

                                                  frac14 7007 kN

                                                  For drained shear 0d frac14 tan1tan 32

                                                  125

                                                  frac14 26

                                                  Nq frac14 12 N frac14 10

                                                  Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                  (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                  Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                  1 2 100 0175 0700qn 0182qn

                                                  2 6 033 0044 0176qn 0046qn

                                                  3 10 020 0017 0068qn 0018qn

                                                  0246qn

                                                  Diameter of equivalent circle B frac14 45m

                                                  H

                                                  Bfrac14 12

                                                  45frac14 27 and A frac14 042

                                                  13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                  64 Bearing capacity

                                                  For sc frac14 30mm

                                                  qn frac14 30

                                                  0147frac14 204 kN=m2

                                                  q frac14 204thorn 21 frac14 225 kN=m2

                                                  Design load frac14 42 225 frac14 3600 kN

                                                  The design load is 3600 kN settlement being the limiting criterion

                                                  87

                                                  D

                                                  Bfrac14 8

                                                  4frac14 20

                                                  From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                  F frac14 cuNc

                                                  Dfrac14 40 71

                                                  20 8frac14 18

                                                  88

                                                  Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                  Design value of 0 frac14 tan1tan 38

                                                  125

                                                  frac14 32

                                                  Figure Q86

                                                  Bearing capacity 65

                                                  For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                  Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                  The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                  Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                  For B frac14 250m qn frac14 3750

                                                  2502 17 frac14 583 kN=m2

                                                  From Figure 510 m frac14 n frac14 126

                                                  6frac14 021

                                                  Ir frac14 0019

                                                  Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                  Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                  The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                  89

                                                  Depth (m) N 0v (kNm2) CN N1

                                                  070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                  Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                  (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                  Cw frac14 05thorn 0530

                                                  47

                                                  frac14 082

                                                  66 Bearing capacity

                                                  Thus

                                                  qa frac14 150 082 frac14 120 kN=m2

                                                  (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                  Thus

                                                  qa frac14 90 15 frac14 135 kN=m2

                                                  (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                  Ic frac14 171

                                                  1014frac14 0068

                                                  From Equation 819(a) with s frac14 25mm

                                                  q frac14 25

                                                  3507 0068frac14 150 kN=m2

                                                  810

                                                  Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                  Izp frac14 05thorn 01130

                                                  16 27

                                                  05

                                                  frac14 067

                                                  Refer to Figure Q810

                                                  E frac14 25qc

                                                  Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                  Ez (mm3MN)

                                                  1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                  0203

                                                  C1 frac14 1 0500qnfrac14 1 05 12 16

                                                  130frac14 093

                                                  C2 frac14 1 ethsayTHORN

                                                  s frac14 C1C2qnX Iz

                                                  Ez frac14 093 1 130 0203 frac14 25mm

                                                  Bearing capacity 67

                                                  811

                                                  At pile base level

                                                  cu frac14 220 kN=m2

                                                  qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                  Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                  00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                  qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                  Then

                                                  Qf frac14 Abqb thorn Asqs

                                                  frac14

                                                  4 32 1980

                                                  thorn eth 105 139 86THORN

                                                  frac14 13 996thorn 3941 frac14 17 937 kN

                                                  0 01 02 03 04 05 06 07

                                                  0 2 4 6 8 10 12 14

                                                  1

                                                  2

                                                  3

                                                  4

                                                  5

                                                  6

                                                  7

                                                  8

                                                  (1)

                                                  (2)

                                                  (3)

                                                  (4)

                                                  (5)

                                                  qc

                                                  qc

                                                  Iz

                                                  Iz

                                                  (MNm2)

                                                  z (m)

                                                  Figure Q810

                                                  68 Bearing capacity

                                                  Allowable load

                                                  ethaTHORN Qf

                                                  2frac14 17 937

                                                  2frac14 8968 kN

                                                  ethbTHORN Abqb

                                                  3thorn Asqs frac14 13 996

                                                  3thorn 3941 frac14 8606 kN

                                                  ie allowable load frac14 8600 kN

                                                  Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                  According to the limit state method

                                                  Characteristic undrained strength at base level cuk frac14 220

                                                  150kN=m2

                                                  Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                  150frac14 1320 kN=m2

                                                  Characteristic shaft resistance qsk frac14 00150

                                                  frac14 86

                                                  150frac14 57 kN=m2

                                                  Characteristic base and shaft resistances

                                                  Rbk frac14

                                                  4 32 1320 frac14 9330 kN

                                                  Rsk frac14 105 139 86

                                                  150frac14 2629 kN

                                                  For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                  Design bearing resistance Rcd frac14 9330

                                                  160thorn 2629

                                                  130

                                                  frac14 5831thorn 2022

                                                  frac14 7850 kN

                                                  Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                  812

                                                  ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                  qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                  For a single pile

                                                  Qf frac14 Abqb thorn Asqs

                                                  frac14

                                                  4 062 1305

                                                  thorn eth 06 15 42THORN

                                                  frac14 369thorn 1187 frac14 1556 kN

                                                  Bearing capacity 69

                                                  Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                  eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                  (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                  (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                  qbkfrac14 9cuk frac14 9 220

                                                  150frac14 1320 kN=m2

                                                  qskfrac14cuk frac14 040 105

                                                  150frac14 28 kN=m2

                                                  Rbkfrac14

                                                  4 0602 1320 frac14 373 kN

                                                  Rskfrac14 060 15 28 frac14 791 kN

                                                  Rcdfrac14 373

                                                  160thorn 791

                                                  130frac14 233thorn 608 frac14 841 kN

                                                  Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                  Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                  (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                  q frac14 21 000

                                                  1762frac14 68 kN=m2

                                                  Immediate settlement

                                                  H

                                                  Bfrac14 15

                                                  176frac14 085

                                                  D

                                                  Bfrac14 13

                                                  176frac14 074

                                                  L

                                                  Bfrac14 1

                                                  Hence from Figure 515

                                                  130 frac14 078 and 131 frac14 041

                                                  70 Bearing capacity

                                                  Thus using Equation 528

                                                  si frac14 078 041 68 176

                                                  65frac14 6mm

                                                  Consolidation settlement

                                                  Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                  1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                  434 (sod)

                                                  Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                  sc frac14 056 434 frac14 24mm

                                                  The total settlement is (6thorn 24) frac14 30mm

                                                  813

                                                  At base level N frac14 26 Then using Equation 830

                                                  qb frac14 40NDb

                                                  Bfrac14 40 26 2

                                                  025frac14 8320 kN=m2

                                                  ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                  Figure Q812

                                                  Bearing capacity 71

                                                  Over the length embedded in sand

                                                  N frac14 21 ie18thorn 24

                                                  2

                                                  Using Equation 831

                                                  qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                  For a single pile

                                                  Qf frac14 Abqb thorn Asqs

                                                  frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                  For the pile group assuming a group efficiency of 12

                                                  XQf frac14 12 9 604 frac14 6523 kN

                                                  Then the load factor is

                                                  F frac14 6523

                                                  2000thorn 1000frac14 21

                                                  (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                  Characteristic base resistance per unit area qbk frac14 8320

                                                  150frac14 5547 kNm2

                                                  Characteristic shaft resistance per unit area qsk frac14 42

                                                  150frac14 28 kNm2

                                                  Characteristic base and shaft resistances for a single pile

                                                  Rbk frac14 0252 5547 frac14 347 kN

                                                  Rsk frac14 4 025 2 28 frac14 56 kN

                                                  For a driven pile the partial factors are b frac14 s frac14 130

                                                  Design bearing resistance Rcd frac14 347

                                                  130thorn 56

                                                  130frac14 310 kN

                                                  For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                  Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                  (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                  From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                  72 Bearing capacity

                                                  N frac14 24thorn 26thorn 34

                                                  3frac14 28

                                                  Ic frac14 171

                                                  2814frac14 0016 ethEquation 818THORN

                                                  s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                  814

                                                  Using Equation 841

                                                  Tf frac14 DLcu thorn

                                                  4ethD2 d2THORNcuNc

                                                  frac14 eth 02 5 06 110THORN thorn

                                                  4eth022 012THORN110 9

                                                  frac14 207thorn 23 frac14 230 kN

                                                  Figure Q813

                                                  Bearing capacity 73

                                                  Chapter 9

                                                  Stability of slopes

                                                  91

                                                  Referring to Figure Q91

                                                  W frac14 417 19 frac14 792 kN=m

                                                  Q frac14 20 28 frac14 56 kN=m

                                                  Arc lengthAB frac14

                                                  180 73 90 frac14 115m

                                                  Arc length BC frac14

                                                  180 28 90 frac14 44m

                                                  The factor of safety is given by

                                                  F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                  90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                  Depth of tension crack z0 frac14 2cu

                                                  frac14 2 20

                                                  19frac14 21m

                                                  Arc length BD frac14

                                                  180 13

                                                  1

                                                  2 90 frac14 21m

                                                  F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                  The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                  Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                  14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                  Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                  The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                  92

                                                  u frac14 0

                                                  Depth factor D frac14 11

                                                  9frac14 122

                                                  Using Equation 92 with F frac14 10

                                                  Ns frac14 cu

                                                  FHfrac14 30

                                                  10 19 9frac14 0175

                                                  Hence from Figure 93

                                                  frac14 50

                                                  For F frac14 12

                                                  Ns frac14 30

                                                  12 19 9frac14 0146

                                                  frac14 27

                                                  93

                                                  Refer to Figure Q93

                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                  1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                  74 m

                                                  214 1deg

                                                  213 1deg

                                                  39 m

                                                  WB

                                                  D

                                                  C

                                                  28 m

                                                  21 m

                                                  A

                                                  Q

                                                  Soil (1)Soil (2)

                                                  73deg

                                                  Figure Q91

                                                  Stability of slopes 75

                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                  9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                  599 256 328 1372

                                                  Figure Q93

                                                  76 Stability of slopes

                                                  XW cos frac14 b

                                                  Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                  W sin frac14 bX

                                                  h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                  Arc length La frac14

                                                  180 57

                                                  1

                                                  2 326 frac14 327m

                                                  The factor of safety is given by

                                                  F frac14 c0La thorn tan0ethW cos ulTHORN

                                                  W sin

                                                  frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                  frac14 091

                                                  According to the limit state method

                                                  0d frac14 tan1tan 32

                                                  125

                                                  frac14 265

                                                  c0 frac14 8

                                                  160frac14 5 kN=m2

                                                  Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                  Design disturbing moment frac14 1075 kN=m

                                                  The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                  94

                                                  F frac14 1

                                                  W sin

                                                  Xfc0bthorn ethW ubTHORN tan0g sec

                                                  1thorn ethtan tan0=FTHORN

                                                  c0 frac14 8 kN=m2

                                                  0 frac14 32

                                                  c0b frac14 8 2 frac14 16 kN=m

                                                  W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                  Try F frac14 100

                                                  tan0

                                                  Ffrac14 0625

                                                  Stability of slopes 77

                                                  Values of u are as obtained in Figure Q93

                                                  SliceNo

                                                  h(m)

                                                  W frac14 bh(kNm)

                                                  W sin(kNm)

                                                  ub(kNm)

                                                  c0bthorn (W ub) tan0(kNm)

                                                  sec

                                                  1thorn (tan tan0)FProduct(kNm)

                                                  1 05 21 6 2 8 24 1078 262 13 55 31

                                                  23 33 30 1042 31

                                                  3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                  224 92 72 0931 67

                                                  6 50 210 11 40 100 85 0907 777 55 231 14

                                                  12 58 112 90 0889 80

                                                  8 60 252 1812

                                                  80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                  10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                  2154 88 116 0853 99

                                                  14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                  1074 1091

                                                  F frac14 1091

                                                  1074frac14 102 (assumed value 100)

                                                  Thus

                                                  F frac14 101

                                                  95

                                                  F frac14 1

                                                  W sin

                                                  XfWeth1 ruTHORN tan0g sec

                                                  1thorn ethtan tan0THORN=F

                                                  0 frac14 33

                                                  ru frac14 020

                                                  W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                  eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                  Try F frac14 110

                                                  tan 0

                                                  Ffrac14 tan 33

                                                  110frac14 0590

                                                  78 Stability of slopes

                                                  Referring to Figure Q95

                                                  SliceNo

                                                  h(m)

                                                  W frac14 bh(kNm)

                                                  W sin(kNm)

                                                  W(1 ru) tan0(kNm)

                                                  sec

                                                  1thorn ( tan tan0)FProduct(kNm)

                                                  1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                  2120 234 0892 209

                                                  4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                  1185 1271

                                                  Figure Q95

                                                  Stability of slopes 79

                                                  F frac14 1271

                                                  1185frac14 107

                                                  The trial value was 110 therefore take F to be 108

                                                  96

                                                  (a) Water table at surface the factor of safety is given by Equation 912

                                                  F frac14 0

                                                  sat

                                                  tan0

                                                  tan

                                                  ptie 15 frac14 92

                                                  19

                                                  tan 36

                                                  tan

                                                  tan frac14 0234

                                                  frac14 13

                                                  Water table well below surface the factor of safety is given by Equation 911

                                                  F frac14 tan0

                                                  tan

                                                  frac14 tan 36

                                                  tan 13

                                                  frac14 31

                                                  (b) 0d frac14 tan1tan 36

                                                  125

                                                  frac14 30

                                                  Depth of potential failure surface frac14 z

                                                  Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                  frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                  frac14 504z kN

                                                  Design disturbing moment per unit area Sd frac14 sat sin cos

                                                  frac14 19 z sin 13 cos 13

                                                  frac14 416z kN

                                                  Rd gtSd therefore the limit state for overall stability is satisfied

                                                  80 Stability of slopes

                                                  • Book Cover
                                                  • Title
                                                  • Contents
                                                  • Basic characteristics of soils
                                                  • Seepage
                                                  • Effective stress
                                                  • Shear strength
                                                  • Stresses and displacements
                                                  • Lateral earth pressure
                                                  • Consolidation theory
                                                  • Bearing capacity
                                                  • Stability of slopes

                                                    The critical hydraulic gradient is given by Equation 39

                                                    ic frac14 0

                                                    wfrac14 102

                                                    98frac14 104

                                                    Therefore factor of safety against lsquoboilingrsquo (Equation 311)

                                                    F frac14 iciefrac14 104

                                                    071frac14 15

                                                    Total head at C

                                                    hC frac14 nd

                                                    Ndh frac14 24

                                                    11 550 frac14 120m

                                                    Elevation head at C

                                                    zC frac14 250m

                                                    Pore water pressure at C

                                                    uC frac14 98eth120thorn 250THORN frac14 36 kN=m2

                                                    Therefore effective vertical stress at C

                                                    0C frac14 C uC frac14 eth25 20THORN 36 frac14 14 kN=m2

                                                    For point D

                                                    hD frac14 73

                                                    11 550 frac14 365m

                                                    zD frac14 450muD frac14 98eth365thorn 450THORN frac14 80 kN=m2

                                                    0D frac14 D uD frac14 eth3 98THORN thorn eth7 20THORN 80 frac14 90 kN=m2

                                                    39

                                                    The flow net is drawn in Figure Q25

                                                    For a soil prism 150 300m adjacent to the piling

                                                    hm frac14 26

                                                    9 500 frac14 145m

                                                    20 Effective stress

                                                    Factor of safety against lsquoheavingrsquo (Equation 310)

                                                    F frac14 ic

                                                    imfrac14 0d

                                                    whmfrac14 97 300

                                                    98 145frac14 20

                                                    With a filter

                                                    F frac14 0d thorn wwhm

                                                    3 frac14 eth97 300THORN thorn w98 145

                                                    w frac14 135 kN=m2

                                                    Depth of filterfrac14 13521frac14 065m (if above water level)

                                                    Effective stress 21

                                                    Chapter 4

                                                    Shear strength

                                                    41

                                                    frac14 295 kN=m2

                                                    u frac14 120 kN=m2

                                                    0 frac14 u frac14 295 120 frac14 175 kN=m2

                                                    f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                                                    42

                                                    03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                                                    100 452 552200 908 1108400 1810 2210800 3624 4424

                                                    The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                                                    Figure Q42

                                                    43

                                                    3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                                                    200 222 422400 218 618600 220 820

                                                    The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                                                    44

                                                    The modified shear strength parameters are

                                                    0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                                                    a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                                                    The coordinates of the stress point representing failure conditions in the test are

                                                    1

                                                    2eth1 2THORN frac14 1

                                                    2 170 frac14 85 kN=m2

                                                    1

                                                    2eth1 thorn 3THORN frac14 1

                                                    2eth270thorn 100THORN frac14 185 kN=m2

                                                    The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                                                    uf frac14 36 kN=m2

                                                    Figure Q43

                                                    Figure Q44

                                                    Shear strength 23

                                                    45

                                                    3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                                    150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                                    The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                                    The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                                    46

                                                    03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                                    200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                                    The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                                    Figure Q45

                                                    24 Shear strength

                                                    47

                                                    The torque required to produce shear failure is given by

                                                    T frac14 dh cud

                                                    2thorn 2

                                                    Z d=2

                                                    0

                                                    2r drcur

                                                    frac14 cud2h

                                                    2thorn 4cu

                                                    Z d=2

                                                    0

                                                    r2dr

                                                    frac14 cud2h

                                                    2thorn d

                                                    3

                                                    6

                                                    Then

                                                    35 frac14 cu52 10

                                                    2thorn 53

                                                    6

                                                    103

                                                    cu frac14 76 kN=m3

                                                    400

                                                    0 400 800 1200 1600

                                                    τ (k

                                                    Nm

                                                    2 )

                                                    σprime (kNm2)

                                                    34deg

                                                    315deg29deg

                                                    (a)

                                                    (b)

                                                    0 400

                                                    400

                                                    800 1200 1600

                                                    Failure envelope

                                                    300 500

                                                    σprime (kNm2)

                                                    τ (k

                                                    Nm

                                                    2 )

                                                    20 (kNm2)

                                                    31deg

                                                    Figure Q46

                                                    Shear strength 25

                                                    48

                                                    The relevant stress values are calculated as follows

                                                    3 frac14 600 kN=m2

                                                    1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                                    2(1 3) 0 40 79 107 139 159

                                                    1

                                                    2(01 thorn 03) 400 411 402 389 351 326

                                                    1

                                                    2(1 thorn 3) 600 640 679 707 739 759

                                                    The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                                    Af frac14 433 200

                                                    319frac14 073

                                                    49

                                                    B frac14 u33

                                                    frac14 144

                                                    350 200frac14 096

                                                    a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                                    0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                                    10 283 65 023

                                                    Figure Q48

                                                    26 Shear strength

                                                    The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                                    Figure Q49

                                                    Shear strength 27

                                                    Chapter 5

                                                    Stresses and displacements

                                                    51

                                                    Vertical stress is given by

                                                    z frac14 Qz2Ip frac14 5000

                                                    52Ip

                                                    Values of Ip are obtained from Table 51

                                                    r (m) rz Ip z (kNm2)

                                                    0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                                    10 20 0009 2

                                                    The variation of z with radial distance (r) is plotted in Figure Q51

                                                    Figure Q51

                                                    52

                                                    Below the centre load (Figure Q52)

                                                    r

                                                    zfrac14 0 for the 7500-kN load

                                                    Ip frac14 0478

                                                    r

                                                    zfrac14 5

                                                    4frac14 125 for the 10 000- and 9000-kN loads

                                                    Ip frac14 0045

                                                    Then

                                                    z frac14X Q

                                                    z2Ip

                                                    frac14 7500 0478

                                                    42thorn 10 000 0045

                                                    42thorn 9000 0045

                                                    42

                                                    frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                                    53

                                                    The vertical stress under a corner of a rectangular area is given by

                                                    z frac14 qIr

                                                    where values of Ir are obtained from Figure 510 In this case

                                                    z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                                    z

                                                    Figure Q52

                                                    Stresses and displacements 29

                                                    z (m) m n Ir z (kNm2)

                                                    0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                                    10 010 0005 5

                                                    z is plotted against z in Figure Q53

                                                    54

                                                    (a)

                                                    m frac14 125

                                                    12frac14 104

                                                    n frac14 18

                                                    12frac14 150

                                                    From Figure 510 Irfrac14 0196

                                                    z frac14 2 175 0196 frac14 68 kN=m2

                                                    Figure Q53

                                                    30 Stresses and displacements

                                                    (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                                    z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                                    55

                                                    Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                                    Px frac14 2Q

                                                    1

                                                    m2 thorn 1frac14 2 150

                                                    125frac14 76 kN=m

                                                    Equation 517 is used to obtain the pressure distribution

                                                    px frac14 4Q

                                                    h

                                                    m2n

                                                    ethm2 thorn n2THORN2 frac14150

                                                    m2n

                                                    ethm2 thorn n2THORN2 ethkN=m2THORN

                                                    Figure Q54

                                                    Stresses and displacements 31

                                                    n m2n

                                                    (m2 thorn n2)2

                                                    px(kNm2)

                                                    0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                    The pressure distribution is plotted in Figure Q55

                                                    56

                                                    H

                                                    Bfrac14 10

                                                    2frac14 5

                                                    L

                                                    Bfrac14 4

                                                    2frac14 2

                                                    D

                                                    Bfrac14 1

                                                    2frac14 05

                                                    Hence from Figure 515

                                                    131 frac14 082

                                                    130 frac14 094

                                                    Figure Q55

                                                    32 Stresses and displacements

                                                    The immediate settlement is given by Equation 528

                                                    si frac14 130131qB

                                                    Eu

                                                    frac14 094 082 200 2

                                                    45frac14 7mm

                                                    Stresses and displacements 33

                                                    Chapter 6

                                                    Lateral earth pressure

                                                    61

                                                    For 0 frac14 37 the active pressure coefficient is given by

                                                    Ka frac14 1 sin 37

                                                    1thorn sin 37frac14 025

                                                    The total active thrust (Equation 66a with c0 frac14 0) is

                                                    Pa frac14 1

                                                    2KaH

                                                    2 frac14 1

                                                    2 025 17 62 frac14 765 kN=m

                                                    If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                    K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                    and the thrust on the wall is

                                                    P0 frac14 1

                                                    2K0H

                                                    2 frac14 1

                                                    2 040 17 62 frac14 122 kN=m

                                                    62

                                                    The active pressure coefficients for the three soil types are as follows

                                                    Ka1 frac141 sin 35

                                                    1thorn sin 35frac14 0271

                                                    Ka2 frac141 sin 27

                                                    1thorn sin 27frac14 0375

                                                    ffiffiffiffiffiffiffiKa2

                                                    p frac14 0613

                                                    Ka3 frac141 sin 42

                                                    1thorn sin 42frac14 0198

                                                    Distribution of active pressure (plotted in Figure Q62)

                                                    Depth (m) Soil Active pressure (kNm2)

                                                    3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                    12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                    At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                    Total thrust frac14 571 kNm

                                                    Point of application is (4893571) m from the top of the wall ie 857m

                                                    Force (kN) Arm (m) Moment (kN m)

                                                    (1)1

                                                    2 0271 16 32 frac14 195 20 390

                                                    (2) 0271 16 3 2 frac14 260 40 1040

                                                    (3)1

                                                    2 0271 92 22 frac14 50 433 217

                                                    (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                    (5)1

                                                    2 0375 102 32 frac14 172 70 1204

                                                    (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                    (7)1

                                                    2 0198 112 42 frac14 177 1067 1889

                                                    (8)1

                                                    2 98 92 frac14 3969 90 35721

                                                    5713 48934

                                                    Figure Q62

                                                    Lateral earth pressure 35

                                                    63

                                                    (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                    Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                    frac14 245

                                                    At the lower end of the piling

                                                    pa frac14 Kaqthorn Kasatz Kaccu

                                                    frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                    frac14 115 kN=m2

                                                    pp frac14 Kpsatzthorn Kpccu

                                                    frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                    frac14 202 kN=m2

                                                    (b) For 0 frac14 26 and frac14 1

                                                    20

                                                    Ka frac14 035

                                                    Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                    pfrac14 145 ethEquation 619THORN

                                                    Kp frac14 37

                                                    Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                    pfrac14 47 ethEquation 624THORN

                                                    At the lower end of the piling

                                                    pa frac14 Kaqthorn Ka0z Kacc

                                                    0

                                                    frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                    frac14 187 kN=m2

                                                    pp frac14 Kp0zthorn Kpcc

                                                    0

                                                    frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                    frac14 198 kN=m2

                                                    36 Lateral earth pressure

                                                    64

                                                    (a) For 0 frac14 38 Ka frac14 024

                                                    0 frac14 20 98 frac14 102 kN=m3

                                                    The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                    Force (kN) Arm (m) Moment (kN m)

                                                    (1) 024 10 66 frac14 159 33 525

                                                    (2)1

                                                    2 024 17 392 frac14 310 400 1240

                                                    (3) 024 17 39 27 frac14 430 135 580

                                                    (4)1

                                                    2 024 102 272 frac14 89 090 80

                                                    (5)1

                                                    2 98 272 frac14 357 090 321

                                                    Hfrac14 1345 MH frac14 2746

                                                    (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                    (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                    XM frac14MV MH frac14 7790 kNm

                                                    Lever arm of base resultant

                                                    M

                                                    Vfrac14 779

                                                    488frac14 160

                                                    Eccentricity of base resultant

                                                    e frac14 200 160 frac14 040m

                                                    39 m

                                                    27 m

                                                    40 m

                                                    04 m

                                                    04 m

                                                    26 m

                                                    (7)

                                                    (9)

                                                    (1)(2)

                                                    (3)

                                                    (4)

                                                    (5)

                                                    (8)(6)

                                                    (10)

                                                    WT

                                                    10 kNm2

                                                    Hydrostatic

                                                    Figure Q64

                                                    Lateral earth pressure 37

                                                    Base pressures (Equation 627)

                                                    p frac14 VB

                                                    1 6e

                                                    B

                                                    frac14 488

                                                    4eth1 060THORN

                                                    frac14 195 kN=m2 and 49 kN=m2

                                                    Factor of safety against sliding (Equation 628)

                                                    F frac14 V tan

                                                    Hfrac14 488 tan 25

                                                    1345frac14 17

                                                    (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                    Hfrac14 1633 kN

                                                    V frac14 4879 kN

                                                    MH frac14 3453 kNm

                                                    MV frac14 10536 kNm

                                                    The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                    65

                                                    For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                    Kp

                                                    Ffrac14 385

                                                    2

                                                    0 frac14 20 98 frac14 102 kN=m3

                                                    The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                    Force (kN) Arm (m) Moment (kN m)

                                                    (1)1

                                                    2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                    (2) 026 17 45 d frac14 199d d2 995d2

                                                    (3)1

                                                    2 026 102 d2 frac14 133d2 d3 044d3

                                                    (4)1

                                                    2 385

                                                    2 17 152 frac14 368 dthorn 05 368d 184

                                                    (5)385

                                                    2 17 15 d frac14 491d d2 2455d2

                                                    (6)1

                                                    2 385

                                                    2 102 d2 frac14 982d2 d3 327d3

                                                    38 Lateral earth pressure

                                                    XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                    d3 thorn 516d2 283d 1724 frac14 0

                                                    d frac14 179m

                                                    Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                    Over additional 20 embedded depth

                                                    pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                    Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                    66

                                                    The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                    Ka frac14 sin 69=sin 105

                                                    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                    pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                    26664

                                                    37775

                                                    2

                                                    frac14 050

                                                    The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                    Pa frac14 1

                                                    2 050 19 7502 frac14 267 kN=m

                                                    Figure Q65

                                                    Lateral earth pressure 39

                                                    Horizontal component

                                                    Ph frac14 267 cos 40 frac14 205 kN=m

                                                    Vertical component

                                                    Pv frac14 267 sin 40 frac14 172 kN=m

                                                    Consider moments about the toe of the wall (Figure Q66) (per m)

                                                    Force (kN) Arm (m) Moment (kN m)

                                                    (1)1

                                                    2 175 650 235 frac14 1337 258 345

                                                    (2) 050 650 235 frac14 764 175 134

                                                    (3)1

                                                    2 070 650 235 frac14 535 127 68

                                                    (4) 100 400 235 frac14 940 200 188

                                                    (5) 1

                                                    2 080 050 235 frac14 47 027 1

                                                    Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                    Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                    Lever arm of base resultant

                                                    M

                                                    Vfrac14 795

                                                    525frac14 151m

                                                    Eccentricity of base resultant

                                                    e frac14 200 151 frac14 049m

                                                    Figure Q66

                                                    40 Lateral earth pressure

                                                    Base pressures (Equation 627)

                                                    p frac14 525

                                                    41 6 049

                                                    4

                                                    frac14 228 kN=m2 and 35 kN=m2

                                                    The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                    The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                    The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                    67

                                                    For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                    Force (kN) Arm (m) Moment (kNm)

                                                    (1)1

                                                    2 027 17 52 frac14 574 183 1050

                                                    (2) 027 17 5 3 frac14 689 500 3445

                                                    (3)1

                                                    2 027 102 32 frac14 124 550 682

                                                    (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                    (5)1

                                                    2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                    (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                    (7) 1

                                                    2 267

                                                    2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                    (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                    2 d frac14 163d d2thorn 650 82d2 1060d

                                                    Tie rod force per m frac14 T 0 0

                                                    XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                    d3 thorn 77d2 269d 1438 frac14 0

                                                    d frac14 467m

                                                    Depth of penetration frac14 12d frac14 560m

                                                    Lateral earth pressure 41

                                                    Algebraic sum of forces for d frac14 467m isX

                                                    F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                    T frac14 905 kN=m

                                                    Force in each tie rod frac14 25T frac14 226 kN

                                                    68

                                                    (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                    0 frac14 21 98 frac14 112 kN=m3

                                                    The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                    uC frac14 150

                                                    165 15 98 frac14 134 kN=m2

                                                    The average seepage pressure is

                                                    j frac14 15

                                                    165 98 frac14 09 kN=m3

                                                    Hence

                                                    0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                    0 j frac14 112 09 frac14 103 kN=m3

                                                    Figure Q67

                                                    42 Lateral earth pressure

                                                    Consider moments about the anchor point A (per m)

                                                    Force (kN) Arm (m) Moment (kN m)

                                                    (1) 10 026 150 frac14 390 60 2340

                                                    (2)1

                                                    2 026 18 452 frac14 474 15 711

                                                    (3) 026 18 45 105 frac14 2211 825 18240

                                                    (4)1

                                                    2 026 121 1052 frac14 1734 100 17340

                                                    (5)1

                                                    2 134 15 frac14 101 40 404

                                                    (6) 134 30 frac14 402 60 2412

                                                    (7)1

                                                    2 134 60 frac14 402 95 3819

                                                    571 4527(8) Ppm

                                                    115 115PPm

                                                    XM frac14 0

                                                    Ppm frac144527

                                                    115frac14 394 kN=m

                                                    Available passive resistance

                                                    Pp frac14 1

                                                    2 385 103 62 frac14 714 kN=m

                                                    Factor of safety

                                                    Fp frac14 Pp

                                                    Ppm

                                                    frac14 714

                                                    394frac14 18

                                                    Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                    Figure Q68

                                                    Lateral earth pressure 43

                                                    (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                    Consider moments (per m) about the tie point A

                                                    Force (kN) Arm (m)

                                                    (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                    (2)1

                                                    2 033 18 452 frac14 601 15

                                                    (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                    (4)1

                                                    2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                    (5)1

                                                    2 134 15 frac14 101 40

                                                    (6) 134 30 frac14 402 60

                                                    (7)1

                                                    2 134 d frac14 67d d3thorn 75

                                                    (8) 1

                                                    2 30 103 d2 frac141545d2 2d3thorn 75

                                                    Moment (kN m)

                                                    (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                    XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                    d3 thorn 827d2 466d 1518 frac14 0

                                                    By trial

                                                    d frac14 544m

                                                    The minimum depth of embedment required is 544m

                                                    69

                                                    For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                    0 frac14 20 98 frac14 102 kN=m3

                                                    The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                    44 Lateral earth pressure

                                                    uC frac14 147

                                                    173 26 98 frac14 216 kN=m2

                                                    and the average seepage pressure around the wall is

                                                    j frac14 26

                                                    173 98 frac14 15 kN=m3

                                                    Consider moments about the prop (A) (per m)

                                                    Force (kN) Arm (m) Moment (kN m)

                                                    (1)1

                                                    2 03 17 272 frac14 186 020 37

                                                    (2) 03 17 27 53 frac14 730 335 2445

                                                    (3)1

                                                    2 03 (102thorn 15) 532 frac14 493 423 2085

                                                    (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                    (5)1

                                                    2 216 26 frac14 281 243 684

                                                    (6) 216 27 frac14 583 465 2712

                                                    (7)1

                                                    2 216 60 frac14 648 800 5184

                                                    3055(8)

                                                    1

                                                    2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                    Factor of safety

                                                    Fr frac14 6885

                                                    3055frac14 225

                                                    Figure Q69

                                                    Lateral earth pressure 45

                                                    610

                                                    For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                    p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                    Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                    Using the recommendations of Twine and Roscoe

                                                    p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                    Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                    611

                                                    frac14 18 kN=m3 0 frac14 34

                                                    H frac14 350m nH frac14 335m mH frac14 185m

                                                    Consider a trial value of F frac14 20 Refer to Figure 635

                                                    0m frac14 tan1tan 34

                                                    20

                                                    frac14 186

                                                    Then

                                                    frac14 45 thorn 0m2frac14 543

                                                    W frac14 1

                                                    2 18 3502 cot 543 frac14 792 kN=m

                                                    Figure Q610

                                                    46 Lateral earth pressure

                                                    P frac14 1

                                                    2 s 3352 frac14 561s kN=m

                                                    U frac14 1

                                                    2 98 1852 cosec 543 frac14 206 kN=m

                                                    Equations 630 and 631 then become

                                                    561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                    792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                    ie

                                                    561s 0616N 405 frac14 0

                                                    792 0857N thorn 563 frac14 0

                                                    N frac14 848

                                                    0857frac14 989 kN=m

                                                    Then

                                                    561s 609 405 frac14 0

                                                    s frac14 649

                                                    561frac14 116 kN=m3

                                                    The calculations for trial values of F of 20 15 and 10 are summarized below

                                                    F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                    20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                    s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                    Figure Q611

                                                    Lateral earth pressure 47

                                                    612

                                                    For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                    45 thorn 0

                                                    2frac14 63

                                                    For the retained material between the surface and a depth of 36m

                                                    Pa frac14 1

                                                    2 030 18 362 frac14 350 kN=m

                                                    Weight of reinforced fill between the surface and a depth of 36m is

                                                    Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                    eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                    Eccentricity of Rv

                                                    e frac14 263 250 frac14 013m

                                                    The average vertical stress at a depth of 36m is

                                                    z frac14 Rv

                                                    L 2efrac14 324

                                                    474frac14 68 kN=m2

                                                    (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                    Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                    Tensile stress in the element frac14 138 103

                                                    65 3frac14 71N=mm2

                                                    Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                    Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                    Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                    The weight of ABC is

                                                    W frac14 1

                                                    2 18 52 265 frac14 124 kN=m

                                                    From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                    48 Lateral earth pressure

                                                    (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                    Tp frac14 032 68 120 065 frac14 170 kN

                                                    Tr frac14 213 420

                                                    418frac14 214 kN

                                                    Again the tensile failure and slipping limit states are satisfied for this element

                                                    Figure Q612

                                                    Lateral earth pressure 49

                                                    Chapter 7

                                                    Consolidation theory

                                                    71

                                                    Total change in thickness

                                                    H frac14 782 602 frac14 180mm

                                                    Average thickness frac14 1530thorn 180

                                                    2frac14 1620mm

                                                    Length of drainage path d frac14 1620

                                                    2frac14 810mm

                                                    Root time plot (Figure Q71a)

                                                    ffiffiffiffiffiffit90p frac14 33

                                                    t90 frac14 109min

                                                    cv frac14 0848d2

                                                    t90frac14 0848 8102

                                                    109 1440 365

                                                    106frac14 27m2=year

                                                    r0 frac14 782 764

                                                    782 602frac14 018

                                                    180frac14 0100

                                                    rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                    10 119

                                                    9 180frac14 0735

                                                    rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                    Log time plot (Figure Q71b)

                                                    t50 frac14 26min

                                                    cv frac14 0196d2

                                                    t50frac14 0196 8102

                                                    26 1440 365

                                                    106frac14 26m2=year

                                                    r0 frac14 782 763

                                                    782 602frac14 019

                                                    180frac14 0106

                                                    rp frac14 763 623

                                                    782 602frac14 140

                                                    180frac14 0778

                                                    rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                    Figure Q71(a)

                                                    Figure Q71(b)

                                                    Final void ratio

                                                    e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                    e

                                                    Hfrac14 1thorn e0

                                                    H0frac14 1thorn e1 thorne

                                                    H0

                                                    ie

                                                    e

                                                    180frac14 1631thorne

                                                    1710

                                                    e frac14 2936

                                                    1530frac14 0192

                                                    Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                    mv frac14 1

                                                    1thorn e0 e0 e101 00

                                                    frac14 1

                                                    1823 0192

                                                    0107frac14 098m2=MN

                                                    k frac14 cvmvw frac14 265 098 98

                                                    60 1440 365 103frac14 81 1010 m=s

                                                    72

                                                    Using Equation 77 (one-dimensional method)

                                                    sc frac14 e0 e11thorn e0 H

                                                    Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                    Figure Q72

                                                    52 Consolidation theory

                                                    Settlement

                                                    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                    1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                    318

                                                    Notes 5 92y 460thorn 84

                                                    Heave

                                                    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                    1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                    38

                                                    73

                                                    U frac14 f ethTvTHORN frac14 f cvt

                                                    d2

                                                    Hence if cv is constant

                                                    t1

                                                    t2frac14 d

                                                    21

                                                    d22

                                                    where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                    d1 frac14 95mm and d2 frac14 2500mm

                                                    for U frac14 050 t2 frac14 t1 d22

                                                    d21

                                                    frac14 20

                                                    60 24 365 25002

                                                    952frac14 263 years

                                                    for U lt 060 Tv frac14

                                                    4U2 (Equation 724(a))

                                                    t030 frac14 t050 0302

                                                    0502

                                                    frac14 263 036 frac14 095 years

                                                    Consolidation theory 53

                                                    74

                                                    The layer is open

                                                    d frac14 8

                                                    2frac14 4m

                                                    Tv frac14 cvtd2frac14 24 3

                                                    42frac14 0450

                                                    ui frac14 frac14 84 kN=m2

                                                    The excess pore water pressure is given by Equation 721

                                                    ue frac14Xmfrac141mfrac140

                                                    2ui

                                                    Msin

                                                    Mz

                                                    d

                                                    expethM2TvTHORN

                                                    In this case z frac14 d

                                                    sinMz

                                                    d

                                                    frac14 sinM

                                                    where

                                                    M frac14

                                                    23

                                                    25

                                                    2

                                                    M sin M M2Tv exp (M2Tv)

                                                    2thorn1 1110 0329

                                                    3

                                                    21 9993 457 105

                                                    ue frac14 2 84 2

                                                    1 0329 ethother terms negligibleTHORN

                                                    frac14 352 kN=m2

                                                    75

                                                    The layer is open

                                                    d frac14 6

                                                    2frac14 3m

                                                    Tv frac14 cvtd2frac14 10 3

                                                    32frac14 0333

                                                    The layer thickness will be divided into six equal parts ie m frac14 6

                                                    54 Consolidation theory

                                                    For an open layer

                                                    Tv frac14 4n

                                                    m2

                                                    n frac14 0333 62

                                                    4frac14 300

                                                    The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                    ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                    i j

                                                    0 1 2 3 4 5 6 7 8 9 10 11 12

                                                    0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                    The initial and 3-year isochrones are plotted in Figure Q75

                                                    Area under initial isochrone frac14 180 units

                                                    Area under 3-year isochrone frac14 63 units

                                                    The average degree of consolidation is given by Equation 725Thus

                                                    U frac14 1 63

                                                    180frac14 065

                                                    Figure Q75

                                                    Consolidation theory 55

                                                    76

                                                    At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                    0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                    The final consolidation settlement (one-dimensional method) is

                                                    sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                    Corrected time t frac14 2 1

                                                    2

                                                    40

                                                    52

                                                    frac14 1615 years

                                                    Tv frac14 cvtd2frac14 44 1615

                                                    42frac14 0444

                                                    From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                    77

                                                    The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                    Figure Q77

                                                    56 Consolidation theory

                                                    Point m n Ir (kNm2) sc (mm)

                                                    13020frac14 15 20

                                                    20frac14 10 0194 (4) 113 124

                                                    260

                                                    20frac14 30

                                                    20

                                                    20frac14 10 0204 (2) 59 65

                                                    360

                                                    20frac14 30

                                                    40

                                                    20frac14 20 0238 (1) 35 38

                                                    430

                                                    20frac14 15

                                                    40

                                                    20frac14 20 0224 (2) 65 72

                                                    Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                    78

                                                    Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                    (a) Immediate settlement

                                                    H

                                                    Bfrac14 30

                                                    35frac14 086

                                                    D

                                                    Bfrac14 2

                                                    35frac14 006

                                                    Figure Q78

                                                    Consolidation theory 57

                                                    From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                    si frac14 130131qB

                                                    Eufrac14 10 032 105 35

                                                    40frac14 30mm

                                                    (b) Consolidation settlement

                                                    Layer z (m) Dz Ic (kNm2) syod (mm)

                                                    1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                    3150

                                                    Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                    Now

                                                    H

                                                    Bfrac14 30

                                                    35frac14 086 and A frac14 065

                                                    from Figure 712 13 frac14 079

                                                    sc frac14 13sod frac14 079 315 frac14 250mm

                                                    Total settlement

                                                    s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                    79

                                                    Without sand drains

                                                    Uv frac14 025

                                                    Tv frac14 0049 ethfrom Figure 718THORN

                                                    t frac14 Tvd2

                                                    cvfrac14 0049 82

                                                    cvWith sand drains

                                                    R frac14 0564S frac14 0564 3 frac14 169m

                                                    n frac14 Rrfrac14 169

                                                    015frac14 113

                                                    Tr frac14 cht

                                                    4R2frac14 ch

                                                    4 1692 0049 82

                                                    cvethand ch frac14 cvTHORN

                                                    frac14 0275

                                                    Ur frac14 073 (from Figure 730)

                                                    58 Consolidation theory

                                                    Using Equation 740

                                                    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                    U frac14 080

                                                    710

                                                    Without sand drains

                                                    Uv frac14 090

                                                    Tv frac14 0848

                                                    t frac14 Tvd2

                                                    cvfrac14 0848 102

                                                    96frac14 88 years

                                                    With sand drains

                                                    R frac14 0564S frac14 0564 4 frac14 226m

                                                    n frac14 Rrfrac14 226

                                                    015frac14 15

                                                    Tr

                                                    Tvfrac14 chcv

                                                    d2

                                                    4R2ethsame tTHORN

                                                    Tr

                                                    Tvfrac14 140

                                                    96 102

                                                    4 2262frac14 714 eth1THORN

                                                    Using Equation 740

                                                    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                    An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                    Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                    040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                    Thus

                                                    Uv frac14 0295 and Ur frac14 086

                                                    t frac14 88 00683

                                                    0848frac14 07 years

                                                    Consolidation theory 59

                                                    Chapter 8

                                                    Bearing capacity

                                                    81

                                                    (a) The ultimate bearing capacity is given by Equation 83

                                                    qf frac14 cNc thorn DNq thorn 1

                                                    2BN

                                                    For u frac14 0

                                                    Nc frac14 514 Nq frac14 1 N frac14 0

                                                    qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                    The net ultimate bearing capacity is

                                                    qnf frac14 qf D frac14 540 kN=m2

                                                    The net foundation pressure is

                                                    qn frac14 q D frac14 425

                                                    2 eth21 1THORN frac14 192 kN=m2

                                                    The factor of safety (Equation 86) is

                                                    F frac14 qnfqnfrac14 540

                                                    192frac14 28

                                                    (b) For 0 frac14 28

                                                    Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                    qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                    2 112 2 13

                                                    frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                    qnf frac14 574 112 frac14 563 kN=m2

                                                    F frac14 563

                                                    192frac14 29

                                                    (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                    82

                                                    For 0 frac14 38

                                                    Nq frac14 49 N frac14 67

                                                    qnf frac14 DethNq 1THORN thorn 1

                                                    2BN ethfrom Equation 83THORN

                                                    frac14 eth18 075 48THORN thorn 1

                                                    2 18 15 67

                                                    frac14 648thorn 905 frac14 1553 kN=m2

                                                    qn frac14 500

                                                    15 eth18 075THORN frac14 320 kN=m2

                                                    F frac14 qnfqnfrac14 1553

                                                    320frac14 48

                                                    0d frac14 tan1tan 38

                                                    125

                                                    frac14 32 therefore Nq frac14 23 and N frac14 25

                                                    Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                    2 18 15 25

                                                    frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                    Design load (action) Vd frac14 500 kN=m

                                                    The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                    83

                                                    D

                                                    Bfrac14 350

                                                    225frac14 155

                                                    From Figure 85 for a square foundation

                                                    Nc frac14 81

                                                    Bearing capacity 61

                                                    For a rectangular foundation (L frac14 450m B frac14 225m)

                                                    Nc frac14 084thorn 016B

                                                    L

                                                    81 frac14 745

                                                    Using Equation 810

                                                    qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                    For F frac14 3

                                                    qn frac14 1006

                                                    3frac14 335 kN=m2

                                                    q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                    Design load frac14 405 450 225 frac14 4100 kN

                                                    Design undrained strength cud frac14 135

                                                    14frac14 96 kN=m2

                                                    Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                    frac14 7241 kN

                                                    Design load Vd frac14 4100 kN

                                                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                    84

                                                    For 0 frac14 40

                                                    Nq frac14 64 N frac14 95

                                                    qnf frac14 DethNq 1THORN thorn 04BN

                                                    (a) Water table 5m below ground level

                                                    qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                    qn frac14 400 17 frac14 383 kN=m2

                                                    F frac14 2686

                                                    383frac14 70

                                                    (b) Water table 1m below ground level (ie at foundation level)

                                                    0 frac14 20 98 frac14 102 kN=m3

                                                    62 Bearing capacity

                                                    qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                    F frac14 2040

                                                    383frac14 53

                                                    (c) Water table at ground level with upward hydraulic gradient 02

                                                    eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                    qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                    F frac14 1296

                                                    392frac14 33

                                                    85

                                                    The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                    Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                    Design value of 0 frac14 tan1tan 39

                                                    125

                                                    frac14 33

                                                    For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                    Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                    86

                                                    (a) Undrained shear for u frac14 0

                                                    Nc frac14 514 Nq frac14 1 N frac14 0

                                                    qnf frac14 12cuNc

                                                    frac14 12 100 514 frac14 617 kN=m2

                                                    qn frac14 qnfFfrac14 617

                                                    3frac14 206 kN=m2

                                                    q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                    Bearing capacity 63

                                                    Drained shear for 0 frac14 32

                                                    Nq frac14 23 N frac14 25

                                                    0 frac14 21 98 frac14 112 kN=m3

                                                    qnf frac14 0DethNq 1THORN thorn 040BN

                                                    frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                    frac14 694 kN=m2

                                                    q frac14 694

                                                    3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                    Design load frac14 42 227 frac14 3632 kN

                                                    (b) Design undrained strength cud frac14 100

                                                    14frac14 71 kNm2

                                                    Design bearing resistance Rd frac14 12cudNe area

                                                    frac14 12 71 514 42

                                                    frac14 7007 kN

                                                    For drained shear 0d frac14 tan1tan 32

                                                    125

                                                    frac14 26

                                                    Nq frac14 12 N frac14 10

                                                    Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                    (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                    Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                    1 2 100 0175 0700qn 0182qn

                                                    2 6 033 0044 0176qn 0046qn

                                                    3 10 020 0017 0068qn 0018qn

                                                    0246qn

                                                    Diameter of equivalent circle B frac14 45m

                                                    H

                                                    Bfrac14 12

                                                    45frac14 27 and A frac14 042

                                                    13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                    64 Bearing capacity

                                                    For sc frac14 30mm

                                                    qn frac14 30

                                                    0147frac14 204 kN=m2

                                                    q frac14 204thorn 21 frac14 225 kN=m2

                                                    Design load frac14 42 225 frac14 3600 kN

                                                    The design load is 3600 kN settlement being the limiting criterion

                                                    87

                                                    D

                                                    Bfrac14 8

                                                    4frac14 20

                                                    From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                    F frac14 cuNc

                                                    Dfrac14 40 71

                                                    20 8frac14 18

                                                    88

                                                    Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                    Design value of 0 frac14 tan1tan 38

                                                    125

                                                    frac14 32

                                                    Figure Q86

                                                    Bearing capacity 65

                                                    For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                    Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                    The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                    Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                    For B frac14 250m qn frac14 3750

                                                    2502 17 frac14 583 kN=m2

                                                    From Figure 510 m frac14 n frac14 126

                                                    6frac14 021

                                                    Ir frac14 0019

                                                    Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                    Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                    The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                    89

                                                    Depth (m) N 0v (kNm2) CN N1

                                                    070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                    Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                    (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                    Cw frac14 05thorn 0530

                                                    47

                                                    frac14 082

                                                    66 Bearing capacity

                                                    Thus

                                                    qa frac14 150 082 frac14 120 kN=m2

                                                    (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                    Thus

                                                    qa frac14 90 15 frac14 135 kN=m2

                                                    (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                    Ic frac14 171

                                                    1014frac14 0068

                                                    From Equation 819(a) with s frac14 25mm

                                                    q frac14 25

                                                    3507 0068frac14 150 kN=m2

                                                    810

                                                    Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                    Izp frac14 05thorn 01130

                                                    16 27

                                                    05

                                                    frac14 067

                                                    Refer to Figure Q810

                                                    E frac14 25qc

                                                    Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                    Ez (mm3MN)

                                                    1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                    0203

                                                    C1 frac14 1 0500qnfrac14 1 05 12 16

                                                    130frac14 093

                                                    C2 frac14 1 ethsayTHORN

                                                    s frac14 C1C2qnX Iz

                                                    Ez frac14 093 1 130 0203 frac14 25mm

                                                    Bearing capacity 67

                                                    811

                                                    At pile base level

                                                    cu frac14 220 kN=m2

                                                    qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                    Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                    00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                    qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                    Then

                                                    Qf frac14 Abqb thorn Asqs

                                                    frac14

                                                    4 32 1980

                                                    thorn eth 105 139 86THORN

                                                    frac14 13 996thorn 3941 frac14 17 937 kN

                                                    0 01 02 03 04 05 06 07

                                                    0 2 4 6 8 10 12 14

                                                    1

                                                    2

                                                    3

                                                    4

                                                    5

                                                    6

                                                    7

                                                    8

                                                    (1)

                                                    (2)

                                                    (3)

                                                    (4)

                                                    (5)

                                                    qc

                                                    qc

                                                    Iz

                                                    Iz

                                                    (MNm2)

                                                    z (m)

                                                    Figure Q810

                                                    68 Bearing capacity

                                                    Allowable load

                                                    ethaTHORN Qf

                                                    2frac14 17 937

                                                    2frac14 8968 kN

                                                    ethbTHORN Abqb

                                                    3thorn Asqs frac14 13 996

                                                    3thorn 3941 frac14 8606 kN

                                                    ie allowable load frac14 8600 kN

                                                    Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                    According to the limit state method

                                                    Characteristic undrained strength at base level cuk frac14 220

                                                    150kN=m2

                                                    Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                    150frac14 1320 kN=m2

                                                    Characteristic shaft resistance qsk frac14 00150

                                                    frac14 86

                                                    150frac14 57 kN=m2

                                                    Characteristic base and shaft resistances

                                                    Rbk frac14

                                                    4 32 1320 frac14 9330 kN

                                                    Rsk frac14 105 139 86

                                                    150frac14 2629 kN

                                                    For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                    Design bearing resistance Rcd frac14 9330

                                                    160thorn 2629

                                                    130

                                                    frac14 5831thorn 2022

                                                    frac14 7850 kN

                                                    Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                    812

                                                    ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                    qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                    For a single pile

                                                    Qf frac14 Abqb thorn Asqs

                                                    frac14

                                                    4 062 1305

                                                    thorn eth 06 15 42THORN

                                                    frac14 369thorn 1187 frac14 1556 kN

                                                    Bearing capacity 69

                                                    Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                    eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                    (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                    (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                    qbkfrac14 9cuk frac14 9 220

                                                    150frac14 1320 kN=m2

                                                    qskfrac14cuk frac14 040 105

                                                    150frac14 28 kN=m2

                                                    Rbkfrac14

                                                    4 0602 1320 frac14 373 kN

                                                    Rskfrac14 060 15 28 frac14 791 kN

                                                    Rcdfrac14 373

                                                    160thorn 791

                                                    130frac14 233thorn 608 frac14 841 kN

                                                    Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                    Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                    (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                    q frac14 21 000

                                                    1762frac14 68 kN=m2

                                                    Immediate settlement

                                                    H

                                                    Bfrac14 15

                                                    176frac14 085

                                                    D

                                                    Bfrac14 13

                                                    176frac14 074

                                                    L

                                                    Bfrac14 1

                                                    Hence from Figure 515

                                                    130 frac14 078 and 131 frac14 041

                                                    70 Bearing capacity

                                                    Thus using Equation 528

                                                    si frac14 078 041 68 176

                                                    65frac14 6mm

                                                    Consolidation settlement

                                                    Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                    1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                    434 (sod)

                                                    Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                    sc frac14 056 434 frac14 24mm

                                                    The total settlement is (6thorn 24) frac14 30mm

                                                    813

                                                    At base level N frac14 26 Then using Equation 830

                                                    qb frac14 40NDb

                                                    Bfrac14 40 26 2

                                                    025frac14 8320 kN=m2

                                                    ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                    Figure Q812

                                                    Bearing capacity 71

                                                    Over the length embedded in sand

                                                    N frac14 21 ie18thorn 24

                                                    2

                                                    Using Equation 831

                                                    qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                    For a single pile

                                                    Qf frac14 Abqb thorn Asqs

                                                    frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                    For the pile group assuming a group efficiency of 12

                                                    XQf frac14 12 9 604 frac14 6523 kN

                                                    Then the load factor is

                                                    F frac14 6523

                                                    2000thorn 1000frac14 21

                                                    (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                    Characteristic base resistance per unit area qbk frac14 8320

                                                    150frac14 5547 kNm2

                                                    Characteristic shaft resistance per unit area qsk frac14 42

                                                    150frac14 28 kNm2

                                                    Characteristic base and shaft resistances for a single pile

                                                    Rbk frac14 0252 5547 frac14 347 kN

                                                    Rsk frac14 4 025 2 28 frac14 56 kN

                                                    For a driven pile the partial factors are b frac14 s frac14 130

                                                    Design bearing resistance Rcd frac14 347

                                                    130thorn 56

                                                    130frac14 310 kN

                                                    For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                    Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                    (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                    From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                    72 Bearing capacity

                                                    N frac14 24thorn 26thorn 34

                                                    3frac14 28

                                                    Ic frac14 171

                                                    2814frac14 0016 ethEquation 818THORN

                                                    s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                    814

                                                    Using Equation 841

                                                    Tf frac14 DLcu thorn

                                                    4ethD2 d2THORNcuNc

                                                    frac14 eth 02 5 06 110THORN thorn

                                                    4eth022 012THORN110 9

                                                    frac14 207thorn 23 frac14 230 kN

                                                    Figure Q813

                                                    Bearing capacity 73

                                                    Chapter 9

                                                    Stability of slopes

                                                    91

                                                    Referring to Figure Q91

                                                    W frac14 417 19 frac14 792 kN=m

                                                    Q frac14 20 28 frac14 56 kN=m

                                                    Arc lengthAB frac14

                                                    180 73 90 frac14 115m

                                                    Arc length BC frac14

                                                    180 28 90 frac14 44m

                                                    The factor of safety is given by

                                                    F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                    90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                    Depth of tension crack z0 frac14 2cu

                                                    frac14 2 20

                                                    19frac14 21m

                                                    Arc length BD frac14

                                                    180 13

                                                    1

                                                    2 90 frac14 21m

                                                    F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                    The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                    Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                    14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                    Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                    The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                    92

                                                    u frac14 0

                                                    Depth factor D frac14 11

                                                    9frac14 122

                                                    Using Equation 92 with F frac14 10

                                                    Ns frac14 cu

                                                    FHfrac14 30

                                                    10 19 9frac14 0175

                                                    Hence from Figure 93

                                                    frac14 50

                                                    For F frac14 12

                                                    Ns frac14 30

                                                    12 19 9frac14 0146

                                                    frac14 27

                                                    93

                                                    Refer to Figure Q93

                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                    1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                    74 m

                                                    214 1deg

                                                    213 1deg

                                                    39 m

                                                    WB

                                                    D

                                                    C

                                                    28 m

                                                    21 m

                                                    A

                                                    Q

                                                    Soil (1)Soil (2)

                                                    73deg

                                                    Figure Q91

                                                    Stability of slopes 75

                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                    9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                    599 256 328 1372

                                                    Figure Q93

                                                    76 Stability of slopes

                                                    XW cos frac14 b

                                                    Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                    W sin frac14 bX

                                                    h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                    Arc length La frac14

                                                    180 57

                                                    1

                                                    2 326 frac14 327m

                                                    The factor of safety is given by

                                                    F frac14 c0La thorn tan0ethW cos ulTHORN

                                                    W sin

                                                    frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                    frac14 091

                                                    According to the limit state method

                                                    0d frac14 tan1tan 32

                                                    125

                                                    frac14 265

                                                    c0 frac14 8

                                                    160frac14 5 kN=m2

                                                    Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                    Design disturbing moment frac14 1075 kN=m

                                                    The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                    94

                                                    F frac14 1

                                                    W sin

                                                    Xfc0bthorn ethW ubTHORN tan0g sec

                                                    1thorn ethtan tan0=FTHORN

                                                    c0 frac14 8 kN=m2

                                                    0 frac14 32

                                                    c0b frac14 8 2 frac14 16 kN=m

                                                    W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                    Try F frac14 100

                                                    tan0

                                                    Ffrac14 0625

                                                    Stability of slopes 77

                                                    Values of u are as obtained in Figure Q93

                                                    SliceNo

                                                    h(m)

                                                    W frac14 bh(kNm)

                                                    W sin(kNm)

                                                    ub(kNm)

                                                    c0bthorn (W ub) tan0(kNm)

                                                    sec

                                                    1thorn (tan tan0)FProduct(kNm)

                                                    1 05 21 6 2 8 24 1078 262 13 55 31

                                                    23 33 30 1042 31

                                                    3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                    224 92 72 0931 67

                                                    6 50 210 11 40 100 85 0907 777 55 231 14

                                                    12 58 112 90 0889 80

                                                    8 60 252 1812

                                                    80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                    10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                    2154 88 116 0853 99

                                                    14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                    1074 1091

                                                    F frac14 1091

                                                    1074frac14 102 (assumed value 100)

                                                    Thus

                                                    F frac14 101

                                                    95

                                                    F frac14 1

                                                    W sin

                                                    XfWeth1 ruTHORN tan0g sec

                                                    1thorn ethtan tan0THORN=F

                                                    0 frac14 33

                                                    ru frac14 020

                                                    W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                    eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                    Try F frac14 110

                                                    tan 0

                                                    Ffrac14 tan 33

                                                    110frac14 0590

                                                    78 Stability of slopes

                                                    Referring to Figure Q95

                                                    SliceNo

                                                    h(m)

                                                    W frac14 bh(kNm)

                                                    W sin(kNm)

                                                    W(1 ru) tan0(kNm)

                                                    sec

                                                    1thorn ( tan tan0)FProduct(kNm)

                                                    1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                    2120 234 0892 209

                                                    4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                    1185 1271

                                                    Figure Q95

                                                    Stability of slopes 79

                                                    F frac14 1271

                                                    1185frac14 107

                                                    The trial value was 110 therefore take F to be 108

                                                    96

                                                    (a) Water table at surface the factor of safety is given by Equation 912

                                                    F frac14 0

                                                    sat

                                                    tan0

                                                    tan

                                                    ptie 15 frac14 92

                                                    19

                                                    tan 36

                                                    tan

                                                    tan frac14 0234

                                                    frac14 13

                                                    Water table well below surface the factor of safety is given by Equation 911

                                                    F frac14 tan0

                                                    tan

                                                    frac14 tan 36

                                                    tan 13

                                                    frac14 31

                                                    (b) 0d frac14 tan1tan 36

                                                    125

                                                    frac14 30

                                                    Depth of potential failure surface frac14 z

                                                    Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                    frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                    frac14 504z kN

                                                    Design disturbing moment per unit area Sd frac14 sat sin cos

                                                    frac14 19 z sin 13 cos 13

                                                    frac14 416z kN

                                                    Rd gtSd therefore the limit state for overall stability is satisfied

                                                    80 Stability of slopes

                                                    • Book Cover
                                                    • Title
                                                    • Contents
                                                    • Basic characteristics of soils
                                                    • Seepage
                                                    • Effective stress
                                                    • Shear strength
                                                    • Stresses and displacements
                                                    • Lateral earth pressure
                                                    • Consolidation theory
                                                    • Bearing capacity
                                                    • Stability of slopes

                                                      Factor of safety against lsquoheavingrsquo (Equation 310)

                                                      F frac14 ic

                                                      imfrac14 0d

                                                      whmfrac14 97 300

                                                      98 145frac14 20

                                                      With a filter

                                                      F frac14 0d thorn wwhm

                                                      3 frac14 eth97 300THORN thorn w98 145

                                                      w frac14 135 kN=m2

                                                      Depth of filterfrac14 13521frac14 065m (if above water level)

                                                      Effective stress 21

                                                      Chapter 4

                                                      Shear strength

                                                      41

                                                      frac14 295 kN=m2

                                                      u frac14 120 kN=m2

                                                      0 frac14 u frac14 295 120 frac14 175 kN=m2

                                                      f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                                                      42

                                                      03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                                                      100 452 552200 908 1108400 1810 2210800 3624 4424

                                                      The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                                                      Figure Q42

                                                      43

                                                      3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                                                      200 222 422400 218 618600 220 820

                                                      The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                                                      44

                                                      The modified shear strength parameters are

                                                      0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                                                      a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                                                      The coordinates of the stress point representing failure conditions in the test are

                                                      1

                                                      2eth1 2THORN frac14 1

                                                      2 170 frac14 85 kN=m2

                                                      1

                                                      2eth1 thorn 3THORN frac14 1

                                                      2eth270thorn 100THORN frac14 185 kN=m2

                                                      The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                                                      uf frac14 36 kN=m2

                                                      Figure Q43

                                                      Figure Q44

                                                      Shear strength 23

                                                      45

                                                      3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                                      150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                                      The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                                      The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                                      46

                                                      03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                                      200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                                      The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                                      Figure Q45

                                                      24 Shear strength

                                                      47

                                                      The torque required to produce shear failure is given by

                                                      T frac14 dh cud

                                                      2thorn 2

                                                      Z d=2

                                                      0

                                                      2r drcur

                                                      frac14 cud2h

                                                      2thorn 4cu

                                                      Z d=2

                                                      0

                                                      r2dr

                                                      frac14 cud2h

                                                      2thorn d

                                                      3

                                                      6

                                                      Then

                                                      35 frac14 cu52 10

                                                      2thorn 53

                                                      6

                                                      103

                                                      cu frac14 76 kN=m3

                                                      400

                                                      0 400 800 1200 1600

                                                      τ (k

                                                      Nm

                                                      2 )

                                                      σprime (kNm2)

                                                      34deg

                                                      315deg29deg

                                                      (a)

                                                      (b)

                                                      0 400

                                                      400

                                                      800 1200 1600

                                                      Failure envelope

                                                      300 500

                                                      σprime (kNm2)

                                                      τ (k

                                                      Nm

                                                      2 )

                                                      20 (kNm2)

                                                      31deg

                                                      Figure Q46

                                                      Shear strength 25

                                                      48

                                                      The relevant stress values are calculated as follows

                                                      3 frac14 600 kN=m2

                                                      1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                                      2(1 3) 0 40 79 107 139 159

                                                      1

                                                      2(01 thorn 03) 400 411 402 389 351 326

                                                      1

                                                      2(1 thorn 3) 600 640 679 707 739 759

                                                      The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                                      Af frac14 433 200

                                                      319frac14 073

                                                      49

                                                      B frac14 u33

                                                      frac14 144

                                                      350 200frac14 096

                                                      a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                                      0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                                      10 283 65 023

                                                      Figure Q48

                                                      26 Shear strength

                                                      The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                                      Figure Q49

                                                      Shear strength 27

                                                      Chapter 5

                                                      Stresses and displacements

                                                      51

                                                      Vertical stress is given by

                                                      z frac14 Qz2Ip frac14 5000

                                                      52Ip

                                                      Values of Ip are obtained from Table 51

                                                      r (m) rz Ip z (kNm2)

                                                      0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                                      10 20 0009 2

                                                      The variation of z with radial distance (r) is plotted in Figure Q51

                                                      Figure Q51

                                                      52

                                                      Below the centre load (Figure Q52)

                                                      r

                                                      zfrac14 0 for the 7500-kN load

                                                      Ip frac14 0478

                                                      r

                                                      zfrac14 5

                                                      4frac14 125 for the 10 000- and 9000-kN loads

                                                      Ip frac14 0045

                                                      Then

                                                      z frac14X Q

                                                      z2Ip

                                                      frac14 7500 0478

                                                      42thorn 10 000 0045

                                                      42thorn 9000 0045

                                                      42

                                                      frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                                      53

                                                      The vertical stress under a corner of a rectangular area is given by

                                                      z frac14 qIr

                                                      where values of Ir are obtained from Figure 510 In this case

                                                      z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                                      z

                                                      Figure Q52

                                                      Stresses and displacements 29

                                                      z (m) m n Ir z (kNm2)

                                                      0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                                      10 010 0005 5

                                                      z is plotted against z in Figure Q53

                                                      54

                                                      (a)

                                                      m frac14 125

                                                      12frac14 104

                                                      n frac14 18

                                                      12frac14 150

                                                      From Figure 510 Irfrac14 0196

                                                      z frac14 2 175 0196 frac14 68 kN=m2

                                                      Figure Q53

                                                      30 Stresses and displacements

                                                      (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                                      z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                                      55

                                                      Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                                      Px frac14 2Q

                                                      1

                                                      m2 thorn 1frac14 2 150

                                                      125frac14 76 kN=m

                                                      Equation 517 is used to obtain the pressure distribution

                                                      px frac14 4Q

                                                      h

                                                      m2n

                                                      ethm2 thorn n2THORN2 frac14150

                                                      m2n

                                                      ethm2 thorn n2THORN2 ethkN=m2THORN

                                                      Figure Q54

                                                      Stresses and displacements 31

                                                      n m2n

                                                      (m2 thorn n2)2

                                                      px(kNm2)

                                                      0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                      The pressure distribution is plotted in Figure Q55

                                                      56

                                                      H

                                                      Bfrac14 10

                                                      2frac14 5

                                                      L

                                                      Bfrac14 4

                                                      2frac14 2

                                                      D

                                                      Bfrac14 1

                                                      2frac14 05

                                                      Hence from Figure 515

                                                      131 frac14 082

                                                      130 frac14 094

                                                      Figure Q55

                                                      32 Stresses and displacements

                                                      The immediate settlement is given by Equation 528

                                                      si frac14 130131qB

                                                      Eu

                                                      frac14 094 082 200 2

                                                      45frac14 7mm

                                                      Stresses and displacements 33

                                                      Chapter 6

                                                      Lateral earth pressure

                                                      61

                                                      For 0 frac14 37 the active pressure coefficient is given by

                                                      Ka frac14 1 sin 37

                                                      1thorn sin 37frac14 025

                                                      The total active thrust (Equation 66a with c0 frac14 0) is

                                                      Pa frac14 1

                                                      2KaH

                                                      2 frac14 1

                                                      2 025 17 62 frac14 765 kN=m

                                                      If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                      K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                      and the thrust on the wall is

                                                      P0 frac14 1

                                                      2K0H

                                                      2 frac14 1

                                                      2 040 17 62 frac14 122 kN=m

                                                      62

                                                      The active pressure coefficients for the three soil types are as follows

                                                      Ka1 frac141 sin 35

                                                      1thorn sin 35frac14 0271

                                                      Ka2 frac141 sin 27

                                                      1thorn sin 27frac14 0375

                                                      ffiffiffiffiffiffiffiKa2

                                                      p frac14 0613

                                                      Ka3 frac141 sin 42

                                                      1thorn sin 42frac14 0198

                                                      Distribution of active pressure (plotted in Figure Q62)

                                                      Depth (m) Soil Active pressure (kNm2)

                                                      3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                      12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                      At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                      Total thrust frac14 571 kNm

                                                      Point of application is (4893571) m from the top of the wall ie 857m

                                                      Force (kN) Arm (m) Moment (kN m)

                                                      (1)1

                                                      2 0271 16 32 frac14 195 20 390

                                                      (2) 0271 16 3 2 frac14 260 40 1040

                                                      (3)1

                                                      2 0271 92 22 frac14 50 433 217

                                                      (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                      (5)1

                                                      2 0375 102 32 frac14 172 70 1204

                                                      (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                      (7)1

                                                      2 0198 112 42 frac14 177 1067 1889

                                                      (8)1

                                                      2 98 92 frac14 3969 90 35721

                                                      5713 48934

                                                      Figure Q62

                                                      Lateral earth pressure 35

                                                      63

                                                      (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                      Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                      frac14 245

                                                      At the lower end of the piling

                                                      pa frac14 Kaqthorn Kasatz Kaccu

                                                      frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                      frac14 115 kN=m2

                                                      pp frac14 Kpsatzthorn Kpccu

                                                      frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                      frac14 202 kN=m2

                                                      (b) For 0 frac14 26 and frac14 1

                                                      20

                                                      Ka frac14 035

                                                      Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                      pfrac14 145 ethEquation 619THORN

                                                      Kp frac14 37

                                                      Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                      pfrac14 47 ethEquation 624THORN

                                                      At the lower end of the piling

                                                      pa frac14 Kaqthorn Ka0z Kacc

                                                      0

                                                      frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                      frac14 187 kN=m2

                                                      pp frac14 Kp0zthorn Kpcc

                                                      0

                                                      frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                      frac14 198 kN=m2

                                                      36 Lateral earth pressure

                                                      64

                                                      (a) For 0 frac14 38 Ka frac14 024

                                                      0 frac14 20 98 frac14 102 kN=m3

                                                      The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                      Force (kN) Arm (m) Moment (kN m)

                                                      (1) 024 10 66 frac14 159 33 525

                                                      (2)1

                                                      2 024 17 392 frac14 310 400 1240

                                                      (3) 024 17 39 27 frac14 430 135 580

                                                      (4)1

                                                      2 024 102 272 frac14 89 090 80

                                                      (5)1

                                                      2 98 272 frac14 357 090 321

                                                      Hfrac14 1345 MH frac14 2746

                                                      (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                      (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                      XM frac14MV MH frac14 7790 kNm

                                                      Lever arm of base resultant

                                                      M

                                                      Vfrac14 779

                                                      488frac14 160

                                                      Eccentricity of base resultant

                                                      e frac14 200 160 frac14 040m

                                                      39 m

                                                      27 m

                                                      40 m

                                                      04 m

                                                      04 m

                                                      26 m

                                                      (7)

                                                      (9)

                                                      (1)(2)

                                                      (3)

                                                      (4)

                                                      (5)

                                                      (8)(6)

                                                      (10)

                                                      WT

                                                      10 kNm2

                                                      Hydrostatic

                                                      Figure Q64

                                                      Lateral earth pressure 37

                                                      Base pressures (Equation 627)

                                                      p frac14 VB

                                                      1 6e

                                                      B

                                                      frac14 488

                                                      4eth1 060THORN

                                                      frac14 195 kN=m2 and 49 kN=m2

                                                      Factor of safety against sliding (Equation 628)

                                                      F frac14 V tan

                                                      Hfrac14 488 tan 25

                                                      1345frac14 17

                                                      (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                      Hfrac14 1633 kN

                                                      V frac14 4879 kN

                                                      MH frac14 3453 kNm

                                                      MV frac14 10536 kNm

                                                      The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                      65

                                                      For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                      Kp

                                                      Ffrac14 385

                                                      2

                                                      0 frac14 20 98 frac14 102 kN=m3

                                                      The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                      Force (kN) Arm (m) Moment (kN m)

                                                      (1)1

                                                      2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                      (2) 026 17 45 d frac14 199d d2 995d2

                                                      (3)1

                                                      2 026 102 d2 frac14 133d2 d3 044d3

                                                      (4)1

                                                      2 385

                                                      2 17 152 frac14 368 dthorn 05 368d 184

                                                      (5)385

                                                      2 17 15 d frac14 491d d2 2455d2

                                                      (6)1

                                                      2 385

                                                      2 102 d2 frac14 982d2 d3 327d3

                                                      38 Lateral earth pressure

                                                      XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                      d3 thorn 516d2 283d 1724 frac14 0

                                                      d frac14 179m

                                                      Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                      Over additional 20 embedded depth

                                                      pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                      Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                      66

                                                      The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                      Ka frac14 sin 69=sin 105

                                                      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                      pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                      26664

                                                      37775

                                                      2

                                                      frac14 050

                                                      The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                      Pa frac14 1

                                                      2 050 19 7502 frac14 267 kN=m

                                                      Figure Q65

                                                      Lateral earth pressure 39

                                                      Horizontal component

                                                      Ph frac14 267 cos 40 frac14 205 kN=m

                                                      Vertical component

                                                      Pv frac14 267 sin 40 frac14 172 kN=m

                                                      Consider moments about the toe of the wall (Figure Q66) (per m)

                                                      Force (kN) Arm (m) Moment (kN m)

                                                      (1)1

                                                      2 175 650 235 frac14 1337 258 345

                                                      (2) 050 650 235 frac14 764 175 134

                                                      (3)1

                                                      2 070 650 235 frac14 535 127 68

                                                      (4) 100 400 235 frac14 940 200 188

                                                      (5) 1

                                                      2 080 050 235 frac14 47 027 1

                                                      Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                      Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                      Lever arm of base resultant

                                                      M

                                                      Vfrac14 795

                                                      525frac14 151m

                                                      Eccentricity of base resultant

                                                      e frac14 200 151 frac14 049m

                                                      Figure Q66

                                                      40 Lateral earth pressure

                                                      Base pressures (Equation 627)

                                                      p frac14 525

                                                      41 6 049

                                                      4

                                                      frac14 228 kN=m2 and 35 kN=m2

                                                      The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                      The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                      The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                      67

                                                      For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                      Force (kN) Arm (m) Moment (kNm)

                                                      (1)1

                                                      2 027 17 52 frac14 574 183 1050

                                                      (2) 027 17 5 3 frac14 689 500 3445

                                                      (3)1

                                                      2 027 102 32 frac14 124 550 682

                                                      (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                      (5)1

                                                      2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                      (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                      (7) 1

                                                      2 267

                                                      2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                      (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                      2 d frac14 163d d2thorn 650 82d2 1060d

                                                      Tie rod force per m frac14 T 0 0

                                                      XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                      d3 thorn 77d2 269d 1438 frac14 0

                                                      d frac14 467m

                                                      Depth of penetration frac14 12d frac14 560m

                                                      Lateral earth pressure 41

                                                      Algebraic sum of forces for d frac14 467m isX

                                                      F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                      T frac14 905 kN=m

                                                      Force in each tie rod frac14 25T frac14 226 kN

                                                      68

                                                      (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                      0 frac14 21 98 frac14 112 kN=m3

                                                      The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                      uC frac14 150

                                                      165 15 98 frac14 134 kN=m2

                                                      The average seepage pressure is

                                                      j frac14 15

                                                      165 98 frac14 09 kN=m3

                                                      Hence

                                                      0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                      0 j frac14 112 09 frac14 103 kN=m3

                                                      Figure Q67

                                                      42 Lateral earth pressure

                                                      Consider moments about the anchor point A (per m)

                                                      Force (kN) Arm (m) Moment (kN m)

                                                      (1) 10 026 150 frac14 390 60 2340

                                                      (2)1

                                                      2 026 18 452 frac14 474 15 711

                                                      (3) 026 18 45 105 frac14 2211 825 18240

                                                      (4)1

                                                      2 026 121 1052 frac14 1734 100 17340

                                                      (5)1

                                                      2 134 15 frac14 101 40 404

                                                      (6) 134 30 frac14 402 60 2412

                                                      (7)1

                                                      2 134 60 frac14 402 95 3819

                                                      571 4527(8) Ppm

                                                      115 115PPm

                                                      XM frac14 0

                                                      Ppm frac144527

                                                      115frac14 394 kN=m

                                                      Available passive resistance

                                                      Pp frac14 1

                                                      2 385 103 62 frac14 714 kN=m

                                                      Factor of safety

                                                      Fp frac14 Pp

                                                      Ppm

                                                      frac14 714

                                                      394frac14 18

                                                      Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                      Figure Q68

                                                      Lateral earth pressure 43

                                                      (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                      Consider moments (per m) about the tie point A

                                                      Force (kN) Arm (m)

                                                      (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                      (2)1

                                                      2 033 18 452 frac14 601 15

                                                      (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                      (4)1

                                                      2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                      (5)1

                                                      2 134 15 frac14 101 40

                                                      (6) 134 30 frac14 402 60

                                                      (7)1

                                                      2 134 d frac14 67d d3thorn 75

                                                      (8) 1

                                                      2 30 103 d2 frac141545d2 2d3thorn 75

                                                      Moment (kN m)

                                                      (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                      XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                      d3 thorn 827d2 466d 1518 frac14 0

                                                      By trial

                                                      d frac14 544m

                                                      The minimum depth of embedment required is 544m

                                                      69

                                                      For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                      0 frac14 20 98 frac14 102 kN=m3

                                                      The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                      44 Lateral earth pressure

                                                      uC frac14 147

                                                      173 26 98 frac14 216 kN=m2

                                                      and the average seepage pressure around the wall is

                                                      j frac14 26

                                                      173 98 frac14 15 kN=m3

                                                      Consider moments about the prop (A) (per m)

                                                      Force (kN) Arm (m) Moment (kN m)

                                                      (1)1

                                                      2 03 17 272 frac14 186 020 37

                                                      (2) 03 17 27 53 frac14 730 335 2445

                                                      (3)1

                                                      2 03 (102thorn 15) 532 frac14 493 423 2085

                                                      (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                      (5)1

                                                      2 216 26 frac14 281 243 684

                                                      (6) 216 27 frac14 583 465 2712

                                                      (7)1

                                                      2 216 60 frac14 648 800 5184

                                                      3055(8)

                                                      1

                                                      2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                      Factor of safety

                                                      Fr frac14 6885

                                                      3055frac14 225

                                                      Figure Q69

                                                      Lateral earth pressure 45

                                                      610

                                                      For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                      p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                      Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                      Using the recommendations of Twine and Roscoe

                                                      p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                      Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                      611

                                                      frac14 18 kN=m3 0 frac14 34

                                                      H frac14 350m nH frac14 335m mH frac14 185m

                                                      Consider a trial value of F frac14 20 Refer to Figure 635

                                                      0m frac14 tan1tan 34

                                                      20

                                                      frac14 186

                                                      Then

                                                      frac14 45 thorn 0m2frac14 543

                                                      W frac14 1

                                                      2 18 3502 cot 543 frac14 792 kN=m

                                                      Figure Q610

                                                      46 Lateral earth pressure

                                                      P frac14 1

                                                      2 s 3352 frac14 561s kN=m

                                                      U frac14 1

                                                      2 98 1852 cosec 543 frac14 206 kN=m

                                                      Equations 630 and 631 then become

                                                      561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                      792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                      ie

                                                      561s 0616N 405 frac14 0

                                                      792 0857N thorn 563 frac14 0

                                                      N frac14 848

                                                      0857frac14 989 kN=m

                                                      Then

                                                      561s 609 405 frac14 0

                                                      s frac14 649

                                                      561frac14 116 kN=m3

                                                      The calculations for trial values of F of 20 15 and 10 are summarized below

                                                      F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                      20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                      s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                      Figure Q611

                                                      Lateral earth pressure 47

                                                      612

                                                      For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                      45 thorn 0

                                                      2frac14 63

                                                      For the retained material between the surface and a depth of 36m

                                                      Pa frac14 1

                                                      2 030 18 362 frac14 350 kN=m

                                                      Weight of reinforced fill between the surface and a depth of 36m is

                                                      Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                      eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                      Eccentricity of Rv

                                                      e frac14 263 250 frac14 013m

                                                      The average vertical stress at a depth of 36m is

                                                      z frac14 Rv

                                                      L 2efrac14 324

                                                      474frac14 68 kN=m2

                                                      (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                      Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                      Tensile stress in the element frac14 138 103

                                                      65 3frac14 71N=mm2

                                                      Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                      Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                      Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                      The weight of ABC is

                                                      W frac14 1

                                                      2 18 52 265 frac14 124 kN=m

                                                      From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                      48 Lateral earth pressure

                                                      (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                      Tp frac14 032 68 120 065 frac14 170 kN

                                                      Tr frac14 213 420

                                                      418frac14 214 kN

                                                      Again the tensile failure and slipping limit states are satisfied for this element

                                                      Figure Q612

                                                      Lateral earth pressure 49

                                                      Chapter 7

                                                      Consolidation theory

                                                      71

                                                      Total change in thickness

                                                      H frac14 782 602 frac14 180mm

                                                      Average thickness frac14 1530thorn 180

                                                      2frac14 1620mm

                                                      Length of drainage path d frac14 1620

                                                      2frac14 810mm

                                                      Root time plot (Figure Q71a)

                                                      ffiffiffiffiffiffit90p frac14 33

                                                      t90 frac14 109min

                                                      cv frac14 0848d2

                                                      t90frac14 0848 8102

                                                      109 1440 365

                                                      106frac14 27m2=year

                                                      r0 frac14 782 764

                                                      782 602frac14 018

                                                      180frac14 0100

                                                      rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                      10 119

                                                      9 180frac14 0735

                                                      rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                      Log time plot (Figure Q71b)

                                                      t50 frac14 26min

                                                      cv frac14 0196d2

                                                      t50frac14 0196 8102

                                                      26 1440 365

                                                      106frac14 26m2=year

                                                      r0 frac14 782 763

                                                      782 602frac14 019

                                                      180frac14 0106

                                                      rp frac14 763 623

                                                      782 602frac14 140

                                                      180frac14 0778

                                                      rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                      Figure Q71(a)

                                                      Figure Q71(b)

                                                      Final void ratio

                                                      e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                      e

                                                      Hfrac14 1thorn e0

                                                      H0frac14 1thorn e1 thorne

                                                      H0

                                                      ie

                                                      e

                                                      180frac14 1631thorne

                                                      1710

                                                      e frac14 2936

                                                      1530frac14 0192

                                                      Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                      mv frac14 1

                                                      1thorn e0 e0 e101 00

                                                      frac14 1

                                                      1823 0192

                                                      0107frac14 098m2=MN

                                                      k frac14 cvmvw frac14 265 098 98

                                                      60 1440 365 103frac14 81 1010 m=s

                                                      72

                                                      Using Equation 77 (one-dimensional method)

                                                      sc frac14 e0 e11thorn e0 H

                                                      Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                      Figure Q72

                                                      52 Consolidation theory

                                                      Settlement

                                                      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                      1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                      318

                                                      Notes 5 92y 460thorn 84

                                                      Heave

                                                      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                      1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                      38

                                                      73

                                                      U frac14 f ethTvTHORN frac14 f cvt

                                                      d2

                                                      Hence if cv is constant

                                                      t1

                                                      t2frac14 d

                                                      21

                                                      d22

                                                      where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                      d1 frac14 95mm and d2 frac14 2500mm

                                                      for U frac14 050 t2 frac14 t1 d22

                                                      d21

                                                      frac14 20

                                                      60 24 365 25002

                                                      952frac14 263 years

                                                      for U lt 060 Tv frac14

                                                      4U2 (Equation 724(a))

                                                      t030 frac14 t050 0302

                                                      0502

                                                      frac14 263 036 frac14 095 years

                                                      Consolidation theory 53

                                                      74

                                                      The layer is open

                                                      d frac14 8

                                                      2frac14 4m

                                                      Tv frac14 cvtd2frac14 24 3

                                                      42frac14 0450

                                                      ui frac14 frac14 84 kN=m2

                                                      The excess pore water pressure is given by Equation 721

                                                      ue frac14Xmfrac141mfrac140

                                                      2ui

                                                      Msin

                                                      Mz

                                                      d

                                                      expethM2TvTHORN

                                                      In this case z frac14 d

                                                      sinMz

                                                      d

                                                      frac14 sinM

                                                      where

                                                      M frac14

                                                      23

                                                      25

                                                      2

                                                      M sin M M2Tv exp (M2Tv)

                                                      2thorn1 1110 0329

                                                      3

                                                      21 9993 457 105

                                                      ue frac14 2 84 2

                                                      1 0329 ethother terms negligibleTHORN

                                                      frac14 352 kN=m2

                                                      75

                                                      The layer is open

                                                      d frac14 6

                                                      2frac14 3m

                                                      Tv frac14 cvtd2frac14 10 3

                                                      32frac14 0333

                                                      The layer thickness will be divided into six equal parts ie m frac14 6

                                                      54 Consolidation theory

                                                      For an open layer

                                                      Tv frac14 4n

                                                      m2

                                                      n frac14 0333 62

                                                      4frac14 300

                                                      The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                      ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                      i j

                                                      0 1 2 3 4 5 6 7 8 9 10 11 12

                                                      0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                      The initial and 3-year isochrones are plotted in Figure Q75

                                                      Area under initial isochrone frac14 180 units

                                                      Area under 3-year isochrone frac14 63 units

                                                      The average degree of consolidation is given by Equation 725Thus

                                                      U frac14 1 63

                                                      180frac14 065

                                                      Figure Q75

                                                      Consolidation theory 55

                                                      76

                                                      At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                      0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                      The final consolidation settlement (one-dimensional method) is

                                                      sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                      Corrected time t frac14 2 1

                                                      2

                                                      40

                                                      52

                                                      frac14 1615 years

                                                      Tv frac14 cvtd2frac14 44 1615

                                                      42frac14 0444

                                                      From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                      77

                                                      The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                      Figure Q77

                                                      56 Consolidation theory

                                                      Point m n Ir (kNm2) sc (mm)

                                                      13020frac14 15 20

                                                      20frac14 10 0194 (4) 113 124

                                                      260

                                                      20frac14 30

                                                      20

                                                      20frac14 10 0204 (2) 59 65

                                                      360

                                                      20frac14 30

                                                      40

                                                      20frac14 20 0238 (1) 35 38

                                                      430

                                                      20frac14 15

                                                      40

                                                      20frac14 20 0224 (2) 65 72

                                                      Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                      78

                                                      Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                      (a) Immediate settlement

                                                      H

                                                      Bfrac14 30

                                                      35frac14 086

                                                      D

                                                      Bfrac14 2

                                                      35frac14 006

                                                      Figure Q78

                                                      Consolidation theory 57

                                                      From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                      si frac14 130131qB

                                                      Eufrac14 10 032 105 35

                                                      40frac14 30mm

                                                      (b) Consolidation settlement

                                                      Layer z (m) Dz Ic (kNm2) syod (mm)

                                                      1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                      3150

                                                      Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                      Now

                                                      H

                                                      Bfrac14 30

                                                      35frac14 086 and A frac14 065

                                                      from Figure 712 13 frac14 079

                                                      sc frac14 13sod frac14 079 315 frac14 250mm

                                                      Total settlement

                                                      s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                      79

                                                      Without sand drains

                                                      Uv frac14 025

                                                      Tv frac14 0049 ethfrom Figure 718THORN

                                                      t frac14 Tvd2

                                                      cvfrac14 0049 82

                                                      cvWith sand drains

                                                      R frac14 0564S frac14 0564 3 frac14 169m

                                                      n frac14 Rrfrac14 169

                                                      015frac14 113

                                                      Tr frac14 cht

                                                      4R2frac14 ch

                                                      4 1692 0049 82

                                                      cvethand ch frac14 cvTHORN

                                                      frac14 0275

                                                      Ur frac14 073 (from Figure 730)

                                                      58 Consolidation theory

                                                      Using Equation 740

                                                      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                      U frac14 080

                                                      710

                                                      Without sand drains

                                                      Uv frac14 090

                                                      Tv frac14 0848

                                                      t frac14 Tvd2

                                                      cvfrac14 0848 102

                                                      96frac14 88 years

                                                      With sand drains

                                                      R frac14 0564S frac14 0564 4 frac14 226m

                                                      n frac14 Rrfrac14 226

                                                      015frac14 15

                                                      Tr

                                                      Tvfrac14 chcv

                                                      d2

                                                      4R2ethsame tTHORN

                                                      Tr

                                                      Tvfrac14 140

                                                      96 102

                                                      4 2262frac14 714 eth1THORN

                                                      Using Equation 740

                                                      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                      An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                      Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                      040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                      Thus

                                                      Uv frac14 0295 and Ur frac14 086

                                                      t frac14 88 00683

                                                      0848frac14 07 years

                                                      Consolidation theory 59

                                                      Chapter 8

                                                      Bearing capacity

                                                      81

                                                      (a) The ultimate bearing capacity is given by Equation 83

                                                      qf frac14 cNc thorn DNq thorn 1

                                                      2BN

                                                      For u frac14 0

                                                      Nc frac14 514 Nq frac14 1 N frac14 0

                                                      qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                      The net ultimate bearing capacity is

                                                      qnf frac14 qf D frac14 540 kN=m2

                                                      The net foundation pressure is

                                                      qn frac14 q D frac14 425

                                                      2 eth21 1THORN frac14 192 kN=m2

                                                      The factor of safety (Equation 86) is

                                                      F frac14 qnfqnfrac14 540

                                                      192frac14 28

                                                      (b) For 0 frac14 28

                                                      Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                      qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                      2 112 2 13

                                                      frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                      qnf frac14 574 112 frac14 563 kN=m2

                                                      F frac14 563

                                                      192frac14 29

                                                      (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                      82

                                                      For 0 frac14 38

                                                      Nq frac14 49 N frac14 67

                                                      qnf frac14 DethNq 1THORN thorn 1

                                                      2BN ethfrom Equation 83THORN

                                                      frac14 eth18 075 48THORN thorn 1

                                                      2 18 15 67

                                                      frac14 648thorn 905 frac14 1553 kN=m2

                                                      qn frac14 500

                                                      15 eth18 075THORN frac14 320 kN=m2

                                                      F frac14 qnfqnfrac14 1553

                                                      320frac14 48

                                                      0d frac14 tan1tan 38

                                                      125

                                                      frac14 32 therefore Nq frac14 23 and N frac14 25

                                                      Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                      2 18 15 25

                                                      frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                      Design load (action) Vd frac14 500 kN=m

                                                      The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                      83

                                                      D

                                                      Bfrac14 350

                                                      225frac14 155

                                                      From Figure 85 for a square foundation

                                                      Nc frac14 81

                                                      Bearing capacity 61

                                                      For a rectangular foundation (L frac14 450m B frac14 225m)

                                                      Nc frac14 084thorn 016B

                                                      L

                                                      81 frac14 745

                                                      Using Equation 810

                                                      qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                      For F frac14 3

                                                      qn frac14 1006

                                                      3frac14 335 kN=m2

                                                      q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                      Design load frac14 405 450 225 frac14 4100 kN

                                                      Design undrained strength cud frac14 135

                                                      14frac14 96 kN=m2

                                                      Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                      frac14 7241 kN

                                                      Design load Vd frac14 4100 kN

                                                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                      84

                                                      For 0 frac14 40

                                                      Nq frac14 64 N frac14 95

                                                      qnf frac14 DethNq 1THORN thorn 04BN

                                                      (a) Water table 5m below ground level

                                                      qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                      qn frac14 400 17 frac14 383 kN=m2

                                                      F frac14 2686

                                                      383frac14 70

                                                      (b) Water table 1m below ground level (ie at foundation level)

                                                      0 frac14 20 98 frac14 102 kN=m3

                                                      62 Bearing capacity

                                                      qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                      F frac14 2040

                                                      383frac14 53

                                                      (c) Water table at ground level with upward hydraulic gradient 02

                                                      eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                      qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                      F frac14 1296

                                                      392frac14 33

                                                      85

                                                      The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                      Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                      Design value of 0 frac14 tan1tan 39

                                                      125

                                                      frac14 33

                                                      For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                      Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                      86

                                                      (a) Undrained shear for u frac14 0

                                                      Nc frac14 514 Nq frac14 1 N frac14 0

                                                      qnf frac14 12cuNc

                                                      frac14 12 100 514 frac14 617 kN=m2

                                                      qn frac14 qnfFfrac14 617

                                                      3frac14 206 kN=m2

                                                      q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                      Bearing capacity 63

                                                      Drained shear for 0 frac14 32

                                                      Nq frac14 23 N frac14 25

                                                      0 frac14 21 98 frac14 112 kN=m3

                                                      qnf frac14 0DethNq 1THORN thorn 040BN

                                                      frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                      frac14 694 kN=m2

                                                      q frac14 694

                                                      3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                      Design load frac14 42 227 frac14 3632 kN

                                                      (b) Design undrained strength cud frac14 100

                                                      14frac14 71 kNm2

                                                      Design bearing resistance Rd frac14 12cudNe area

                                                      frac14 12 71 514 42

                                                      frac14 7007 kN

                                                      For drained shear 0d frac14 tan1tan 32

                                                      125

                                                      frac14 26

                                                      Nq frac14 12 N frac14 10

                                                      Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                      (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                      Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                      1 2 100 0175 0700qn 0182qn

                                                      2 6 033 0044 0176qn 0046qn

                                                      3 10 020 0017 0068qn 0018qn

                                                      0246qn

                                                      Diameter of equivalent circle B frac14 45m

                                                      H

                                                      Bfrac14 12

                                                      45frac14 27 and A frac14 042

                                                      13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                      64 Bearing capacity

                                                      For sc frac14 30mm

                                                      qn frac14 30

                                                      0147frac14 204 kN=m2

                                                      q frac14 204thorn 21 frac14 225 kN=m2

                                                      Design load frac14 42 225 frac14 3600 kN

                                                      The design load is 3600 kN settlement being the limiting criterion

                                                      87

                                                      D

                                                      Bfrac14 8

                                                      4frac14 20

                                                      From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                      F frac14 cuNc

                                                      Dfrac14 40 71

                                                      20 8frac14 18

                                                      88

                                                      Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                      Design value of 0 frac14 tan1tan 38

                                                      125

                                                      frac14 32

                                                      Figure Q86

                                                      Bearing capacity 65

                                                      For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                      Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                      The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                      Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                      For B frac14 250m qn frac14 3750

                                                      2502 17 frac14 583 kN=m2

                                                      From Figure 510 m frac14 n frac14 126

                                                      6frac14 021

                                                      Ir frac14 0019

                                                      Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                      Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                      The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                      89

                                                      Depth (m) N 0v (kNm2) CN N1

                                                      070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                      Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                      (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                      Cw frac14 05thorn 0530

                                                      47

                                                      frac14 082

                                                      66 Bearing capacity

                                                      Thus

                                                      qa frac14 150 082 frac14 120 kN=m2

                                                      (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                      Thus

                                                      qa frac14 90 15 frac14 135 kN=m2

                                                      (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                      Ic frac14 171

                                                      1014frac14 0068

                                                      From Equation 819(a) with s frac14 25mm

                                                      q frac14 25

                                                      3507 0068frac14 150 kN=m2

                                                      810

                                                      Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                      Izp frac14 05thorn 01130

                                                      16 27

                                                      05

                                                      frac14 067

                                                      Refer to Figure Q810

                                                      E frac14 25qc

                                                      Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                      Ez (mm3MN)

                                                      1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                      0203

                                                      C1 frac14 1 0500qnfrac14 1 05 12 16

                                                      130frac14 093

                                                      C2 frac14 1 ethsayTHORN

                                                      s frac14 C1C2qnX Iz

                                                      Ez frac14 093 1 130 0203 frac14 25mm

                                                      Bearing capacity 67

                                                      811

                                                      At pile base level

                                                      cu frac14 220 kN=m2

                                                      qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                      Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                      00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                      qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                      Then

                                                      Qf frac14 Abqb thorn Asqs

                                                      frac14

                                                      4 32 1980

                                                      thorn eth 105 139 86THORN

                                                      frac14 13 996thorn 3941 frac14 17 937 kN

                                                      0 01 02 03 04 05 06 07

                                                      0 2 4 6 8 10 12 14

                                                      1

                                                      2

                                                      3

                                                      4

                                                      5

                                                      6

                                                      7

                                                      8

                                                      (1)

                                                      (2)

                                                      (3)

                                                      (4)

                                                      (5)

                                                      qc

                                                      qc

                                                      Iz

                                                      Iz

                                                      (MNm2)

                                                      z (m)

                                                      Figure Q810

                                                      68 Bearing capacity

                                                      Allowable load

                                                      ethaTHORN Qf

                                                      2frac14 17 937

                                                      2frac14 8968 kN

                                                      ethbTHORN Abqb

                                                      3thorn Asqs frac14 13 996

                                                      3thorn 3941 frac14 8606 kN

                                                      ie allowable load frac14 8600 kN

                                                      Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                      According to the limit state method

                                                      Characteristic undrained strength at base level cuk frac14 220

                                                      150kN=m2

                                                      Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                      150frac14 1320 kN=m2

                                                      Characteristic shaft resistance qsk frac14 00150

                                                      frac14 86

                                                      150frac14 57 kN=m2

                                                      Characteristic base and shaft resistances

                                                      Rbk frac14

                                                      4 32 1320 frac14 9330 kN

                                                      Rsk frac14 105 139 86

                                                      150frac14 2629 kN

                                                      For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                      Design bearing resistance Rcd frac14 9330

                                                      160thorn 2629

                                                      130

                                                      frac14 5831thorn 2022

                                                      frac14 7850 kN

                                                      Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                      812

                                                      ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                      qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                      For a single pile

                                                      Qf frac14 Abqb thorn Asqs

                                                      frac14

                                                      4 062 1305

                                                      thorn eth 06 15 42THORN

                                                      frac14 369thorn 1187 frac14 1556 kN

                                                      Bearing capacity 69

                                                      Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                      eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                      (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                      (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                      qbkfrac14 9cuk frac14 9 220

                                                      150frac14 1320 kN=m2

                                                      qskfrac14cuk frac14 040 105

                                                      150frac14 28 kN=m2

                                                      Rbkfrac14

                                                      4 0602 1320 frac14 373 kN

                                                      Rskfrac14 060 15 28 frac14 791 kN

                                                      Rcdfrac14 373

                                                      160thorn 791

                                                      130frac14 233thorn 608 frac14 841 kN

                                                      Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                      Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                      (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                      q frac14 21 000

                                                      1762frac14 68 kN=m2

                                                      Immediate settlement

                                                      H

                                                      Bfrac14 15

                                                      176frac14 085

                                                      D

                                                      Bfrac14 13

                                                      176frac14 074

                                                      L

                                                      Bfrac14 1

                                                      Hence from Figure 515

                                                      130 frac14 078 and 131 frac14 041

                                                      70 Bearing capacity

                                                      Thus using Equation 528

                                                      si frac14 078 041 68 176

                                                      65frac14 6mm

                                                      Consolidation settlement

                                                      Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                      1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                      434 (sod)

                                                      Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                      sc frac14 056 434 frac14 24mm

                                                      The total settlement is (6thorn 24) frac14 30mm

                                                      813

                                                      At base level N frac14 26 Then using Equation 830

                                                      qb frac14 40NDb

                                                      Bfrac14 40 26 2

                                                      025frac14 8320 kN=m2

                                                      ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                      Figure Q812

                                                      Bearing capacity 71

                                                      Over the length embedded in sand

                                                      N frac14 21 ie18thorn 24

                                                      2

                                                      Using Equation 831

                                                      qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                      For a single pile

                                                      Qf frac14 Abqb thorn Asqs

                                                      frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                      For the pile group assuming a group efficiency of 12

                                                      XQf frac14 12 9 604 frac14 6523 kN

                                                      Then the load factor is

                                                      F frac14 6523

                                                      2000thorn 1000frac14 21

                                                      (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                      Characteristic base resistance per unit area qbk frac14 8320

                                                      150frac14 5547 kNm2

                                                      Characteristic shaft resistance per unit area qsk frac14 42

                                                      150frac14 28 kNm2

                                                      Characteristic base and shaft resistances for a single pile

                                                      Rbk frac14 0252 5547 frac14 347 kN

                                                      Rsk frac14 4 025 2 28 frac14 56 kN

                                                      For a driven pile the partial factors are b frac14 s frac14 130

                                                      Design bearing resistance Rcd frac14 347

                                                      130thorn 56

                                                      130frac14 310 kN

                                                      For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                      Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                      (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                      From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                      72 Bearing capacity

                                                      N frac14 24thorn 26thorn 34

                                                      3frac14 28

                                                      Ic frac14 171

                                                      2814frac14 0016 ethEquation 818THORN

                                                      s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                      814

                                                      Using Equation 841

                                                      Tf frac14 DLcu thorn

                                                      4ethD2 d2THORNcuNc

                                                      frac14 eth 02 5 06 110THORN thorn

                                                      4eth022 012THORN110 9

                                                      frac14 207thorn 23 frac14 230 kN

                                                      Figure Q813

                                                      Bearing capacity 73

                                                      Chapter 9

                                                      Stability of slopes

                                                      91

                                                      Referring to Figure Q91

                                                      W frac14 417 19 frac14 792 kN=m

                                                      Q frac14 20 28 frac14 56 kN=m

                                                      Arc lengthAB frac14

                                                      180 73 90 frac14 115m

                                                      Arc length BC frac14

                                                      180 28 90 frac14 44m

                                                      The factor of safety is given by

                                                      F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                      90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                      Depth of tension crack z0 frac14 2cu

                                                      frac14 2 20

                                                      19frac14 21m

                                                      Arc length BD frac14

                                                      180 13

                                                      1

                                                      2 90 frac14 21m

                                                      F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                      The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                      Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                      14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                      Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                      The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                      92

                                                      u frac14 0

                                                      Depth factor D frac14 11

                                                      9frac14 122

                                                      Using Equation 92 with F frac14 10

                                                      Ns frac14 cu

                                                      FHfrac14 30

                                                      10 19 9frac14 0175

                                                      Hence from Figure 93

                                                      frac14 50

                                                      For F frac14 12

                                                      Ns frac14 30

                                                      12 19 9frac14 0146

                                                      frac14 27

                                                      93

                                                      Refer to Figure Q93

                                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                      1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                      74 m

                                                      214 1deg

                                                      213 1deg

                                                      39 m

                                                      WB

                                                      D

                                                      C

                                                      28 m

                                                      21 m

                                                      A

                                                      Q

                                                      Soil (1)Soil (2)

                                                      73deg

                                                      Figure Q91

                                                      Stability of slopes 75

                                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                      9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                      599 256 328 1372

                                                      Figure Q93

                                                      76 Stability of slopes

                                                      XW cos frac14 b

                                                      Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                      W sin frac14 bX

                                                      h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                      Arc length La frac14

                                                      180 57

                                                      1

                                                      2 326 frac14 327m

                                                      The factor of safety is given by

                                                      F frac14 c0La thorn tan0ethW cos ulTHORN

                                                      W sin

                                                      frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                      frac14 091

                                                      According to the limit state method

                                                      0d frac14 tan1tan 32

                                                      125

                                                      frac14 265

                                                      c0 frac14 8

                                                      160frac14 5 kN=m2

                                                      Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                      Design disturbing moment frac14 1075 kN=m

                                                      The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                      94

                                                      F frac14 1

                                                      W sin

                                                      Xfc0bthorn ethW ubTHORN tan0g sec

                                                      1thorn ethtan tan0=FTHORN

                                                      c0 frac14 8 kN=m2

                                                      0 frac14 32

                                                      c0b frac14 8 2 frac14 16 kN=m

                                                      W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                      Try F frac14 100

                                                      tan0

                                                      Ffrac14 0625

                                                      Stability of slopes 77

                                                      Values of u are as obtained in Figure Q93

                                                      SliceNo

                                                      h(m)

                                                      W frac14 bh(kNm)

                                                      W sin(kNm)

                                                      ub(kNm)

                                                      c0bthorn (W ub) tan0(kNm)

                                                      sec

                                                      1thorn (tan tan0)FProduct(kNm)

                                                      1 05 21 6 2 8 24 1078 262 13 55 31

                                                      23 33 30 1042 31

                                                      3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                      224 92 72 0931 67

                                                      6 50 210 11 40 100 85 0907 777 55 231 14

                                                      12 58 112 90 0889 80

                                                      8 60 252 1812

                                                      80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                      10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                      2154 88 116 0853 99

                                                      14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                      1074 1091

                                                      F frac14 1091

                                                      1074frac14 102 (assumed value 100)

                                                      Thus

                                                      F frac14 101

                                                      95

                                                      F frac14 1

                                                      W sin

                                                      XfWeth1 ruTHORN tan0g sec

                                                      1thorn ethtan tan0THORN=F

                                                      0 frac14 33

                                                      ru frac14 020

                                                      W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                      eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                      Try F frac14 110

                                                      tan 0

                                                      Ffrac14 tan 33

                                                      110frac14 0590

                                                      78 Stability of slopes

                                                      Referring to Figure Q95

                                                      SliceNo

                                                      h(m)

                                                      W frac14 bh(kNm)

                                                      W sin(kNm)

                                                      W(1 ru) tan0(kNm)

                                                      sec

                                                      1thorn ( tan tan0)FProduct(kNm)

                                                      1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                      2120 234 0892 209

                                                      4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                      1185 1271

                                                      Figure Q95

                                                      Stability of slopes 79

                                                      F frac14 1271

                                                      1185frac14 107

                                                      The trial value was 110 therefore take F to be 108

                                                      96

                                                      (a) Water table at surface the factor of safety is given by Equation 912

                                                      F frac14 0

                                                      sat

                                                      tan0

                                                      tan

                                                      ptie 15 frac14 92

                                                      19

                                                      tan 36

                                                      tan

                                                      tan frac14 0234

                                                      frac14 13

                                                      Water table well below surface the factor of safety is given by Equation 911

                                                      F frac14 tan0

                                                      tan

                                                      frac14 tan 36

                                                      tan 13

                                                      frac14 31

                                                      (b) 0d frac14 tan1tan 36

                                                      125

                                                      frac14 30

                                                      Depth of potential failure surface frac14 z

                                                      Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                      frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                      frac14 504z kN

                                                      Design disturbing moment per unit area Sd frac14 sat sin cos

                                                      frac14 19 z sin 13 cos 13

                                                      frac14 416z kN

                                                      Rd gtSd therefore the limit state for overall stability is satisfied

                                                      80 Stability of slopes

                                                      • Book Cover
                                                      • Title
                                                      • Contents
                                                      • Basic characteristics of soils
                                                      • Seepage
                                                      • Effective stress
                                                      • Shear strength
                                                      • Stresses and displacements
                                                      • Lateral earth pressure
                                                      • Consolidation theory
                                                      • Bearing capacity
                                                      • Stability of slopes

                                                        Chapter 4

                                                        Shear strength

                                                        41

                                                        frac14 295 kN=m2

                                                        u frac14 120 kN=m2

                                                        0 frac14 u frac14 295 120 frac14 175 kN=m2

                                                        f frac14 c0 thorn 0 tan 0 frac14 12thorn 175 tan 30 frac14 113 kN=m2

                                                        42

                                                        03 (kNm2) 1 3 (kNm2) 01 (kNm2)

                                                        100 452 552200 908 1108400 1810 2210800 3624 4424

                                                        The Mohr circles are drawn in Figure Q42 together with the failure envelope fromwhich 0 frac14 44

                                                        Figure Q42

                                                        43

                                                        3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                                                        200 222 422400 218 618600 220 820

                                                        The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                                                        44

                                                        The modified shear strength parameters are

                                                        0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                                                        a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                                                        The coordinates of the stress point representing failure conditions in the test are

                                                        1

                                                        2eth1 2THORN frac14 1

                                                        2 170 frac14 85 kN=m2

                                                        1

                                                        2eth1 thorn 3THORN frac14 1

                                                        2eth270thorn 100THORN frac14 185 kN=m2

                                                        The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                                                        uf frac14 36 kN=m2

                                                        Figure Q43

                                                        Figure Q44

                                                        Shear strength 23

                                                        45

                                                        3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                                        150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                                        The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                                        The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                                        46

                                                        03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                                        200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                                        The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                                        Figure Q45

                                                        24 Shear strength

                                                        47

                                                        The torque required to produce shear failure is given by

                                                        T frac14 dh cud

                                                        2thorn 2

                                                        Z d=2

                                                        0

                                                        2r drcur

                                                        frac14 cud2h

                                                        2thorn 4cu

                                                        Z d=2

                                                        0

                                                        r2dr

                                                        frac14 cud2h

                                                        2thorn d

                                                        3

                                                        6

                                                        Then

                                                        35 frac14 cu52 10

                                                        2thorn 53

                                                        6

                                                        103

                                                        cu frac14 76 kN=m3

                                                        400

                                                        0 400 800 1200 1600

                                                        τ (k

                                                        Nm

                                                        2 )

                                                        σprime (kNm2)

                                                        34deg

                                                        315deg29deg

                                                        (a)

                                                        (b)

                                                        0 400

                                                        400

                                                        800 1200 1600

                                                        Failure envelope

                                                        300 500

                                                        σprime (kNm2)

                                                        τ (k

                                                        Nm

                                                        2 )

                                                        20 (kNm2)

                                                        31deg

                                                        Figure Q46

                                                        Shear strength 25

                                                        48

                                                        The relevant stress values are calculated as follows

                                                        3 frac14 600 kN=m2

                                                        1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                                        2(1 3) 0 40 79 107 139 159

                                                        1

                                                        2(01 thorn 03) 400 411 402 389 351 326

                                                        1

                                                        2(1 thorn 3) 600 640 679 707 739 759

                                                        The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                                        Af frac14 433 200

                                                        319frac14 073

                                                        49

                                                        B frac14 u33

                                                        frac14 144

                                                        350 200frac14 096

                                                        a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                                        0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                                        10 283 65 023

                                                        Figure Q48

                                                        26 Shear strength

                                                        The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                                        Figure Q49

                                                        Shear strength 27

                                                        Chapter 5

                                                        Stresses and displacements

                                                        51

                                                        Vertical stress is given by

                                                        z frac14 Qz2Ip frac14 5000

                                                        52Ip

                                                        Values of Ip are obtained from Table 51

                                                        r (m) rz Ip z (kNm2)

                                                        0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                                        10 20 0009 2

                                                        The variation of z with radial distance (r) is plotted in Figure Q51

                                                        Figure Q51

                                                        52

                                                        Below the centre load (Figure Q52)

                                                        r

                                                        zfrac14 0 for the 7500-kN load

                                                        Ip frac14 0478

                                                        r

                                                        zfrac14 5

                                                        4frac14 125 for the 10 000- and 9000-kN loads

                                                        Ip frac14 0045

                                                        Then

                                                        z frac14X Q

                                                        z2Ip

                                                        frac14 7500 0478

                                                        42thorn 10 000 0045

                                                        42thorn 9000 0045

                                                        42

                                                        frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                                        53

                                                        The vertical stress under a corner of a rectangular area is given by

                                                        z frac14 qIr

                                                        where values of Ir are obtained from Figure 510 In this case

                                                        z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                                        z

                                                        Figure Q52

                                                        Stresses and displacements 29

                                                        z (m) m n Ir z (kNm2)

                                                        0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                                        10 010 0005 5

                                                        z is plotted against z in Figure Q53

                                                        54

                                                        (a)

                                                        m frac14 125

                                                        12frac14 104

                                                        n frac14 18

                                                        12frac14 150

                                                        From Figure 510 Irfrac14 0196

                                                        z frac14 2 175 0196 frac14 68 kN=m2

                                                        Figure Q53

                                                        30 Stresses and displacements

                                                        (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                                        z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                                        55

                                                        Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                                        Px frac14 2Q

                                                        1

                                                        m2 thorn 1frac14 2 150

                                                        125frac14 76 kN=m

                                                        Equation 517 is used to obtain the pressure distribution

                                                        px frac14 4Q

                                                        h

                                                        m2n

                                                        ethm2 thorn n2THORN2 frac14150

                                                        m2n

                                                        ethm2 thorn n2THORN2 ethkN=m2THORN

                                                        Figure Q54

                                                        Stresses and displacements 31

                                                        n m2n

                                                        (m2 thorn n2)2

                                                        px(kNm2)

                                                        0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                        The pressure distribution is plotted in Figure Q55

                                                        56

                                                        H

                                                        Bfrac14 10

                                                        2frac14 5

                                                        L

                                                        Bfrac14 4

                                                        2frac14 2

                                                        D

                                                        Bfrac14 1

                                                        2frac14 05

                                                        Hence from Figure 515

                                                        131 frac14 082

                                                        130 frac14 094

                                                        Figure Q55

                                                        32 Stresses and displacements

                                                        The immediate settlement is given by Equation 528

                                                        si frac14 130131qB

                                                        Eu

                                                        frac14 094 082 200 2

                                                        45frac14 7mm

                                                        Stresses and displacements 33

                                                        Chapter 6

                                                        Lateral earth pressure

                                                        61

                                                        For 0 frac14 37 the active pressure coefficient is given by

                                                        Ka frac14 1 sin 37

                                                        1thorn sin 37frac14 025

                                                        The total active thrust (Equation 66a with c0 frac14 0) is

                                                        Pa frac14 1

                                                        2KaH

                                                        2 frac14 1

                                                        2 025 17 62 frac14 765 kN=m

                                                        If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                        K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                        and the thrust on the wall is

                                                        P0 frac14 1

                                                        2K0H

                                                        2 frac14 1

                                                        2 040 17 62 frac14 122 kN=m

                                                        62

                                                        The active pressure coefficients for the three soil types are as follows

                                                        Ka1 frac141 sin 35

                                                        1thorn sin 35frac14 0271

                                                        Ka2 frac141 sin 27

                                                        1thorn sin 27frac14 0375

                                                        ffiffiffiffiffiffiffiKa2

                                                        p frac14 0613

                                                        Ka3 frac141 sin 42

                                                        1thorn sin 42frac14 0198

                                                        Distribution of active pressure (plotted in Figure Q62)

                                                        Depth (m) Soil Active pressure (kNm2)

                                                        3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                        12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                        At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                        Total thrust frac14 571 kNm

                                                        Point of application is (4893571) m from the top of the wall ie 857m

                                                        Force (kN) Arm (m) Moment (kN m)

                                                        (1)1

                                                        2 0271 16 32 frac14 195 20 390

                                                        (2) 0271 16 3 2 frac14 260 40 1040

                                                        (3)1

                                                        2 0271 92 22 frac14 50 433 217

                                                        (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                        (5)1

                                                        2 0375 102 32 frac14 172 70 1204

                                                        (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                        (7)1

                                                        2 0198 112 42 frac14 177 1067 1889

                                                        (8)1

                                                        2 98 92 frac14 3969 90 35721

                                                        5713 48934

                                                        Figure Q62

                                                        Lateral earth pressure 35

                                                        63

                                                        (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                        Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                        frac14 245

                                                        At the lower end of the piling

                                                        pa frac14 Kaqthorn Kasatz Kaccu

                                                        frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                        frac14 115 kN=m2

                                                        pp frac14 Kpsatzthorn Kpccu

                                                        frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                        frac14 202 kN=m2

                                                        (b) For 0 frac14 26 and frac14 1

                                                        20

                                                        Ka frac14 035

                                                        Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                        pfrac14 145 ethEquation 619THORN

                                                        Kp frac14 37

                                                        Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                        pfrac14 47 ethEquation 624THORN

                                                        At the lower end of the piling

                                                        pa frac14 Kaqthorn Ka0z Kacc

                                                        0

                                                        frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                        frac14 187 kN=m2

                                                        pp frac14 Kp0zthorn Kpcc

                                                        0

                                                        frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                        frac14 198 kN=m2

                                                        36 Lateral earth pressure

                                                        64

                                                        (a) For 0 frac14 38 Ka frac14 024

                                                        0 frac14 20 98 frac14 102 kN=m3

                                                        The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                        Force (kN) Arm (m) Moment (kN m)

                                                        (1) 024 10 66 frac14 159 33 525

                                                        (2)1

                                                        2 024 17 392 frac14 310 400 1240

                                                        (3) 024 17 39 27 frac14 430 135 580

                                                        (4)1

                                                        2 024 102 272 frac14 89 090 80

                                                        (5)1

                                                        2 98 272 frac14 357 090 321

                                                        Hfrac14 1345 MH frac14 2746

                                                        (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                        (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                        XM frac14MV MH frac14 7790 kNm

                                                        Lever arm of base resultant

                                                        M

                                                        Vfrac14 779

                                                        488frac14 160

                                                        Eccentricity of base resultant

                                                        e frac14 200 160 frac14 040m

                                                        39 m

                                                        27 m

                                                        40 m

                                                        04 m

                                                        04 m

                                                        26 m

                                                        (7)

                                                        (9)

                                                        (1)(2)

                                                        (3)

                                                        (4)

                                                        (5)

                                                        (8)(6)

                                                        (10)

                                                        WT

                                                        10 kNm2

                                                        Hydrostatic

                                                        Figure Q64

                                                        Lateral earth pressure 37

                                                        Base pressures (Equation 627)

                                                        p frac14 VB

                                                        1 6e

                                                        B

                                                        frac14 488

                                                        4eth1 060THORN

                                                        frac14 195 kN=m2 and 49 kN=m2

                                                        Factor of safety against sliding (Equation 628)

                                                        F frac14 V tan

                                                        Hfrac14 488 tan 25

                                                        1345frac14 17

                                                        (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                        Hfrac14 1633 kN

                                                        V frac14 4879 kN

                                                        MH frac14 3453 kNm

                                                        MV frac14 10536 kNm

                                                        The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                        65

                                                        For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                        Kp

                                                        Ffrac14 385

                                                        2

                                                        0 frac14 20 98 frac14 102 kN=m3

                                                        The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                        Force (kN) Arm (m) Moment (kN m)

                                                        (1)1

                                                        2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                        (2) 026 17 45 d frac14 199d d2 995d2

                                                        (3)1

                                                        2 026 102 d2 frac14 133d2 d3 044d3

                                                        (4)1

                                                        2 385

                                                        2 17 152 frac14 368 dthorn 05 368d 184

                                                        (5)385

                                                        2 17 15 d frac14 491d d2 2455d2

                                                        (6)1

                                                        2 385

                                                        2 102 d2 frac14 982d2 d3 327d3

                                                        38 Lateral earth pressure

                                                        XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                        d3 thorn 516d2 283d 1724 frac14 0

                                                        d frac14 179m

                                                        Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                        Over additional 20 embedded depth

                                                        pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                        Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                        66

                                                        The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                        Ka frac14 sin 69=sin 105

                                                        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                        pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                        26664

                                                        37775

                                                        2

                                                        frac14 050

                                                        The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                        Pa frac14 1

                                                        2 050 19 7502 frac14 267 kN=m

                                                        Figure Q65

                                                        Lateral earth pressure 39

                                                        Horizontal component

                                                        Ph frac14 267 cos 40 frac14 205 kN=m

                                                        Vertical component

                                                        Pv frac14 267 sin 40 frac14 172 kN=m

                                                        Consider moments about the toe of the wall (Figure Q66) (per m)

                                                        Force (kN) Arm (m) Moment (kN m)

                                                        (1)1

                                                        2 175 650 235 frac14 1337 258 345

                                                        (2) 050 650 235 frac14 764 175 134

                                                        (3)1

                                                        2 070 650 235 frac14 535 127 68

                                                        (4) 100 400 235 frac14 940 200 188

                                                        (5) 1

                                                        2 080 050 235 frac14 47 027 1

                                                        Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                        Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                        Lever arm of base resultant

                                                        M

                                                        Vfrac14 795

                                                        525frac14 151m

                                                        Eccentricity of base resultant

                                                        e frac14 200 151 frac14 049m

                                                        Figure Q66

                                                        40 Lateral earth pressure

                                                        Base pressures (Equation 627)

                                                        p frac14 525

                                                        41 6 049

                                                        4

                                                        frac14 228 kN=m2 and 35 kN=m2

                                                        The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                        The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                        The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                        67

                                                        For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                        Force (kN) Arm (m) Moment (kNm)

                                                        (1)1

                                                        2 027 17 52 frac14 574 183 1050

                                                        (2) 027 17 5 3 frac14 689 500 3445

                                                        (3)1

                                                        2 027 102 32 frac14 124 550 682

                                                        (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                        (5)1

                                                        2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                        (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                        (7) 1

                                                        2 267

                                                        2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                        (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                        2 d frac14 163d d2thorn 650 82d2 1060d

                                                        Tie rod force per m frac14 T 0 0

                                                        XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                        d3 thorn 77d2 269d 1438 frac14 0

                                                        d frac14 467m

                                                        Depth of penetration frac14 12d frac14 560m

                                                        Lateral earth pressure 41

                                                        Algebraic sum of forces for d frac14 467m isX

                                                        F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                        T frac14 905 kN=m

                                                        Force in each tie rod frac14 25T frac14 226 kN

                                                        68

                                                        (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                        0 frac14 21 98 frac14 112 kN=m3

                                                        The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                        uC frac14 150

                                                        165 15 98 frac14 134 kN=m2

                                                        The average seepage pressure is

                                                        j frac14 15

                                                        165 98 frac14 09 kN=m3

                                                        Hence

                                                        0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                        0 j frac14 112 09 frac14 103 kN=m3

                                                        Figure Q67

                                                        42 Lateral earth pressure

                                                        Consider moments about the anchor point A (per m)

                                                        Force (kN) Arm (m) Moment (kN m)

                                                        (1) 10 026 150 frac14 390 60 2340

                                                        (2)1

                                                        2 026 18 452 frac14 474 15 711

                                                        (3) 026 18 45 105 frac14 2211 825 18240

                                                        (4)1

                                                        2 026 121 1052 frac14 1734 100 17340

                                                        (5)1

                                                        2 134 15 frac14 101 40 404

                                                        (6) 134 30 frac14 402 60 2412

                                                        (7)1

                                                        2 134 60 frac14 402 95 3819

                                                        571 4527(8) Ppm

                                                        115 115PPm

                                                        XM frac14 0

                                                        Ppm frac144527

                                                        115frac14 394 kN=m

                                                        Available passive resistance

                                                        Pp frac14 1

                                                        2 385 103 62 frac14 714 kN=m

                                                        Factor of safety

                                                        Fp frac14 Pp

                                                        Ppm

                                                        frac14 714

                                                        394frac14 18

                                                        Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                        Figure Q68

                                                        Lateral earth pressure 43

                                                        (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                        Consider moments (per m) about the tie point A

                                                        Force (kN) Arm (m)

                                                        (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                        (2)1

                                                        2 033 18 452 frac14 601 15

                                                        (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                        (4)1

                                                        2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                        (5)1

                                                        2 134 15 frac14 101 40

                                                        (6) 134 30 frac14 402 60

                                                        (7)1

                                                        2 134 d frac14 67d d3thorn 75

                                                        (8) 1

                                                        2 30 103 d2 frac141545d2 2d3thorn 75

                                                        Moment (kN m)

                                                        (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                        XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                        d3 thorn 827d2 466d 1518 frac14 0

                                                        By trial

                                                        d frac14 544m

                                                        The minimum depth of embedment required is 544m

                                                        69

                                                        For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                        0 frac14 20 98 frac14 102 kN=m3

                                                        The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                        44 Lateral earth pressure

                                                        uC frac14 147

                                                        173 26 98 frac14 216 kN=m2

                                                        and the average seepage pressure around the wall is

                                                        j frac14 26

                                                        173 98 frac14 15 kN=m3

                                                        Consider moments about the prop (A) (per m)

                                                        Force (kN) Arm (m) Moment (kN m)

                                                        (1)1

                                                        2 03 17 272 frac14 186 020 37

                                                        (2) 03 17 27 53 frac14 730 335 2445

                                                        (3)1

                                                        2 03 (102thorn 15) 532 frac14 493 423 2085

                                                        (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                        (5)1

                                                        2 216 26 frac14 281 243 684

                                                        (6) 216 27 frac14 583 465 2712

                                                        (7)1

                                                        2 216 60 frac14 648 800 5184

                                                        3055(8)

                                                        1

                                                        2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                        Factor of safety

                                                        Fr frac14 6885

                                                        3055frac14 225

                                                        Figure Q69

                                                        Lateral earth pressure 45

                                                        610

                                                        For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                        p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                        Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                        Using the recommendations of Twine and Roscoe

                                                        p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                        Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                        611

                                                        frac14 18 kN=m3 0 frac14 34

                                                        H frac14 350m nH frac14 335m mH frac14 185m

                                                        Consider a trial value of F frac14 20 Refer to Figure 635

                                                        0m frac14 tan1tan 34

                                                        20

                                                        frac14 186

                                                        Then

                                                        frac14 45 thorn 0m2frac14 543

                                                        W frac14 1

                                                        2 18 3502 cot 543 frac14 792 kN=m

                                                        Figure Q610

                                                        46 Lateral earth pressure

                                                        P frac14 1

                                                        2 s 3352 frac14 561s kN=m

                                                        U frac14 1

                                                        2 98 1852 cosec 543 frac14 206 kN=m

                                                        Equations 630 and 631 then become

                                                        561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                        792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                        ie

                                                        561s 0616N 405 frac14 0

                                                        792 0857N thorn 563 frac14 0

                                                        N frac14 848

                                                        0857frac14 989 kN=m

                                                        Then

                                                        561s 609 405 frac14 0

                                                        s frac14 649

                                                        561frac14 116 kN=m3

                                                        The calculations for trial values of F of 20 15 and 10 are summarized below

                                                        F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                        20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                        s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                        Figure Q611

                                                        Lateral earth pressure 47

                                                        612

                                                        For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                        45 thorn 0

                                                        2frac14 63

                                                        For the retained material between the surface and a depth of 36m

                                                        Pa frac14 1

                                                        2 030 18 362 frac14 350 kN=m

                                                        Weight of reinforced fill between the surface and a depth of 36m is

                                                        Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                        eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                        Eccentricity of Rv

                                                        e frac14 263 250 frac14 013m

                                                        The average vertical stress at a depth of 36m is

                                                        z frac14 Rv

                                                        L 2efrac14 324

                                                        474frac14 68 kN=m2

                                                        (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                        Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                        Tensile stress in the element frac14 138 103

                                                        65 3frac14 71N=mm2

                                                        Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                        Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                        Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                        The weight of ABC is

                                                        W frac14 1

                                                        2 18 52 265 frac14 124 kN=m

                                                        From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                        48 Lateral earth pressure

                                                        (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                        Tp frac14 032 68 120 065 frac14 170 kN

                                                        Tr frac14 213 420

                                                        418frac14 214 kN

                                                        Again the tensile failure and slipping limit states are satisfied for this element

                                                        Figure Q612

                                                        Lateral earth pressure 49

                                                        Chapter 7

                                                        Consolidation theory

                                                        71

                                                        Total change in thickness

                                                        H frac14 782 602 frac14 180mm

                                                        Average thickness frac14 1530thorn 180

                                                        2frac14 1620mm

                                                        Length of drainage path d frac14 1620

                                                        2frac14 810mm

                                                        Root time plot (Figure Q71a)

                                                        ffiffiffiffiffiffit90p frac14 33

                                                        t90 frac14 109min

                                                        cv frac14 0848d2

                                                        t90frac14 0848 8102

                                                        109 1440 365

                                                        106frac14 27m2=year

                                                        r0 frac14 782 764

                                                        782 602frac14 018

                                                        180frac14 0100

                                                        rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                        10 119

                                                        9 180frac14 0735

                                                        rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                        Log time plot (Figure Q71b)

                                                        t50 frac14 26min

                                                        cv frac14 0196d2

                                                        t50frac14 0196 8102

                                                        26 1440 365

                                                        106frac14 26m2=year

                                                        r0 frac14 782 763

                                                        782 602frac14 019

                                                        180frac14 0106

                                                        rp frac14 763 623

                                                        782 602frac14 140

                                                        180frac14 0778

                                                        rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                        Figure Q71(a)

                                                        Figure Q71(b)

                                                        Final void ratio

                                                        e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                        e

                                                        Hfrac14 1thorn e0

                                                        H0frac14 1thorn e1 thorne

                                                        H0

                                                        ie

                                                        e

                                                        180frac14 1631thorne

                                                        1710

                                                        e frac14 2936

                                                        1530frac14 0192

                                                        Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                        mv frac14 1

                                                        1thorn e0 e0 e101 00

                                                        frac14 1

                                                        1823 0192

                                                        0107frac14 098m2=MN

                                                        k frac14 cvmvw frac14 265 098 98

                                                        60 1440 365 103frac14 81 1010 m=s

                                                        72

                                                        Using Equation 77 (one-dimensional method)

                                                        sc frac14 e0 e11thorn e0 H

                                                        Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                        Figure Q72

                                                        52 Consolidation theory

                                                        Settlement

                                                        Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                        1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                        318

                                                        Notes 5 92y 460thorn 84

                                                        Heave

                                                        Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                        1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                        38

                                                        73

                                                        U frac14 f ethTvTHORN frac14 f cvt

                                                        d2

                                                        Hence if cv is constant

                                                        t1

                                                        t2frac14 d

                                                        21

                                                        d22

                                                        where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                        d1 frac14 95mm and d2 frac14 2500mm

                                                        for U frac14 050 t2 frac14 t1 d22

                                                        d21

                                                        frac14 20

                                                        60 24 365 25002

                                                        952frac14 263 years

                                                        for U lt 060 Tv frac14

                                                        4U2 (Equation 724(a))

                                                        t030 frac14 t050 0302

                                                        0502

                                                        frac14 263 036 frac14 095 years

                                                        Consolidation theory 53

                                                        74

                                                        The layer is open

                                                        d frac14 8

                                                        2frac14 4m

                                                        Tv frac14 cvtd2frac14 24 3

                                                        42frac14 0450

                                                        ui frac14 frac14 84 kN=m2

                                                        The excess pore water pressure is given by Equation 721

                                                        ue frac14Xmfrac141mfrac140

                                                        2ui

                                                        Msin

                                                        Mz

                                                        d

                                                        expethM2TvTHORN

                                                        In this case z frac14 d

                                                        sinMz

                                                        d

                                                        frac14 sinM

                                                        where

                                                        M frac14

                                                        23

                                                        25

                                                        2

                                                        M sin M M2Tv exp (M2Tv)

                                                        2thorn1 1110 0329

                                                        3

                                                        21 9993 457 105

                                                        ue frac14 2 84 2

                                                        1 0329 ethother terms negligibleTHORN

                                                        frac14 352 kN=m2

                                                        75

                                                        The layer is open

                                                        d frac14 6

                                                        2frac14 3m

                                                        Tv frac14 cvtd2frac14 10 3

                                                        32frac14 0333

                                                        The layer thickness will be divided into six equal parts ie m frac14 6

                                                        54 Consolidation theory

                                                        For an open layer

                                                        Tv frac14 4n

                                                        m2

                                                        n frac14 0333 62

                                                        4frac14 300

                                                        The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                        ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                        i j

                                                        0 1 2 3 4 5 6 7 8 9 10 11 12

                                                        0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                        The initial and 3-year isochrones are plotted in Figure Q75

                                                        Area under initial isochrone frac14 180 units

                                                        Area under 3-year isochrone frac14 63 units

                                                        The average degree of consolidation is given by Equation 725Thus

                                                        U frac14 1 63

                                                        180frac14 065

                                                        Figure Q75

                                                        Consolidation theory 55

                                                        76

                                                        At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                        0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                        The final consolidation settlement (one-dimensional method) is

                                                        sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                        Corrected time t frac14 2 1

                                                        2

                                                        40

                                                        52

                                                        frac14 1615 years

                                                        Tv frac14 cvtd2frac14 44 1615

                                                        42frac14 0444

                                                        From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                        77

                                                        The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                        Figure Q77

                                                        56 Consolidation theory

                                                        Point m n Ir (kNm2) sc (mm)

                                                        13020frac14 15 20

                                                        20frac14 10 0194 (4) 113 124

                                                        260

                                                        20frac14 30

                                                        20

                                                        20frac14 10 0204 (2) 59 65

                                                        360

                                                        20frac14 30

                                                        40

                                                        20frac14 20 0238 (1) 35 38

                                                        430

                                                        20frac14 15

                                                        40

                                                        20frac14 20 0224 (2) 65 72

                                                        Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                        78

                                                        Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                        (a) Immediate settlement

                                                        H

                                                        Bfrac14 30

                                                        35frac14 086

                                                        D

                                                        Bfrac14 2

                                                        35frac14 006

                                                        Figure Q78

                                                        Consolidation theory 57

                                                        From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                        si frac14 130131qB

                                                        Eufrac14 10 032 105 35

                                                        40frac14 30mm

                                                        (b) Consolidation settlement

                                                        Layer z (m) Dz Ic (kNm2) syod (mm)

                                                        1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                        3150

                                                        Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                        Now

                                                        H

                                                        Bfrac14 30

                                                        35frac14 086 and A frac14 065

                                                        from Figure 712 13 frac14 079

                                                        sc frac14 13sod frac14 079 315 frac14 250mm

                                                        Total settlement

                                                        s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                        79

                                                        Without sand drains

                                                        Uv frac14 025

                                                        Tv frac14 0049 ethfrom Figure 718THORN

                                                        t frac14 Tvd2

                                                        cvfrac14 0049 82

                                                        cvWith sand drains

                                                        R frac14 0564S frac14 0564 3 frac14 169m

                                                        n frac14 Rrfrac14 169

                                                        015frac14 113

                                                        Tr frac14 cht

                                                        4R2frac14 ch

                                                        4 1692 0049 82

                                                        cvethand ch frac14 cvTHORN

                                                        frac14 0275

                                                        Ur frac14 073 (from Figure 730)

                                                        58 Consolidation theory

                                                        Using Equation 740

                                                        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                        U frac14 080

                                                        710

                                                        Without sand drains

                                                        Uv frac14 090

                                                        Tv frac14 0848

                                                        t frac14 Tvd2

                                                        cvfrac14 0848 102

                                                        96frac14 88 years

                                                        With sand drains

                                                        R frac14 0564S frac14 0564 4 frac14 226m

                                                        n frac14 Rrfrac14 226

                                                        015frac14 15

                                                        Tr

                                                        Tvfrac14 chcv

                                                        d2

                                                        4R2ethsame tTHORN

                                                        Tr

                                                        Tvfrac14 140

                                                        96 102

                                                        4 2262frac14 714 eth1THORN

                                                        Using Equation 740

                                                        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                        An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                        Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                        040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                        Thus

                                                        Uv frac14 0295 and Ur frac14 086

                                                        t frac14 88 00683

                                                        0848frac14 07 years

                                                        Consolidation theory 59

                                                        Chapter 8

                                                        Bearing capacity

                                                        81

                                                        (a) The ultimate bearing capacity is given by Equation 83

                                                        qf frac14 cNc thorn DNq thorn 1

                                                        2BN

                                                        For u frac14 0

                                                        Nc frac14 514 Nq frac14 1 N frac14 0

                                                        qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                        The net ultimate bearing capacity is

                                                        qnf frac14 qf D frac14 540 kN=m2

                                                        The net foundation pressure is

                                                        qn frac14 q D frac14 425

                                                        2 eth21 1THORN frac14 192 kN=m2

                                                        The factor of safety (Equation 86) is

                                                        F frac14 qnfqnfrac14 540

                                                        192frac14 28

                                                        (b) For 0 frac14 28

                                                        Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                        qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                        2 112 2 13

                                                        frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                        qnf frac14 574 112 frac14 563 kN=m2

                                                        F frac14 563

                                                        192frac14 29

                                                        (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                        82

                                                        For 0 frac14 38

                                                        Nq frac14 49 N frac14 67

                                                        qnf frac14 DethNq 1THORN thorn 1

                                                        2BN ethfrom Equation 83THORN

                                                        frac14 eth18 075 48THORN thorn 1

                                                        2 18 15 67

                                                        frac14 648thorn 905 frac14 1553 kN=m2

                                                        qn frac14 500

                                                        15 eth18 075THORN frac14 320 kN=m2

                                                        F frac14 qnfqnfrac14 1553

                                                        320frac14 48

                                                        0d frac14 tan1tan 38

                                                        125

                                                        frac14 32 therefore Nq frac14 23 and N frac14 25

                                                        Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                        2 18 15 25

                                                        frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                        Design load (action) Vd frac14 500 kN=m

                                                        The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                        83

                                                        D

                                                        Bfrac14 350

                                                        225frac14 155

                                                        From Figure 85 for a square foundation

                                                        Nc frac14 81

                                                        Bearing capacity 61

                                                        For a rectangular foundation (L frac14 450m B frac14 225m)

                                                        Nc frac14 084thorn 016B

                                                        L

                                                        81 frac14 745

                                                        Using Equation 810

                                                        qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                        For F frac14 3

                                                        qn frac14 1006

                                                        3frac14 335 kN=m2

                                                        q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                        Design load frac14 405 450 225 frac14 4100 kN

                                                        Design undrained strength cud frac14 135

                                                        14frac14 96 kN=m2

                                                        Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                        frac14 7241 kN

                                                        Design load Vd frac14 4100 kN

                                                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                        84

                                                        For 0 frac14 40

                                                        Nq frac14 64 N frac14 95

                                                        qnf frac14 DethNq 1THORN thorn 04BN

                                                        (a) Water table 5m below ground level

                                                        qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                        qn frac14 400 17 frac14 383 kN=m2

                                                        F frac14 2686

                                                        383frac14 70

                                                        (b) Water table 1m below ground level (ie at foundation level)

                                                        0 frac14 20 98 frac14 102 kN=m3

                                                        62 Bearing capacity

                                                        qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                        F frac14 2040

                                                        383frac14 53

                                                        (c) Water table at ground level with upward hydraulic gradient 02

                                                        eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                        qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                        F frac14 1296

                                                        392frac14 33

                                                        85

                                                        The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                        Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                        Design value of 0 frac14 tan1tan 39

                                                        125

                                                        frac14 33

                                                        For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                        Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                        86

                                                        (a) Undrained shear for u frac14 0

                                                        Nc frac14 514 Nq frac14 1 N frac14 0

                                                        qnf frac14 12cuNc

                                                        frac14 12 100 514 frac14 617 kN=m2

                                                        qn frac14 qnfFfrac14 617

                                                        3frac14 206 kN=m2

                                                        q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                        Bearing capacity 63

                                                        Drained shear for 0 frac14 32

                                                        Nq frac14 23 N frac14 25

                                                        0 frac14 21 98 frac14 112 kN=m3

                                                        qnf frac14 0DethNq 1THORN thorn 040BN

                                                        frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                        frac14 694 kN=m2

                                                        q frac14 694

                                                        3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                        Design load frac14 42 227 frac14 3632 kN

                                                        (b) Design undrained strength cud frac14 100

                                                        14frac14 71 kNm2

                                                        Design bearing resistance Rd frac14 12cudNe area

                                                        frac14 12 71 514 42

                                                        frac14 7007 kN

                                                        For drained shear 0d frac14 tan1tan 32

                                                        125

                                                        frac14 26

                                                        Nq frac14 12 N frac14 10

                                                        Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                        (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                        Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                        1 2 100 0175 0700qn 0182qn

                                                        2 6 033 0044 0176qn 0046qn

                                                        3 10 020 0017 0068qn 0018qn

                                                        0246qn

                                                        Diameter of equivalent circle B frac14 45m

                                                        H

                                                        Bfrac14 12

                                                        45frac14 27 and A frac14 042

                                                        13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                        64 Bearing capacity

                                                        For sc frac14 30mm

                                                        qn frac14 30

                                                        0147frac14 204 kN=m2

                                                        q frac14 204thorn 21 frac14 225 kN=m2

                                                        Design load frac14 42 225 frac14 3600 kN

                                                        The design load is 3600 kN settlement being the limiting criterion

                                                        87

                                                        D

                                                        Bfrac14 8

                                                        4frac14 20

                                                        From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                        F frac14 cuNc

                                                        Dfrac14 40 71

                                                        20 8frac14 18

                                                        88

                                                        Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                        Design value of 0 frac14 tan1tan 38

                                                        125

                                                        frac14 32

                                                        Figure Q86

                                                        Bearing capacity 65

                                                        For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                        Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                        The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                        Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                        For B frac14 250m qn frac14 3750

                                                        2502 17 frac14 583 kN=m2

                                                        From Figure 510 m frac14 n frac14 126

                                                        6frac14 021

                                                        Ir frac14 0019

                                                        Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                        Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                        The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                        89

                                                        Depth (m) N 0v (kNm2) CN N1

                                                        070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                        Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                        (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                        Cw frac14 05thorn 0530

                                                        47

                                                        frac14 082

                                                        66 Bearing capacity

                                                        Thus

                                                        qa frac14 150 082 frac14 120 kN=m2

                                                        (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                        Thus

                                                        qa frac14 90 15 frac14 135 kN=m2

                                                        (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                        Ic frac14 171

                                                        1014frac14 0068

                                                        From Equation 819(a) with s frac14 25mm

                                                        q frac14 25

                                                        3507 0068frac14 150 kN=m2

                                                        810

                                                        Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                        Izp frac14 05thorn 01130

                                                        16 27

                                                        05

                                                        frac14 067

                                                        Refer to Figure Q810

                                                        E frac14 25qc

                                                        Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                        Ez (mm3MN)

                                                        1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                        0203

                                                        C1 frac14 1 0500qnfrac14 1 05 12 16

                                                        130frac14 093

                                                        C2 frac14 1 ethsayTHORN

                                                        s frac14 C1C2qnX Iz

                                                        Ez frac14 093 1 130 0203 frac14 25mm

                                                        Bearing capacity 67

                                                        811

                                                        At pile base level

                                                        cu frac14 220 kN=m2

                                                        qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                        Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                        00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                        qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                        Then

                                                        Qf frac14 Abqb thorn Asqs

                                                        frac14

                                                        4 32 1980

                                                        thorn eth 105 139 86THORN

                                                        frac14 13 996thorn 3941 frac14 17 937 kN

                                                        0 01 02 03 04 05 06 07

                                                        0 2 4 6 8 10 12 14

                                                        1

                                                        2

                                                        3

                                                        4

                                                        5

                                                        6

                                                        7

                                                        8

                                                        (1)

                                                        (2)

                                                        (3)

                                                        (4)

                                                        (5)

                                                        qc

                                                        qc

                                                        Iz

                                                        Iz

                                                        (MNm2)

                                                        z (m)

                                                        Figure Q810

                                                        68 Bearing capacity

                                                        Allowable load

                                                        ethaTHORN Qf

                                                        2frac14 17 937

                                                        2frac14 8968 kN

                                                        ethbTHORN Abqb

                                                        3thorn Asqs frac14 13 996

                                                        3thorn 3941 frac14 8606 kN

                                                        ie allowable load frac14 8600 kN

                                                        Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                        According to the limit state method

                                                        Characteristic undrained strength at base level cuk frac14 220

                                                        150kN=m2

                                                        Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                        150frac14 1320 kN=m2

                                                        Characteristic shaft resistance qsk frac14 00150

                                                        frac14 86

                                                        150frac14 57 kN=m2

                                                        Characteristic base and shaft resistances

                                                        Rbk frac14

                                                        4 32 1320 frac14 9330 kN

                                                        Rsk frac14 105 139 86

                                                        150frac14 2629 kN

                                                        For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                        Design bearing resistance Rcd frac14 9330

                                                        160thorn 2629

                                                        130

                                                        frac14 5831thorn 2022

                                                        frac14 7850 kN

                                                        Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                        812

                                                        ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                        qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                        For a single pile

                                                        Qf frac14 Abqb thorn Asqs

                                                        frac14

                                                        4 062 1305

                                                        thorn eth 06 15 42THORN

                                                        frac14 369thorn 1187 frac14 1556 kN

                                                        Bearing capacity 69

                                                        Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                        eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                        (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                        (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                        qbkfrac14 9cuk frac14 9 220

                                                        150frac14 1320 kN=m2

                                                        qskfrac14cuk frac14 040 105

                                                        150frac14 28 kN=m2

                                                        Rbkfrac14

                                                        4 0602 1320 frac14 373 kN

                                                        Rskfrac14 060 15 28 frac14 791 kN

                                                        Rcdfrac14 373

                                                        160thorn 791

                                                        130frac14 233thorn 608 frac14 841 kN

                                                        Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                        Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                        (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                        q frac14 21 000

                                                        1762frac14 68 kN=m2

                                                        Immediate settlement

                                                        H

                                                        Bfrac14 15

                                                        176frac14 085

                                                        D

                                                        Bfrac14 13

                                                        176frac14 074

                                                        L

                                                        Bfrac14 1

                                                        Hence from Figure 515

                                                        130 frac14 078 and 131 frac14 041

                                                        70 Bearing capacity

                                                        Thus using Equation 528

                                                        si frac14 078 041 68 176

                                                        65frac14 6mm

                                                        Consolidation settlement

                                                        Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                        1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                        434 (sod)

                                                        Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                        sc frac14 056 434 frac14 24mm

                                                        The total settlement is (6thorn 24) frac14 30mm

                                                        813

                                                        At base level N frac14 26 Then using Equation 830

                                                        qb frac14 40NDb

                                                        Bfrac14 40 26 2

                                                        025frac14 8320 kN=m2

                                                        ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                        Figure Q812

                                                        Bearing capacity 71

                                                        Over the length embedded in sand

                                                        N frac14 21 ie18thorn 24

                                                        2

                                                        Using Equation 831

                                                        qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                        For a single pile

                                                        Qf frac14 Abqb thorn Asqs

                                                        frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                        For the pile group assuming a group efficiency of 12

                                                        XQf frac14 12 9 604 frac14 6523 kN

                                                        Then the load factor is

                                                        F frac14 6523

                                                        2000thorn 1000frac14 21

                                                        (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                        Characteristic base resistance per unit area qbk frac14 8320

                                                        150frac14 5547 kNm2

                                                        Characteristic shaft resistance per unit area qsk frac14 42

                                                        150frac14 28 kNm2

                                                        Characteristic base and shaft resistances for a single pile

                                                        Rbk frac14 0252 5547 frac14 347 kN

                                                        Rsk frac14 4 025 2 28 frac14 56 kN

                                                        For a driven pile the partial factors are b frac14 s frac14 130

                                                        Design bearing resistance Rcd frac14 347

                                                        130thorn 56

                                                        130frac14 310 kN

                                                        For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                        Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                        (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                        From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                        72 Bearing capacity

                                                        N frac14 24thorn 26thorn 34

                                                        3frac14 28

                                                        Ic frac14 171

                                                        2814frac14 0016 ethEquation 818THORN

                                                        s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                        814

                                                        Using Equation 841

                                                        Tf frac14 DLcu thorn

                                                        4ethD2 d2THORNcuNc

                                                        frac14 eth 02 5 06 110THORN thorn

                                                        4eth022 012THORN110 9

                                                        frac14 207thorn 23 frac14 230 kN

                                                        Figure Q813

                                                        Bearing capacity 73

                                                        Chapter 9

                                                        Stability of slopes

                                                        91

                                                        Referring to Figure Q91

                                                        W frac14 417 19 frac14 792 kN=m

                                                        Q frac14 20 28 frac14 56 kN=m

                                                        Arc lengthAB frac14

                                                        180 73 90 frac14 115m

                                                        Arc length BC frac14

                                                        180 28 90 frac14 44m

                                                        The factor of safety is given by

                                                        F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                        90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                        Depth of tension crack z0 frac14 2cu

                                                        frac14 2 20

                                                        19frac14 21m

                                                        Arc length BD frac14

                                                        180 13

                                                        1

                                                        2 90 frac14 21m

                                                        F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                        The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                        Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                        14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                        Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                        The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                        92

                                                        u frac14 0

                                                        Depth factor D frac14 11

                                                        9frac14 122

                                                        Using Equation 92 with F frac14 10

                                                        Ns frac14 cu

                                                        FHfrac14 30

                                                        10 19 9frac14 0175

                                                        Hence from Figure 93

                                                        frac14 50

                                                        For F frac14 12

                                                        Ns frac14 30

                                                        12 19 9frac14 0146

                                                        frac14 27

                                                        93

                                                        Refer to Figure Q93

                                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                        1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                        74 m

                                                        214 1deg

                                                        213 1deg

                                                        39 m

                                                        WB

                                                        D

                                                        C

                                                        28 m

                                                        21 m

                                                        A

                                                        Q

                                                        Soil (1)Soil (2)

                                                        73deg

                                                        Figure Q91

                                                        Stability of slopes 75

                                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                        9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                        599 256 328 1372

                                                        Figure Q93

                                                        76 Stability of slopes

                                                        XW cos frac14 b

                                                        Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                        W sin frac14 bX

                                                        h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                        Arc length La frac14

                                                        180 57

                                                        1

                                                        2 326 frac14 327m

                                                        The factor of safety is given by

                                                        F frac14 c0La thorn tan0ethW cos ulTHORN

                                                        W sin

                                                        frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                        frac14 091

                                                        According to the limit state method

                                                        0d frac14 tan1tan 32

                                                        125

                                                        frac14 265

                                                        c0 frac14 8

                                                        160frac14 5 kN=m2

                                                        Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                        Design disturbing moment frac14 1075 kN=m

                                                        The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                        94

                                                        F frac14 1

                                                        W sin

                                                        Xfc0bthorn ethW ubTHORN tan0g sec

                                                        1thorn ethtan tan0=FTHORN

                                                        c0 frac14 8 kN=m2

                                                        0 frac14 32

                                                        c0b frac14 8 2 frac14 16 kN=m

                                                        W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                        Try F frac14 100

                                                        tan0

                                                        Ffrac14 0625

                                                        Stability of slopes 77

                                                        Values of u are as obtained in Figure Q93

                                                        SliceNo

                                                        h(m)

                                                        W frac14 bh(kNm)

                                                        W sin(kNm)

                                                        ub(kNm)

                                                        c0bthorn (W ub) tan0(kNm)

                                                        sec

                                                        1thorn (tan tan0)FProduct(kNm)

                                                        1 05 21 6 2 8 24 1078 262 13 55 31

                                                        23 33 30 1042 31

                                                        3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                        224 92 72 0931 67

                                                        6 50 210 11 40 100 85 0907 777 55 231 14

                                                        12 58 112 90 0889 80

                                                        8 60 252 1812

                                                        80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                        10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                        2154 88 116 0853 99

                                                        14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                        1074 1091

                                                        F frac14 1091

                                                        1074frac14 102 (assumed value 100)

                                                        Thus

                                                        F frac14 101

                                                        95

                                                        F frac14 1

                                                        W sin

                                                        XfWeth1 ruTHORN tan0g sec

                                                        1thorn ethtan tan0THORN=F

                                                        0 frac14 33

                                                        ru frac14 020

                                                        W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                        eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                        Try F frac14 110

                                                        tan 0

                                                        Ffrac14 tan 33

                                                        110frac14 0590

                                                        78 Stability of slopes

                                                        Referring to Figure Q95

                                                        SliceNo

                                                        h(m)

                                                        W frac14 bh(kNm)

                                                        W sin(kNm)

                                                        W(1 ru) tan0(kNm)

                                                        sec

                                                        1thorn ( tan tan0)FProduct(kNm)

                                                        1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                        2120 234 0892 209

                                                        4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                        1185 1271

                                                        Figure Q95

                                                        Stability of slopes 79

                                                        F frac14 1271

                                                        1185frac14 107

                                                        The trial value was 110 therefore take F to be 108

                                                        96

                                                        (a) Water table at surface the factor of safety is given by Equation 912

                                                        F frac14 0

                                                        sat

                                                        tan0

                                                        tan

                                                        ptie 15 frac14 92

                                                        19

                                                        tan 36

                                                        tan

                                                        tan frac14 0234

                                                        frac14 13

                                                        Water table well below surface the factor of safety is given by Equation 911

                                                        F frac14 tan0

                                                        tan

                                                        frac14 tan 36

                                                        tan 13

                                                        frac14 31

                                                        (b) 0d frac14 tan1tan 36

                                                        125

                                                        frac14 30

                                                        Depth of potential failure surface frac14 z

                                                        Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                        frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                        frac14 504z kN

                                                        Design disturbing moment per unit area Sd frac14 sat sin cos

                                                        frac14 19 z sin 13 cos 13

                                                        frac14 416z kN

                                                        Rd gtSd therefore the limit state for overall stability is satisfied

                                                        80 Stability of slopes

                                                        • Book Cover
                                                        • Title
                                                        • Contents
                                                        • Basic characteristics of soils
                                                        • Seepage
                                                        • Effective stress
                                                        • Shear strength
                                                        • Stresses and displacements
                                                        • Lateral earth pressure
                                                        • Consolidation theory
                                                        • Bearing capacity
                                                        • Stability of slopes

                                                          43

                                                          3 (kNm2) 1 3 (kNm2) 1 (kNm2)

                                                          200 222 422400 218 618600 220 820

                                                          The Mohr circles and failure envelope are drawn in Figure Q43 from whichcufrac14 110 kNm2 and ufrac14 0

                                                          44

                                                          The modified shear strength parameters are

                                                          0 frac14 tan1ethsin 0THORN frac14 tan1ethsin 29THORN frac14 26

                                                          a0 frac14 c0 cos 0 frac14 15 cos 29 frac14 13 kN=m2

                                                          The coordinates of the stress point representing failure conditions in the test are

                                                          1

                                                          2eth1 2THORN frac14 1

                                                          2 170 frac14 85 kN=m2

                                                          1

                                                          2eth1 thorn 3THORN frac14 1

                                                          2eth270thorn 100THORN frac14 185 kN=m2

                                                          The pore water pressure at failure is given by the horizontal distance between thisstress point and the modified failure envelope Thus from Figure Q44

                                                          uf frac14 36 kN=m2

                                                          Figure Q43

                                                          Figure Q44

                                                          Shear strength 23

                                                          45

                                                          3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                                          150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                                          The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                                          The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                                          46

                                                          03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                                          200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                                          The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                                          Figure Q45

                                                          24 Shear strength

                                                          47

                                                          The torque required to produce shear failure is given by

                                                          T frac14 dh cud

                                                          2thorn 2

                                                          Z d=2

                                                          0

                                                          2r drcur

                                                          frac14 cud2h

                                                          2thorn 4cu

                                                          Z d=2

                                                          0

                                                          r2dr

                                                          frac14 cud2h

                                                          2thorn d

                                                          3

                                                          6

                                                          Then

                                                          35 frac14 cu52 10

                                                          2thorn 53

                                                          6

                                                          103

                                                          cu frac14 76 kN=m3

                                                          400

                                                          0 400 800 1200 1600

                                                          τ (k

                                                          Nm

                                                          2 )

                                                          σprime (kNm2)

                                                          34deg

                                                          315deg29deg

                                                          (a)

                                                          (b)

                                                          0 400

                                                          400

                                                          800 1200 1600

                                                          Failure envelope

                                                          300 500

                                                          σprime (kNm2)

                                                          τ (k

                                                          Nm

                                                          2 )

                                                          20 (kNm2)

                                                          31deg

                                                          Figure Q46

                                                          Shear strength 25

                                                          48

                                                          The relevant stress values are calculated as follows

                                                          3 frac14 600 kN=m2

                                                          1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                                          2(1 3) 0 40 79 107 139 159

                                                          1

                                                          2(01 thorn 03) 400 411 402 389 351 326

                                                          1

                                                          2(1 thorn 3) 600 640 679 707 739 759

                                                          The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                                          Af frac14 433 200

                                                          319frac14 073

                                                          49

                                                          B frac14 u33

                                                          frac14 144

                                                          350 200frac14 096

                                                          a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                                          0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                                          10 283 65 023

                                                          Figure Q48

                                                          26 Shear strength

                                                          The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                                          Figure Q49

                                                          Shear strength 27

                                                          Chapter 5

                                                          Stresses and displacements

                                                          51

                                                          Vertical stress is given by

                                                          z frac14 Qz2Ip frac14 5000

                                                          52Ip

                                                          Values of Ip are obtained from Table 51

                                                          r (m) rz Ip z (kNm2)

                                                          0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                                          10 20 0009 2

                                                          The variation of z with radial distance (r) is plotted in Figure Q51

                                                          Figure Q51

                                                          52

                                                          Below the centre load (Figure Q52)

                                                          r

                                                          zfrac14 0 for the 7500-kN load

                                                          Ip frac14 0478

                                                          r

                                                          zfrac14 5

                                                          4frac14 125 for the 10 000- and 9000-kN loads

                                                          Ip frac14 0045

                                                          Then

                                                          z frac14X Q

                                                          z2Ip

                                                          frac14 7500 0478

                                                          42thorn 10 000 0045

                                                          42thorn 9000 0045

                                                          42

                                                          frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                                          53

                                                          The vertical stress under a corner of a rectangular area is given by

                                                          z frac14 qIr

                                                          where values of Ir are obtained from Figure 510 In this case

                                                          z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                                          z

                                                          Figure Q52

                                                          Stresses and displacements 29

                                                          z (m) m n Ir z (kNm2)

                                                          0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                                          10 010 0005 5

                                                          z is plotted against z in Figure Q53

                                                          54

                                                          (a)

                                                          m frac14 125

                                                          12frac14 104

                                                          n frac14 18

                                                          12frac14 150

                                                          From Figure 510 Irfrac14 0196

                                                          z frac14 2 175 0196 frac14 68 kN=m2

                                                          Figure Q53

                                                          30 Stresses and displacements

                                                          (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                                          z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                                          55

                                                          Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                                          Px frac14 2Q

                                                          1

                                                          m2 thorn 1frac14 2 150

                                                          125frac14 76 kN=m

                                                          Equation 517 is used to obtain the pressure distribution

                                                          px frac14 4Q

                                                          h

                                                          m2n

                                                          ethm2 thorn n2THORN2 frac14150

                                                          m2n

                                                          ethm2 thorn n2THORN2 ethkN=m2THORN

                                                          Figure Q54

                                                          Stresses and displacements 31

                                                          n m2n

                                                          (m2 thorn n2)2

                                                          px(kNm2)

                                                          0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                          The pressure distribution is plotted in Figure Q55

                                                          56

                                                          H

                                                          Bfrac14 10

                                                          2frac14 5

                                                          L

                                                          Bfrac14 4

                                                          2frac14 2

                                                          D

                                                          Bfrac14 1

                                                          2frac14 05

                                                          Hence from Figure 515

                                                          131 frac14 082

                                                          130 frac14 094

                                                          Figure Q55

                                                          32 Stresses and displacements

                                                          The immediate settlement is given by Equation 528

                                                          si frac14 130131qB

                                                          Eu

                                                          frac14 094 082 200 2

                                                          45frac14 7mm

                                                          Stresses and displacements 33

                                                          Chapter 6

                                                          Lateral earth pressure

                                                          61

                                                          For 0 frac14 37 the active pressure coefficient is given by

                                                          Ka frac14 1 sin 37

                                                          1thorn sin 37frac14 025

                                                          The total active thrust (Equation 66a with c0 frac14 0) is

                                                          Pa frac14 1

                                                          2KaH

                                                          2 frac14 1

                                                          2 025 17 62 frac14 765 kN=m

                                                          If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                          K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                          and the thrust on the wall is

                                                          P0 frac14 1

                                                          2K0H

                                                          2 frac14 1

                                                          2 040 17 62 frac14 122 kN=m

                                                          62

                                                          The active pressure coefficients for the three soil types are as follows

                                                          Ka1 frac141 sin 35

                                                          1thorn sin 35frac14 0271

                                                          Ka2 frac141 sin 27

                                                          1thorn sin 27frac14 0375

                                                          ffiffiffiffiffiffiffiKa2

                                                          p frac14 0613

                                                          Ka3 frac141 sin 42

                                                          1thorn sin 42frac14 0198

                                                          Distribution of active pressure (plotted in Figure Q62)

                                                          Depth (m) Soil Active pressure (kNm2)

                                                          3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                          12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                          At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                          Total thrust frac14 571 kNm

                                                          Point of application is (4893571) m from the top of the wall ie 857m

                                                          Force (kN) Arm (m) Moment (kN m)

                                                          (1)1

                                                          2 0271 16 32 frac14 195 20 390

                                                          (2) 0271 16 3 2 frac14 260 40 1040

                                                          (3)1

                                                          2 0271 92 22 frac14 50 433 217

                                                          (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                          (5)1

                                                          2 0375 102 32 frac14 172 70 1204

                                                          (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                          (7)1

                                                          2 0198 112 42 frac14 177 1067 1889

                                                          (8)1

                                                          2 98 92 frac14 3969 90 35721

                                                          5713 48934

                                                          Figure Q62

                                                          Lateral earth pressure 35

                                                          63

                                                          (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                          Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                          frac14 245

                                                          At the lower end of the piling

                                                          pa frac14 Kaqthorn Kasatz Kaccu

                                                          frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                          frac14 115 kN=m2

                                                          pp frac14 Kpsatzthorn Kpccu

                                                          frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                          frac14 202 kN=m2

                                                          (b) For 0 frac14 26 and frac14 1

                                                          20

                                                          Ka frac14 035

                                                          Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                          pfrac14 145 ethEquation 619THORN

                                                          Kp frac14 37

                                                          Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                          pfrac14 47 ethEquation 624THORN

                                                          At the lower end of the piling

                                                          pa frac14 Kaqthorn Ka0z Kacc

                                                          0

                                                          frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                          frac14 187 kN=m2

                                                          pp frac14 Kp0zthorn Kpcc

                                                          0

                                                          frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                          frac14 198 kN=m2

                                                          36 Lateral earth pressure

                                                          64

                                                          (a) For 0 frac14 38 Ka frac14 024

                                                          0 frac14 20 98 frac14 102 kN=m3

                                                          The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                          Force (kN) Arm (m) Moment (kN m)

                                                          (1) 024 10 66 frac14 159 33 525

                                                          (2)1

                                                          2 024 17 392 frac14 310 400 1240

                                                          (3) 024 17 39 27 frac14 430 135 580

                                                          (4)1

                                                          2 024 102 272 frac14 89 090 80

                                                          (5)1

                                                          2 98 272 frac14 357 090 321

                                                          Hfrac14 1345 MH frac14 2746

                                                          (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                          (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                          XM frac14MV MH frac14 7790 kNm

                                                          Lever arm of base resultant

                                                          M

                                                          Vfrac14 779

                                                          488frac14 160

                                                          Eccentricity of base resultant

                                                          e frac14 200 160 frac14 040m

                                                          39 m

                                                          27 m

                                                          40 m

                                                          04 m

                                                          04 m

                                                          26 m

                                                          (7)

                                                          (9)

                                                          (1)(2)

                                                          (3)

                                                          (4)

                                                          (5)

                                                          (8)(6)

                                                          (10)

                                                          WT

                                                          10 kNm2

                                                          Hydrostatic

                                                          Figure Q64

                                                          Lateral earth pressure 37

                                                          Base pressures (Equation 627)

                                                          p frac14 VB

                                                          1 6e

                                                          B

                                                          frac14 488

                                                          4eth1 060THORN

                                                          frac14 195 kN=m2 and 49 kN=m2

                                                          Factor of safety against sliding (Equation 628)

                                                          F frac14 V tan

                                                          Hfrac14 488 tan 25

                                                          1345frac14 17

                                                          (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                          Hfrac14 1633 kN

                                                          V frac14 4879 kN

                                                          MH frac14 3453 kNm

                                                          MV frac14 10536 kNm

                                                          The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                          65

                                                          For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                          Kp

                                                          Ffrac14 385

                                                          2

                                                          0 frac14 20 98 frac14 102 kN=m3

                                                          The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                          Force (kN) Arm (m) Moment (kN m)

                                                          (1)1

                                                          2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                          (2) 026 17 45 d frac14 199d d2 995d2

                                                          (3)1

                                                          2 026 102 d2 frac14 133d2 d3 044d3

                                                          (4)1

                                                          2 385

                                                          2 17 152 frac14 368 dthorn 05 368d 184

                                                          (5)385

                                                          2 17 15 d frac14 491d d2 2455d2

                                                          (6)1

                                                          2 385

                                                          2 102 d2 frac14 982d2 d3 327d3

                                                          38 Lateral earth pressure

                                                          XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                          d3 thorn 516d2 283d 1724 frac14 0

                                                          d frac14 179m

                                                          Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                          Over additional 20 embedded depth

                                                          pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                          Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                          66

                                                          The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                          Ka frac14 sin 69=sin 105

                                                          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                          pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                          26664

                                                          37775

                                                          2

                                                          frac14 050

                                                          The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                          Pa frac14 1

                                                          2 050 19 7502 frac14 267 kN=m

                                                          Figure Q65

                                                          Lateral earth pressure 39

                                                          Horizontal component

                                                          Ph frac14 267 cos 40 frac14 205 kN=m

                                                          Vertical component

                                                          Pv frac14 267 sin 40 frac14 172 kN=m

                                                          Consider moments about the toe of the wall (Figure Q66) (per m)

                                                          Force (kN) Arm (m) Moment (kN m)

                                                          (1)1

                                                          2 175 650 235 frac14 1337 258 345

                                                          (2) 050 650 235 frac14 764 175 134

                                                          (3)1

                                                          2 070 650 235 frac14 535 127 68

                                                          (4) 100 400 235 frac14 940 200 188

                                                          (5) 1

                                                          2 080 050 235 frac14 47 027 1

                                                          Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                          Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                          Lever arm of base resultant

                                                          M

                                                          Vfrac14 795

                                                          525frac14 151m

                                                          Eccentricity of base resultant

                                                          e frac14 200 151 frac14 049m

                                                          Figure Q66

                                                          40 Lateral earth pressure

                                                          Base pressures (Equation 627)

                                                          p frac14 525

                                                          41 6 049

                                                          4

                                                          frac14 228 kN=m2 and 35 kN=m2

                                                          The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                          The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                          The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                          67

                                                          For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                          Force (kN) Arm (m) Moment (kNm)

                                                          (1)1

                                                          2 027 17 52 frac14 574 183 1050

                                                          (2) 027 17 5 3 frac14 689 500 3445

                                                          (3)1

                                                          2 027 102 32 frac14 124 550 682

                                                          (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                          (5)1

                                                          2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                          (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                          (7) 1

                                                          2 267

                                                          2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                          (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                          2 d frac14 163d d2thorn 650 82d2 1060d

                                                          Tie rod force per m frac14 T 0 0

                                                          XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                          d3 thorn 77d2 269d 1438 frac14 0

                                                          d frac14 467m

                                                          Depth of penetration frac14 12d frac14 560m

                                                          Lateral earth pressure 41

                                                          Algebraic sum of forces for d frac14 467m isX

                                                          F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                          T frac14 905 kN=m

                                                          Force in each tie rod frac14 25T frac14 226 kN

                                                          68

                                                          (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                          0 frac14 21 98 frac14 112 kN=m3

                                                          The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                          uC frac14 150

                                                          165 15 98 frac14 134 kN=m2

                                                          The average seepage pressure is

                                                          j frac14 15

                                                          165 98 frac14 09 kN=m3

                                                          Hence

                                                          0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                          0 j frac14 112 09 frac14 103 kN=m3

                                                          Figure Q67

                                                          42 Lateral earth pressure

                                                          Consider moments about the anchor point A (per m)

                                                          Force (kN) Arm (m) Moment (kN m)

                                                          (1) 10 026 150 frac14 390 60 2340

                                                          (2)1

                                                          2 026 18 452 frac14 474 15 711

                                                          (3) 026 18 45 105 frac14 2211 825 18240

                                                          (4)1

                                                          2 026 121 1052 frac14 1734 100 17340

                                                          (5)1

                                                          2 134 15 frac14 101 40 404

                                                          (6) 134 30 frac14 402 60 2412

                                                          (7)1

                                                          2 134 60 frac14 402 95 3819

                                                          571 4527(8) Ppm

                                                          115 115PPm

                                                          XM frac14 0

                                                          Ppm frac144527

                                                          115frac14 394 kN=m

                                                          Available passive resistance

                                                          Pp frac14 1

                                                          2 385 103 62 frac14 714 kN=m

                                                          Factor of safety

                                                          Fp frac14 Pp

                                                          Ppm

                                                          frac14 714

                                                          394frac14 18

                                                          Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                          Figure Q68

                                                          Lateral earth pressure 43

                                                          (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                          Consider moments (per m) about the tie point A

                                                          Force (kN) Arm (m)

                                                          (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                          (2)1

                                                          2 033 18 452 frac14 601 15

                                                          (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                          (4)1

                                                          2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                          (5)1

                                                          2 134 15 frac14 101 40

                                                          (6) 134 30 frac14 402 60

                                                          (7)1

                                                          2 134 d frac14 67d d3thorn 75

                                                          (8) 1

                                                          2 30 103 d2 frac141545d2 2d3thorn 75

                                                          Moment (kN m)

                                                          (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                          XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                          d3 thorn 827d2 466d 1518 frac14 0

                                                          By trial

                                                          d frac14 544m

                                                          The minimum depth of embedment required is 544m

                                                          69

                                                          For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                          0 frac14 20 98 frac14 102 kN=m3

                                                          The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                          44 Lateral earth pressure

                                                          uC frac14 147

                                                          173 26 98 frac14 216 kN=m2

                                                          and the average seepage pressure around the wall is

                                                          j frac14 26

                                                          173 98 frac14 15 kN=m3

                                                          Consider moments about the prop (A) (per m)

                                                          Force (kN) Arm (m) Moment (kN m)

                                                          (1)1

                                                          2 03 17 272 frac14 186 020 37

                                                          (2) 03 17 27 53 frac14 730 335 2445

                                                          (3)1

                                                          2 03 (102thorn 15) 532 frac14 493 423 2085

                                                          (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                          (5)1

                                                          2 216 26 frac14 281 243 684

                                                          (6) 216 27 frac14 583 465 2712

                                                          (7)1

                                                          2 216 60 frac14 648 800 5184

                                                          3055(8)

                                                          1

                                                          2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                          Factor of safety

                                                          Fr frac14 6885

                                                          3055frac14 225

                                                          Figure Q69

                                                          Lateral earth pressure 45

                                                          610

                                                          For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                          p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                          Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                          Using the recommendations of Twine and Roscoe

                                                          p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                          Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                          611

                                                          frac14 18 kN=m3 0 frac14 34

                                                          H frac14 350m nH frac14 335m mH frac14 185m

                                                          Consider a trial value of F frac14 20 Refer to Figure 635

                                                          0m frac14 tan1tan 34

                                                          20

                                                          frac14 186

                                                          Then

                                                          frac14 45 thorn 0m2frac14 543

                                                          W frac14 1

                                                          2 18 3502 cot 543 frac14 792 kN=m

                                                          Figure Q610

                                                          46 Lateral earth pressure

                                                          P frac14 1

                                                          2 s 3352 frac14 561s kN=m

                                                          U frac14 1

                                                          2 98 1852 cosec 543 frac14 206 kN=m

                                                          Equations 630 and 631 then become

                                                          561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                          792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                          ie

                                                          561s 0616N 405 frac14 0

                                                          792 0857N thorn 563 frac14 0

                                                          N frac14 848

                                                          0857frac14 989 kN=m

                                                          Then

                                                          561s 609 405 frac14 0

                                                          s frac14 649

                                                          561frac14 116 kN=m3

                                                          The calculations for trial values of F of 20 15 and 10 are summarized below

                                                          F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                          20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                          s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                          Figure Q611

                                                          Lateral earth pressure 47

                                                          612

                                                          For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                          45 thorn 0

                                                          2frac14 63

                                                          For the retained material between the surface and a depth of 36m

                                                          Pa frac14 1

                                                          2 030 18 362 frac14 350 kN=m

                                                          Weight of reinforced fill between the surface and a depth of 36m is

                                                          Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                          eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                          Eccentricity of Rv

                                                          e frac14 263 250 frac14 013m

                                                          The average vertical stress at a depth of 36m is

                                                          z frac14 Rv

                                                          L 2efrac14 324

                                                          474frac14 68 kN=m2

                                                          (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                          Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                          Tensile stress in the element frac14 138 103

                                                          65 3frac14 71N=mm2

                                                          Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                          Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                          Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                          The weight of ABC is

                                                          W frac14 1

                                                          2 18 52 265 frac14 124 kN=m

                                                          From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                          48 Lateral earth pressure

                                                          (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                          Tp frac14 032 68 120 065 frac14 170 kN

                                                          Tr frac14 213 420

                                                          418frac14 214 kN

                                                          Again the tensile failure and slipping limit states are satisfied for this element

                                                          Figure Q612

                                                          Lateral earth pressure 49

                                                          Chapter 7

                                                          Consolidation theory

                                                          71

                                                          Total change in thickness

                                                          H frac14 782 602 frac14 180mm

                                                          Average thickness frac14 1530thorn 180

                                                          2frac14 1620mm

                                                          Length of drainage path d frac14 1620

                                                          2frac14 810mm

                                                          Root time plot (Figure Q71a)

                                                          ffiffiffiffiffiffit90p frac14 33

                                                          t90 frac14 109min

                                                          cv frac14 0848d2

                                                          t90frac14 0848 8102

                                                          109 1440 365

                                                          106frac14 27m2=year

                                                          r0 frac14 782 764

                                                          782 602frac14 018

                                                          180frac14 0100

                                                          rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                          10 119

                                                          9 180frac14 0735

                                                          rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                          Log time plot (Figure Q71b)

                                                          t50 frac14 26min

                                                          cv frac14 0196d2

                                                          t50frac14 0196 8102

                                                          26 1440 365

                                                          106frac14 26m2=year

                                                          r0 frac14 782 763

                                                          782 602frac14 019

                                                          180frac14 0106

                                                          rp frac14 763 623

                                                          782 602frac14 140

                                                          180frac14 0778

                                                          rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                          Figure Q71(a)

                                                          Figure Q71(b)

                                                          Final void ratio

                                                          e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                          e

                                                          Hfrac14 1thorn e0

                                                          H0frac14 1thorn e1 thorne

                                                          H0

                                                          ie

                                                          e

                                                          180frac14 1631thorne

                                                          1710

                                                          e frac14 2936

                                                          1530frac14 0192

                                                          Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                          mv frac14 1

                                                          1thorn e0 e0 e101 00

                                                          frac14 1

                                                          1823 0192

                                                          0107frac14 098m2=MN

                                                          k frac14 cvmvw frac14 265 098 98

                                                          60 1440 365 103frac14 81 1010 m=s

                                                          72

                                                          Using Equation 77 (one-dimensional method)

                                                          sc frac14 e0 e11thorn e0 H

                                                          Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                          Figure Q72

                                                          52 Consolidation theory

                                                          Settlement

                                                          Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                          1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                          318

                                                          Notes 5 92y 460thorn 84

                                                          Heave

                                                          Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                          1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                          38

                                                          73

                                                          U frac14 f ethTvTHORN frac14 f cvt

                                                          d2

                                                          Hence if cv is constant

                                                          t1

                                                          t2frac14 d

                                                          21

                                                          d22

                                                          where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                          d1 frac14 95mm and d2 frac14 2500mm

                                                          for U frac14 050 t2 frac14 t1 d22

                                                          d21

                                                          frac14 20

                                                          60 24 365 25002

                                                          952frac14 263 years

                                                          for U lt 060 Tv frac14

                                                          4U2 (Equation 724(a))

                                                          t030 frac14 t050 0302

                                                          0502

                                                          frac14 263 036 frac14 095 years

                                                          Consolidation theory 53

                                                          74

                                                          The layer is open

                                                          d frac14 8

                                                          2frac14 4m

                                                          Tv frac14 cvtd2frac14 24 3

                                                          42frac14 0450

                                                          ui frac14 frac14 84 kN=m2

                                                          The excess pore water pressure is given by Equation 721

                                                          ue frac14Xmfrac141mfrac140

                                                          2ui

                                                          Msin

                                                          Mz

                                                          d

                                                          expethM2TvTHORN

                                                          In this case z frac14 d

                                                          sinMz

                                                          d

                                                          frac14 sinM

                                                          where

                                                          M frac14

                                                          23

                                                          25

                                                          2

                                                          M sin M M2Tv exp (M2Tv)

                                                          2thorn1 1110 0329

                                                          3

                                                          21 9993 457 105

                                                          ue frac14 2 84 2

                                                          1 0329 ethother terms negligibleTHORN

                                                          frac14 352 kN=m2

                                                          75

                                                          The layer is open

                                                          d frac14 6

                                                          2frac14 3m

                                                          Tv frac14 cvtd2frac14 10 3

                                                          32frac14 0333

                                                          The layer thickness will be divided into six equal parts ie m frac14 6

                                                          54 Consolidation theory

                                                          For an open layer

                                                          Tv frac14 4n

                                                          m2

                                                          n frac14 0333 62

                                                          4frac14 300

                                                          The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                          ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                          i j

                                                          0 1 2 3 4 5 6 7 8 9 10 11 12

                                                          0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                          The initial and 3-year isochrones are plotted in Figure Q75

                                                          Area under initial isochrone frac14 180 units

                                                          Area under 3-year isochrone frac14 63 units

                                                          The average degree of consolidation is given by Equation 725Thus

                                                          U frac14 1 63

                                                          180frac14 065

                                                          Figure Q75

                                                          Consolidation theory 55

                                                          76

                                                          At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                          0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                          The final consolidation settlement (one-dimensional method) is

                                                          sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                          Corrected time t frac14 2 1

                                                          2

                                                          40

                                                          52

                                                          frac14 1615 years

                                                          Tv frac14 cvtd2frac14 44 1615

                                                          42frac14 0444

                                                          From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                          77

                                                          The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                          Figure Q77

                                                          56 Consolidation theory

                                                          Point m n Ir (kNm2) sc (mm)

                                                          13020frac14 15 20

                                                          20frac14 10 0194 (4) 113 124

                                                          260

                                                          20frac14 30

                                                          20

                                                          20frac14 10 0204 (2) 59 65

                                                          360

                                                          20frac14 30

                                                          40

                                                          20frac14 20 0238 (1) 35 38

                                                          430

                                                          20frac14 15

                                                          40

                                                          20frac14 20 0224 (2) 65 72

                                                          Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                          78

                                                          Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                          (a) Immediate settlement

                                                          H

                                                          Bfrac14 30

                                                          35frac14 086

                                                          D

                                                          Bfrac14 2

                                                          35frac14 006

                                                          Figure Q78

                                                          Consolidation theory 57

                                                          From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                          si frac14 130131qB

                                                          Eufrac14 10 032 105 35

                                                          40frac14 30mm

                                                          (b) Consolidation settlement

                                                          Layer z (m) Dz Ic (kNm2) syod (mm)

                                                          1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                          3150

                                                          Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                          Now

                                                          H

                                                          Bfrac14 30

                                                          35frac14 086 and A frac14 065

                                                          from Figure 712 13 frac14 079

                                                          sc frac14 13sod frac14 079 315 frac14 250mm

                                                          Total settlement

                                                          s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                          79

                                                          Without sand drains

                                                          Uv frac14 025

                                                          Tv frac14 0049 ethfrom Figure 718THORN

                                                          t frac14 Tvd2

                                                          cvfrac14 0049 82

                                                          cvWith sand drains

                                                          R frac14 0564S frac14 0564 3 frac14 169m

                                                          n frac14 Rrfrac14 169

                                                          015frac14 113

                                                          Tr frac14 cht

                                                          4R2frac14 ch

                                                          4 1692 0049 82

                                                          cvethand ch frac14 cvTHORN

                                                          frac14 0275

                                                          Ur frac14 073 (from Figure 730)

                                                          58 Consolidation theory

                                                          Using Equation 740

                                                          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                          U frac14 080

                                                          710

                                                          Without sand drains

                                                          Uv frac14 090

                                                          Tv frac14 0848

                                                          t frac14 Tvd2

                                                          cvfrac14 0848 102

                                                          96frac14 88 years

                                                          With sand drains

                                                          R frac14 0564S frac14 0564 4 frac14 226m

                                                          n frac14 Rrfrac14 226

                                                          015frac14 15

                                                          Tr

                                                          Tvfrac14 chcv

                                                          d2

                                                          4R2ethsame tTHORN

                                                          Tr

                                                          Tvfrac14 140

                                                          96 102

                                                          4 2262frac14 714 eth1THORN

                                                          Using Equation 740

                                                          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                          An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                          Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                          040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                          Thus

                                                          Uv frac14 0295 and Ur frac14 086

                                                          t frac14 88 00683

                                                          0848frac14 07 years

                                                          Consolidation theory 59

                                                          Chapter 8

                                                          Bearing capacity

                                                          81

                                                          (a) The ultimate bearing capacity is given by Equation 83

                                                          qf frac14 cNc thorn DNq thorn 1

                                                          2BN

                                                          For u frac14 0

                                                          Nc frac14 514 Nq frac14 1 N frac14 0

                                                          qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                          The net ultimate bearing capacity is

                                                          qnf frac14 qf D frac14 540 kN=m2

                                                          The net foundation pressure is

                                                          qn frac14 q D frac14 425

                                                          2 eth21 1THORN frac14 192 kN=m2

                                                          The factor of safety (Equation 86) is

                                                          F frac14 qnfqnfrac14 540

                                                          192frac14 28

                                                          (b) For 0 frac14 28

                                                          Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                          qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                          2 112 2 13

                                                          frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                          qnf frac14 574 112 frac14 563 kN=m2

                                                          F frac14 563

                                                          192frac14 29

                                                          (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                          82

                                                          For 0 frac14 38

                                                          Nq frac14 49 N frac14 67

                                                          qnf frac14 DethNq 1THORN thorn 1

                                                          2BN ethfrom Equation 83THORN

                                                          frac14 eth18 075 48THORN thorn 1

                                                          2 18 15 67

                                                          frac14 648thorn 905 frac14 1553 kN=m2

                                                          qn frac14 500

                                                          15 eth18 075THORN frac14 320 kN=m2

                                                          F frac14 qnfqnfrac14 1553

                                                          320frac14 48

                                                          0d frac14 tan1tan 38

                                                          125

                                                          frac14 32 therefore Nq frac14 23 and N frac14 25

                                                          Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                          2 18 15 25

                                                          frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                          Design load (action) Vd frac14 500 kN=m

                                                          The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                          83

                                                          D

                                                          Bfrac14 350

                                                          225frac14 155

                                                          From Figure 85 for a square foundation

                                                          Nc frac14 81

                                                          Bearing capacity 61

                                                          For a rectangular foundation (L frac14 450m B frac14 225m)

                                                          Nc frac14 084thorn 016B

                                                          L

                                                          81 frac14 745

                                                          Using Equation 810

                                                          qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                          For F frac14 3

                                                          qn frac14 1006

                                                          3frac14 335 kN=m2

                                                          q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                          Design load frac14 405 450 225 frac14 4100 kN

                                                          Design undrained strength cud frac14 135

                                                          14frac14 96 kN=m2

                                                          Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                          frac14 7241 kN

                                                          Design load Vd frac14 4100 kN

                                                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                          84

                                                          For 0 frac14 40

                                                          Nq frac14 64 N frac14 95

                                                          qnf frac14 DethNq 1THORN thorn 04BN

                                                          (a) Water table 5m below ground level

                                                          qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                          qn frac14 400 17 frac14 383 kN=m2

                                                          F frac14 2686

                                                          383frac14 70

                                                          (b) Water table 1m below ground level (ie at foundation level)

                                                          0 frac14 20 98 frac14 102 kN=m3

                                                          62 Bearing capacity

                                                          qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                          F frac14 2040

                                                          383frac14 53

                                                          (c) Water table at ground level with upward hydraulic gradient 02

                                                          eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                          qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                          F frac14 1296

                                                          392frac14 33

                                                          85

                                                          The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                          Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                          Design value of 0 frac14 tan1tan 39

                                                          125

                                                          frac14 33

                                                          For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                          Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                          86

                                                          (a) Undrained shear for u frac14 0

                                                          Nc frac14 514 Nq frac14 1 N frac14 0

                                                          qnf frac14 12cuNc

                                                          frac14 12 100 514 frac14 617 kN=m2

                                                          qn frac14 qnfFfrac14 617

                                                          3frac14 206 kN=m2

                                                          q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                          Bearing capacity 63

                                                          Drained shear for 0 frac14 32

                                                          Nq frac14 23 N frac14 25

                                                          0 frac14 21 98 frac14 112 kN=m3

                                                          qnf frac14 0DethNq 1THORN thorn 040BN

                                                          frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                          frac14 694 kN=m2

                                                          q frac14 694

                                                          3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                          Design load frac14 42 227 frac14 3632 kN

                                                          (b) Design undrained strength cud frac14 100

                                                          14frac14 71 kNm2

                                                          Design bearing resistance Rd frac14 12cudNe area

                                                          frac14 12 71 514 42

                                                          frac14 7007 kN

                                                          For drained shear 0d frac14 tan1tan 32

                                                          125

                                                          frac14 26

                                                          Nq frac14 12 N frac14 10

                                                          Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                          (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                          Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                          1 2 100 0175 0700qn 0182qn

                                                          2 6 033 0044 0176qn 0046qn

                                                          3 10 020 0017 0068qn 0018qn

                                                          0246qn

                                                          Diameter of equivalent circle B frac14 45m

                                                          H

                                                          Bfrac14 12

                                                          45frac14 27 and A frac14 042

                                                          13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                          64 Bearing capacity

                                                          For sc frac14 30mm

                                                          qn frac14 30

                                                          0147frac14 204 kN=m2

                                                          q frac14 204thorn 21 frac14 225 kN=m2

                                                          Design load frac14 42 225 frac14 3600 kN

                                                          The design load is 3600 kN settlement being the limiting criterion

                                                          87

                                                          D

                                                          Bfrac14 8

                                                          4frac14 20

                                                          From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                          F frac14 cuNc

                                                          Dfrac14 40 71

                                                          20 8frac14 18

                                                          88

                                                          Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                          Design value of 0 frac14 tan1tan 38

                                                          125

                                                          frac14 32

                                                          Figure Q86

                                                          Bearing capacity 65

                                                          For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                          Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                          The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                          Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                          For B frac14 250m qn frac14 3750

                                                          2502 17 frac14 583 kN=m2

                                                          From Figure 510 m frac14 n frac14 126

                                                          6frac14 021

                                                          Ir frac14 0019

                                                          Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                          Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                          The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                          89

                                                          Depth (m) N 0v (kNm2) CN N1

                                                          070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                          Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                          (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                          Cw frac14 05thorn 0530

                                                          47

                                                          frac14 082

                                                          66 Bearing capacity

                                                          Thus

                                                          qa frac14 150 082 frac14 120 kN=m2

                                                          (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                          Thus

                                                          qa frac14 90 15 frac14 135 kN=m2

                                                          (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                          Ic frac14 171

                                                          1014frac14 0068

                                                          From Equation 819(a) with s frac14 25mm

                                                          q frac14 25

                                                          3507 0068frac14 150 kN=m2

                                                          810

                                                          Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                          Izp frac14 05thorn 01130

                                                          16 27

                                                          05

                                                          frac14 067

                                                          Refer to Figure Q810

                                                          E frac14 25qc

                                                          Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                          Ez (mm3MN)

                                                          1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                          0203

                                                          C1 frac14 1 0500qnfrac14 1 05 12 16

                                                          130frac14 093

                                                          C2 frac14 1 ethsayTHORN

                                                          s frac14 C1C2qnX Iz

                                                          Ez frac14 093 1 130 0203 frac14 25mm

                                                          Bearing capacity 67

                                                          811

                                                          At pile base level

                                                          cu frac14 220 kN=m2

                                                          qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                          Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                          00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                          qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                          Then

                                                          Qf frac14 Abqb thorn Asqs

                                                          frac14

                                                          4 32 1980

                                                          thorn eth 105 139 86THORN

                                                          frac14 13 996thorn 3941 frac14 17 937 kN

                                                          0 01 02 03 04 05 06 07

                                                          0 2 4 6 8 10 12 14

                                                          1

                                                          2

                                                          3

                                                          4

                                                          5

                                                          6

                                                          7

                                                          8

                                                          (1)

                                                          (2)

                                                          (3)

                                                          (4)

                                                          (5)

                                                          qc

                                                          qc

                                                          Iz

                                                          Iz

                                                          (MNm2)

                                                          z (m)

                                                          Figure Q810

                                                          68 Bearing capacity

                                                          Allowable load

                                                          ethaTHORN Qf

                                                          2frac14 17 937

                                                          2frac14 8968 kN

                                                          ethbTHORN Abqb

                                                          3thorn Asqs frac14 13 996

                                                          3thorn 3941 frac14 8606 kN

                                                          ie allowable load frac14 8600 kN

                                                          Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                          According to the limit state method

                                                          Characteristic undrained strength at base level cuk frac14 220

                                                          150kN=m2

                                                          Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                          150frac14 1320 kN=m2

                                                          Characteristic shaft resistance qsk frac14 00150

                                                          frac14 86

                                                          150frac14 57 kN=m2

                                                          Characteristic base and shaft resistances

                                                          Rbk frac14

                                                          4 32 1320 frac14 9330 kN

                                                          Rsk frac14 105 139 86

                                                          150frac14 2629 kN

                                                          For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                          Design bearing resistance Rcd frac14 9330

                                                          160thorn 2629

                                                          130

                                                          frac14 5831thorn 2022

                                                          frac14 7850 kN

                                                          Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                          812

                                                          ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                          qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                          For a single pile

                                                          Qf frac14 Abqb thorn Asqs

                                                          frac14

                                                          4 062 1305

                                                          thorn eth 06 15 42THORN

                                                          frac14 369thorn 1187 frac14 1556 kN

                                                          Bearing capacity 69

                                                          Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                          eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                          (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                          (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                          qbkfrac14 9cuk frac14 9 220

                                                          150frac14 1320 kN=m2

                                                          qskfrac14cuk frac14 040 105

                                                          150frac14 28 kN=m2

                                                          Rbkfrac14

                                                          4 0602 1320 frac14 373 kN

                                                          Rskfrac14 060 15 28 frac14 791 kN

                                                          Rcdfrac14 373

                                                          160thorn 791

                                                          130frac14 233thorn 608 frac14 841 kN

                                                          Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                          Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                          (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                          q frac14 21 000

                                                          1762frac14 68 kN=m2

                                                          Immediate settlement

                                                          H

                                                          Bfrac14 15

                                                          176frac14 085

                                                          D

                                                          Bfrac14 13

                                                          176frac14 074

                                                          L

                                                          Bfrac14 1

                                                          Hence from Figure 515

                                                          130 frac14 078 and 131 frac14 041

                                                          70 Bearing capacity

                                                          Thus using Equation 528

                                                          si frac14 078 041 68 176

                                                          65frac14 6mm

                                                          Consolidation settlement

                                                          Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                          1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                          434 (sod)

                                                          Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                          sc frac14 056 434 frac14 24mm

                                                          The total settlement is (6thorn 24) frac14 30mm

                                                          813

                                                          At base level N frac14 26 Then using Equation 830

                                                          qb frac14 40NDb

                                                          Bfrac14 40 26 2

                                                          025frac14 8320 kN=m2

                                                          ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                          Figure Q812

                                                          Bearing capacity 71

                                                          Over the length embedded in sand

                                                          N frac14 21 ie18thorn 24

                                                          2

                                                          Using Equation 831

                                                          qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                          For a single pile

                                                          Qf frac14 Abqb thorn Asqs

                                                          frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                          For the pile group assuming a group efficiency of 12

                                                          XQf frac14 12 9 604 frac14 6523 kN

                                                          Then the load factor is

                                                          F frac14 6523

                                                          2000thorn 1000frac14 21

                                                          (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                          Characteristic base resistance per unit area qbk frac14 8320

                                                          150frac14 5547 kNm2

                                                          Characteristic shaft resistance per unit area qsk frac14 42

                                                          150frac14 28 kNm2

                                                          Characteristic base and shaft resistances for a single pile

                                                          Rbk frac14 0252 5547 frac14 347 kN

                                                          Rsk frac14 4 025 2 28 frac14 56 kN

                                                          For a driven pile the partial factors are b frac14 s frac14 130

                                                          Design bearing resistance Rcd frac14 347

                                                          130thorn 56

                                                          130frac14 310 kN

                                                          For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                          Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                          (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                          From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                          72 Bearing capacity

                                                          N frac14 24thorn 26thorn 34

                                                          3frac14 28

                                                          Ic frac14 171

                                                          2814frac14 0016 ethEquation 818THORN

                                                          s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                          814

                                                          Using Equation 841

                                                          Tf frac14 DLcu thorn

                                                          4ethD2 d2THORNcuNc

                                                          frac14 eth 02 5 06 110THORN thorn

                                                          4eth022 012THORN110 9

                                                          frac14 207thorn 23 frac14 230 kN

                                                          Figure Q813

                                                          Bearing capacity 73

                                                          Chapter 9

                                                          Stability of slopes

                                                          91

                                                          Referring to Figure Q91

                                                          W frac14 417 19 frac14 792 kN=m

                                                          Q frac14 20 28 frac14 56 kN=m

                                                          Arc lengthAB frac14

                                                          180 73 90 frac14 115m

                                                          Arc length BC frac14

                                                          180 28 90 frac14 44m

                                                          The factor of safety is given by

                                                          F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                          90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                          Depth of tension crack z0 frac14 2cu

                                                          frac14 2 20

                                                          19frac14 21m

                                                          Arc length BD frac14

                                                          180 13

                                                          1

                                                          2 90 frac14 21m

                                                          F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                          The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                          Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                          14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                          Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                          The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                          92

                                                          u frac14 0

                                                          Depth factor D frac14 11

                                                          9frac14 122

                                                          Using Equation 92 with F frac14 10

                                                          Ns frac14 cu

                                                          FHfrac14 30

                                                          10 19 9frac14 0175

                                                          Hence from Figure 93

                                                          frac14 50

                                                          For F frac14 12

                                                          Ns frac14 30

                                                          12 19 9frac14 0146

                                                          frac14 27

                                                          93

                                                          Refer to Figure Q93

                                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                          1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                          74 m

                                                          214 1deg

                                                          213 1deg

                                                          39 m

                                                          WB

                                                          D

                                                          C

                                                          28 m

                                                          21 m

                                                          A

                                                          Q

                                                          Soil (1)Soil (2)

                                                          73deg

                                                          Figure Q91

                                                          Stability of slopes 75

                                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                          9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                          599 256 328 1372

                                                          Figure Q93

                                                          76 Stability of slopes

                                                          XW cos frac14 b

                                                          Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                          W sin frac14 bX

                                                          h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                          Arc length La frac14

                                                          180 57

                                                          1

                                                          2 326 frac14 327m

                                                          The factor of safety is given by

                                                          F frac14 c0La thorn tan0ethW cos ulTHORN

                                                          W sin

                                                          frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                          frac14 091

                                                          According to the limit state method

                                                          0d frac14 tan1tan 32

                                                          125

                                                          frac14 265

                                                          c0 frac14 8

                                                          160frac14 5 kN=m2

                                                          Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                          Design disturbing moment frac14 1075 kN=m

                                                          The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                          94

                                                          F frac14 1

                                                          W sin

                                                          Xfc0bthorn ethW ubTHORN tan0g sec

                                                          1thorn ethtan tan0=FTHORN

                                                          c0 frac14 8 kN=m2

                                                          0 frac14 32

                                                          c0b frac14 8 2 frac14 16 kN=m

                                                          W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                          Try F frac14 100

                                                          tan0

                                                          Ffrac14 0625

                                                          Stability of slopes 77

                                                          Values of u are as obtained in Figure Q93

                                                          SliceNo

                                                          h(m)

                                                          W frac14 bh(kNm)

                                                          W sin(kNm)

                                                          ub(kNm)

                                                          c0bthorn (W ub) tan0(kNm)

                                                          sec

                                                          1thorn (tan tan0)FProduct(kNm)

                                                          1 05 21 6 2 8 24 1078 262 13 55 31

                                                          23 33 30 1042 31

                                                          3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                          224 92 72 0931 67

                                                          6 50 210 11 40 100 85 0907 777 55 231 14

                                                          12 58 112 90 0889 80

                                                          8 60 252 1812

                                                          80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                          10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                          2154 88 116 0853 99

                                                          14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                          1074 1091

                                                          F frac14 1091

                                                          1074frac14 102 (assumed value 100)

                                                          Thus

                                                          F frac14 101

                                                          95

                                                          F frac14 1

                                                          W sin

                                                          XfWeth1 ruTHORN tan0g sec

                                                          1thorn ethtan tan0THORN=F

                                                          0 frac14 33

                                                          ru frac14 020

                                                          W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                          eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                          Try F frac14 110

                                                          tan 0

                                                          Ffrac14 tan 33

                                                          110frac14 0590

                                                          78 Stability of slopes

                                                          Referring to Figure Q95

                                                          SliceNo

                                                          h(m)

                                                          W frac14 bh(kNm)

                                                          W sin(kNm)

                                                          W(1 ru) tan0(kNm)

                                                          sec

                                                          1thorn ( tan tan0)FProduct(kNm)

                                                          1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                          2120 234 0892 209

                                                          4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                          1185 1271

                                                          Figure Q95

                                                          Stability of slopes 79

                                                          F frac14 1271

                                                          1185frac14 107

                                                          The trial value was 110 therefore take F to be 108

                                                          96

                                                          (a) Water table at surface the factor of safety is given by Equation 912

                                                          F frac14 0

                                                          sat

                                                          tan0

                                                          tan

                                                          ptie 15 frac14 92

                                                          19

                                                          tan 36

                                                          tan

                                                          tan frac14 0234

                                                          frac14 13

                                                          Water table well below surface the factor of safety is given by Equation 911

                                                          F frac14 tan0

                                                          tan

                                                          frac14 tan 36

                                                          tan 13

                                                          frac14 31

                                                          (b) 0d frac14 tan1tan 36

                                                          125

                                                          frac14 30

                                                          Depth of potential failure surface frac14 z

                                                          Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                          frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                          frac14 504z kN

                                                          Design disturbing moment per unit area Sd frac14 sat sin cos

                                                          frac14 19 z sin 13 cos 13

                                                          frac14 416z kN

                                                          Rd gtSd therefore the limit state for overall stability is satisfied

                                                          80 Stability of slopes

                                                          • Book Cover
                                                          • Title
                                                          • Contents
                                                          • Basic characteristics of soils
                                                          • Seepage
                                                          • Effective stress
                                                          • Shear strength
                                                          • Stresses and displacements
                                                          • Lateral earth pressure
                                                          • Consolidation theory
                                                          • Bearing capacity
                                                          • Stability of slopes

                                                            45

                                                            3 (kNm2) 1 3 (kNm2) 1 (kNm2) u (kNm2) 03 (kNm2) 01 (kNm2)

                                                            150 103 253 82 68 171300 202 502 169 131 333450 305 755 252 198 503600 410 1010 331 269 679

                                                            The Mohr circles and failure envelope are drawn in Figure Q45 from which c0 frac14 0 and0 frac14 25 1frasl2

                                                            The principal stress difference at failure depends only on the value of all-roundpressure under which consolidation took place ie 250 kNm2 Hence by proportionthe expected value of (1 3)ffrac14 170 kNm2

                                                            46

                                                            03 (kNm2) VV0 ll0 Area (mm2) Load (N) 1 3 (kNm2) 01 (kNm2)

                                                            200 0061 0095 1177 565 480 680400 0086 0110 1165 1015 871 1271600 0108 0124 1155 1321 1144 1744

                                                            The average cross-sectional area of each specimen is obtained from Equation 410 theoriginal values of A l and V are A0frac14 1134mm2 l0frac14 76mm V0frac14 86 200mm3 TheMohr circles are drawn in Figure Q46(a) and (b) From (a) the secant parameters aremeasured as 34 315 and 29 The failure envelope shown in (b) exhibits a curvatureand between 300 and 500 kNm2 is approximated to a straight line from whichc0 frac14 20 kNm2 and 0 frac14 31

                                                            Figure Q45

                                                            24 Shear strength

                                                            47

                                                            The torque required to produce shear failure is given by

                                                            T frac14 dh cud

                                                            2thorn 2

                                                            Z d=2

                                                            0

                                                            2r drcur

                                                            frac14 cud2h

                                                            2thorn 4cu

                                                            Z d=2

                                                            0

                                                            r2dr

                                                            frac14 cud2h

                                                            2thorn d

                                                            3

                                                            6

                                                            Then

                                                            35 frac14 cu52 10

                                                            2thorn 53

                                                            6

                                                            103

                                                            cu frac14 76 kN=m3

                                                            400

                                                            0 400 800 1200 1600

                                                            τ (k

                                                            Nm

                                                            2 )

                                                            σprime (kNm2)

                                                            34deg

                                                            315deg29deg

                                                            (a)

                                                            (b)

                                                            0 400

                                                            400

                                                            800 1200 1600

                                                            Failure envelope

                                                            300 500

                                                            σprime (kNm2)

                                                            τ (k

                                                            Nm

                                                            2 )

                                                            20 (kNm2)

                                                            31deg

                                                            Figure Q46

                                                            Shear strength 25

                                                            48

                                                            The relevant stress values are calculated as follows

                                                            3 frac14 600 kN=m2

                                                            1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                                            2(1 3) 0 40 79 107 139 159

                                                            1

                                                            2(01 thorn 03) 400 411 402 389 351 326

                                                            1

                                                            2(1 thorn 3) 600 640 679 707 739 759

                                                            The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                                            Af frac14 433 200

                                                            319frac14 073

                                                            49

                                                            B frac14 u33

                                                            frac14 144

                                                            350 200frac14 096

                                                            a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                                            0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                                            10 283 65 023

                                                            Figure Q48

                                                            26 Shear strength

                                                            The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                                            Figure Q49

                                                            Shear strength 27

                                                            Chapter 5

                                                            Stresses and displacements

                                                            51

                                                            Vertical stress is given by

                                                            z frac14 Qz2Ip frac14 5000

                                                            52Ip

                                                            Values of Ip are obtained from Table 51

                                                            r (m) rz Ip z (kNm2)

                                                            0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                                            10 20 0009 2

                                                            The variation of z with radial distance (r) is plotted in Figure Q51

                                                            Figure Q51

                                                            52

                                                            Below the centre load (Figure Q52)

                                                            r

                                                            zfrac14 0 for the 7500-kN load

                                                            Ip frac14 0478

                                                            r

                                                            zfrac14 5

                                                            4frac14 125 for the 10 000- and 9000-kN loads

                                                            Ip frac14 0045

                                                            Then

                                                            z frac14X Q

                                                            z2Ip

                                                            frac14 7500 0478

                                                            42thorn 10 000 0045

                                                            42thorn 9000 0045

                                                            42

                                                            frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                                            53

                                                            The vertical stress under a corner of a rectangular area is given by

                                                            z frac14 qIr

                                                            where values of Ir are obtained from Figure 510 In this case

                                                            z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                                            z

                                                            Figure Q52

                                                            Stresses and displacements 29

                                                            z (m) m n Ir z (kNm2)

                                                            0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                                            10 010 0005 5

                                                            z is plotted against z in Figure Q53

                                                            54

                                                            (a)

                                                            m frac14 125

                                                            12frac14 104

                                                            n frac14 18

                                                            12frac14 150

                                                            From Figure 510 Irfrac14 0196

                                                            z frac14 2 175 0196 frac14 68 kN=m2

                                                            Figure Q53

                                                            30 Stresses and displacements

                                                            (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                                            z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                                            55

                                                            Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                                            Px frac14 2Q

                                                            1

                                                            m2 thorn 1frac14 2 150

                                                            125frac14 76 kN=m

                                                            Equation 517 is used to obtain the pressure distribution

                                                            px frac14 4Q

                                                            h

                                                            m2n

                                                            ethm2 thorn n2THORN2 frac14150

                                                            m2n

                                                            ethm2 thorn n2THORN2 ethkN=m2THORN

                                                            Figure Q54

                                                            Stresses and displacements 31

                                                            n m2n

                                                            (m2 thorn n2)2

                                                            px(kNm2)

                                                            0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                            The pressure distribution is plotted in Figure Q55

                                                            56

                                                            H

                                                            Bfrac14 10

                                                            2frac14 5

                                                            L

                                                            Bfrac14 4

                                                            2frac14 2

                                                            D

                                                            Bfrac14 1

                                                            2frac14 05

                                                            Hence from Figure 515

                                                            131 frac14 082

                                                            130 frac14 094

                                                            Figure Q55

                                                            32 Stresses and displacements

                                                            The immediate settlement is given by Equation 528

                                                            si frac14 130131qB

                                                            Eu

                                                            frac14 094 082 200 2

                                                            45frac14 7mm

                                                            Stresses and displacements 33

                                                            Chapter 6

                                                            Lateral earth pressure

                                                            61

                                                            For 0 frac14 37 the active pressure coefficient is given by

                                                            Ka frac14 1 sin 37

                                                            1thorn sin 37frac14 025

                                                            The total active thrust (Equation 66a with c0 frac14 0) is

                                                            Pa frac14 1

                                                            2KaH

                                                            2 frac14 1

                                                            2 025 17 62 frac14 765 kN=m

                                                            If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                            K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                            and the thrust on the wall is

                                                            P0 frac14 1

                                                            2K0H

                                                            2 frac14 1

                                                            2 040 17 62 frac14 122 kN=m

                                                            62

                                                            The active pressure coefficients for the three soil types are as follows

                                                            Ka1 frac141 sin 35

                                                            1thorn sin 35frac14 0271

                                                            Ka2 frac141 sin 27

                                                            1thorn sin 27frac14 0375

                                                            ffiffiffiffiffiffiffiKa2

                                                            p frac14 0613

                                                            Ka3 frac141 sin 42

                                                            1thorn sin 42frac14 0198

                                                            Distribution of active pressure (plotted in Figure Q62)

                                                            Depth (m) Soil Active pressure (kNm2)

                                                            3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                            12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                            At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                            Total thrust frac14 571 kNm

                                                            Point of application is (4893571) m from the top of the wall ie 857m

                                                            Force (kN) Arm (m) Moment (kN m)

                                                            (1)1

                                                            2 0271 16 32 frac14 195 20 390

                                                            (2) 0271 16 3 2 frac14 260 40 1040

                                                            (3)1

                                                            2 0271 92 22 frac14 50 433 217

                                                            (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                            (5)1

                                                            2 0375 102 32 frac14 172 70 1204

                                                            (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                            (7)1

                                                            2 0198 112 42 frac14 177 1067 1889

                                                            (8)1

                                                            2 98 92 frac14 3969 90 35721

                                                            5713 48934

                                                            Figure Q62

                                                            Lateral earth pressure 35

                                                            63

                                                            (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                            Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                            frac14 245

                                                            At the lower end of the piling

                                                            pa frac14 Kaqthorn Kasatz Kaccu

                                                            frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                            frac14 115 kN=m2

                                                            pp frac14 Kpsatzthorn Kpccu

                                                            frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                            frac14 202 kN=m2

                                                            (b) For 0 frac14 26 and frac14 1

                                                            20

                                                            Ka frac14 035

                                                            Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                            pfrac14 145 ethEquation 619THORN

                                                            Kp frac14 37

                                                            Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                            pfrac14 47 ethEquation 624THORN

                                                            At the lower end of the piling

                                                            pa frac14 Kaqthorn Ka0z Kacc

                                                            0

                                                            frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                            frac14 187 kN=m2

                                                            pp frac14 Kp0zthorn Kpcc

                                                            0

                                                            frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                            frac14 198 kN=m2

                                                            36 Lateral earth pressure

                                                            64

                                                            (a) For 0 frac14 38 Ka frac14 024

                                                            0 frac14 20 98 frac14 102 kN=m3

                                                            The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                            Force (kN) Arm (m) Moment (kN m)

                                                            (1) 024 10 66 frac14 159 33 525

                                                            (2)1

                                                            2 024 17 392 frac14 310 400 1240

                                                            (3) 024 17 39 27 frac14 430 135 580

                                                            (4)1

                                                            2 024 102 272 frac14 89 090 80

                                                            (5)1

                                                            2 98 272 frac14 357 090 321

                                                            Hfrac14 1345 MH frac14 2746

                                                            (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                            (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                            XM frac14MV MH frac14 7790 kNm

                                                            Lever arm of base resultant

                                                            M

                                                            Vfrac14 779

                                                            488frac14 160

                                                            Eccentricity of base resultant

                                                            e frac14 200 160 frac14 040m

                                                            39 m

                                                            27 m

                                                            40 m

                                                            04 m

                                                            04 m

                                                            26 m

                                                            (7)

                                                            (9)

                                                            (1)(2)

                                                            (3)

                                                            (4)

                                                            (5)

                                                            (8)(6)

                                                            (10)

                                                            WT

                                                            10 kNm2

                                                            Hydrostatic

                                                            Figure Q64

                                                            Lateral earth pressure 37

                                                            Base pressures (Equation 627)

                                                            p frac14 VB

                                                            1 6e

                                                            B

                                                            frac14 488

                                                            4eth1 060THORN

                                                            frac14 195 kN=m2 and 49 kN=m2

                                                            Factor of safety against sliding (Equation 628)

                                                            F frac14 V tan

                                                            Hfrac14 488 tan 25

                                                            1345frac14 17

                                                            (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                            Hfrac14 1633 kN

                                                            V frac14 4879 kN

                                                            MH frac14 3453 kNm

                                                            MV frac14 10536 kNm

                                                            The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                            65

                                                            For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                            Kp

                                                            Ffrac14 385

                                                            2

                                                            0 frac14 20 98 frac14 102 kN=m3

                                                            The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                            Force (kN) Arm (m) Moment (kN m)

                                                            (1)1

                                                            2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                            (2) 026 17 45 d frac14 199d d2 995d2

                                                            (3)1

                                                            2 026 102 d2 frac14 133d2 d3 044d3

                                                            (4)1

                                                            2 385

                                                            2 17 152 frac14 368 dthorn 05 368d 184

                                                            (5)385

                                                            2 17 15 d frac14 491d d2 2455d2

                                                            (6)1

                                                            2 385

                                                            2 102 d2 frac14 982d2 d3 327d3

                                                            38 Lateral earth pressure

                                                            XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                            d3 thorn 516d2 283d 1724 frac14 0

                                                            d frac14 179m

                                                            Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                            Over additional 20 embedded depth

                                                            pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                            Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                            66

                                                            The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                            Ka frac14 sin 69=sin 105

                                                            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                            pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                            26664

                                                            37775

                                                            2

                                                            frac14 050

                                                            The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                            Pa frac14 1

                                                            2 050 19 7502 frac14 267 kN=m

                                                            Figure Q65

                                                            Lateral earth pressure 39

                                                            Horizontal component

                                                            Ph frac14 267 cos 40 frac14 205 kN=m

                                                            Vertical component

                                                            Pv frac14 267 sin 40 frac14 172 kN=m

                                                            Consider moments about the toe of the wall (Figure Q66) (per m)

                                                            Force (kN) Arm (m) Moment (kN m)

                                                            (1)1

                                                            2 175 650 235 frac14 1337 258 345

                                                            (2) 050 650 235 frac14 764 175 134

                                                            (3)1

                                                            2 070 650 235 frac14 535 127 68

                                                            (4) 100 400 235 frac14 940 200 188

                                                            (5) 1

                                                            2 080 050 235 frac14 47 027 1

                                                            Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                            Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                            Lever arm of base resultant

                                                            M

                                                            Vfrac14 795

                                                            525frac14 151m

                                                            Eccentricity of base resultant

                                                            e frac14 200 151 frac14 049m

                                                            Figure Q66

                                                            40 Lateral earth pressure

                                                            Base pressures (Equation 627)

                                                            p frac14 525

                                                            41 6 049

                                                            4

                                                            frac14 228 kN=m2 and 35 kN=m2

                                                            The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                            The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                            The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                            67

                                                            For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                            Force (kN) Arm (m) Moment (kNm)

                                                            (1)1

                                                            2 027 17 52 frac14 574 183 1050

                                                            (2) 027 17 5 3 frac14 689 500 3445

                                                            (3)1

                                                            2 027 102 32 frac14 124 550 682

                                                            (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                            (5)1

                                                            2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                            (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                            (7) 1

                                                            2 267

                                                            2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                            (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                            2 d frac14 163d d2thorn 650 82d2 1060d

                                                            Tie rod force per m frac14 T 0 0

                                                            XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                            d3 thorn 77d2 269d 1438 frac14 0

                                                            d frac14 467m

                                                            Depth of penetration frac14 12d frac14 560m

                                                            Lateral earth pressure 41

                                                            Algebraic sum of forces for d frac14 467m isX

                                                            F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                            T frac14 905 kN=m

                                                            Force in each tie rod frac14 25T frac14 226 kN

                                                            68

                                                            (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                            0 frac14 21 98 frac14 112 kN=m3

                                                            The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                            uC frac14 150

                                                            165 15 98 frac14 134 kN=m2

                                                            The average seepage pressure is

                                                            j frac14 15

                                                            165 98 frac14 09 kN=m3

                                                            Hence

                                                            0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                            0 j frac14 112 09 frac14 103 kN=m3

                                                            Figure Q67

                                                            42 Lateral earth pressure

                                                            Consider moments about the anchor point A (per m)

                                                            Force (kN) Arm (m) Moment (kN m)

                                                            (1) 10 026 150 frac14 390 60 2340

                                                            (2)1

                                                            2 026 18 452 frac14 474 15 711

                                                            (3) 026 18 45 105 frac14 2211 825 18240

                                                            (4)1

                                                            2 026 121 1052 frac14 1734 100 17340

                                                            (5)1

                                                            2 134 15 frac14 101 40 404

                                                            (6) 134 30 frac14 402 60 2412

                                                            (7)1

                                                            2 134 60 frac14 402 95 3819

                                                            571 4527(8) Ppm

                                                            115 115PPm

                                                            XM frac14 0

                                                            Ppm frac144527

                                                            115frac14 394 kN=m

                                                            Available passive resistance

                                                            Pp frac14 1

                                                            2 385 103 62 frac14 714 kN=m

                                                            Factor of safety

                                                            Fp frac14 Pp

                                                            Ppm

                                                            frac14 714

                                                            394frac14 18

                                                            Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                            Figure Q68

                                                            Lateral earth pressure 43

                                                            (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                            Consider moments (per m) about the tie point A

                                                            Force (kN) Arm (m)

                                                            (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                            (2)1

                                                            2 033 18 452 frac14 601 15

                                                            (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                            (4)1

                                                            2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                            (5)1

                                                            2 134 15 frac14 101 40

                                                            (6) 134 30 frac14 402 60

                                                            (7)1

                                                            2 134 d frac14 67d d3thorn 75

                                                            (8) 1

                                                            2 30 103 d2 frac141545d2 2d3thorn 75

                                                            Moment (kN m)

                                                            (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                            XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                            d3 thorn 827d2 466d 1518 frac14 0

                                                            By trial

                                                            d frac14 544m

                                                            The minimum depth of embedment required is 544m

                                                            69

                                                            For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                            0 frac14 20 98 frac14 102 kN=m3

                                                            The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                            44 Lateral earth pressure

                                                            uC frac14 147

                                                            173 26 98 frac14 216 kN=m2

                                                            and the average seepage pressure around the wall is

                                                            j frac14 26

                                                            173 98 frac14 15 kN=m3

                                                            Consider moments about the prop (A) (per m)

                                                            Force (kN) Arm (m) Moment (kN m)

                                                            (1)1

                                                            2 03 17 272 frac14 186 020 37

                                                            (2) 03 17 27 53 frac14 730 335 2445

                                                            (3)1

                                                            2 03 (102thorn 15) 532 frac14 493 423 2085

                                                            (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                            (5)1

                                                            2 216 26 frac14 281 243 684

                                                            (6) 216 27 frac14 583 465 2712

                                                            (7)1

                                                            2 216 60 frac14 648 800 5184

                                                            3055(8)

                                                            1

                                                            2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                            Factor of safety

                                                            Fr frac14 6885

                                                            3055frac14 225

                                                            Figure Q69

                                                            Lateral earth pressure 45

                                                            610

                                                            For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                            p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                            Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                            Using the recommendations of Twine and Roscoe

                                                            p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                            Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                            611

                                                            frac14 18 kN=m3 0 frac14 34

                                                            H frac14 350m nH frac14 335m mH frac14 185m

                                                            Consider a trial value of F frac14 20 Refer to Figure 635

                                                            0m frac14 tan1tan 34

                                                            20

                                                            frac14 186

                                                            Then

                                                            frac14 45 thorn 0m2frac14 543

                                                            W frac14 1

                                                            2 18 3502 cot 543 frac14 792 kN=m

                                                            Figure Q610

                                                            46 Lateral earth pressure

                                                            P frac14 1

                                                            2 s 3352 frac14 561s kN=m

                                                            U frac14 1

                                                            2 98 1852 cosec 543 frac14 206 kN=m

                                                            Equations 630 and 631 then become

                                                            561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                            792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                            ie

                                                            561s 0616N 405 frac14 0

                                                            792 0857N thorn 563 frac14 0

                                                            N frac14 848

                                                            0857frac14 989 kN=m

                                                            Then

                                                            561s 609 405 frac14 0

                                                            s frac14 649

                                                            561frac14 116 kN=m3

                                                            The calculations for trial values of F of 20 15 and 10 are summarized below

                                                            F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                            20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                            s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                            Figure Q611

                                                            Lateral earth pressure 47

                                                            612

                                                            For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                            45 thorn 0

                                                            2frac14 63

                                                            For the retained material between the surface and a depth of 36m

                                                            Pa frac14 1

                                                            2 030 18 362 frac14 350 kN=m

                                                            Weight of reinforced fill between the surface and a depth of 36m is

                                                            Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                            eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                            Eccentricity of Rv

                                                            e frac14 263 250 frac14 013m

                                                            The average vertical stress at a depth of 36m is

                                                            z frac14 Rv

                                                            L 2efrac14 324

                                                            474frac14 68 kN=m2

                                                            (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                            Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                            Tensile stress in the element frac14 138 103

                                                            65 3frac14 71N=mm2

                                                            Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                            Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                            Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                            The weight of ABC is

                                                            W frac14 1

                                                            2 18 52 265 frac14 124 kN=m

                                                            From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                            48 Lateral earth pressure

                                                            (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                            Tp frac14 032 68 120 065 frac14 170 kN

                                                            Tr frac14 213 420

                                                            418frac14 214 kN

                                                            Again the tensile failure and slipping limit states are satisfied for this element

                                                            Figure Q612

                                                            Lateral earth pressure 49

                                                            Chapter 7

                                                            Consolidation theory

                                                            71

                                                            Total change in thickness

                                                            H frac14 782 602 frac14 180mm

                                                            Average thickness frac14 1530thorn 180

                                                            2frac14 1620mm

                                                            Length of drainage path d frac14 1620

                                                            2frac14 810mm

                                                            Root time plot (Figure Q71a)

                                                            ffiffiffiffiffiffit90p frac14 33

                                                            t90 frac14 109min

                                                            cv frac14 0848d2

                                                            t90frac14 0848 8102

                                                            109 1440 365

                                                            106frac14 27m2=year

                                                            r0 frac14 782 764

                                                            782 602frac14 018

                                                            180frac14 0100

                                                            rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                            10 119

                                                            9 180frac14 0735

                                                            rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                            Log time plot (Figure Q71b)

                                                            t50 frac14 26min

                                                            cv frac14 0196d2

                                                            t50frac14 0196 8102

                                                            26 1440 365

                                                            106frac14 26m2=year

                                                            r0 frac14 782 763

                                                            782 602frac14 019

                                                            180frac14 0106

                                                            rp frac14 763 623

                                                            782 602frac14 140

                                                            180frac14 0778

                                                            rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                            Figure Q71(a)

                                                            Figure Q71(b)

                                                            Final void ratio

                                                            e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                            e

                                                            Hfrac14 1thorn e0

                                                            H0frac14 1thorn e1 thorne

                                                            H0

                                                            ie

                                                            e

                                                            180frac14 1631thorne

                                                            1710

                                                            e frac14 2936

                                                            1530frac14 0192

                                                            Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                            mv frac14 1

                                                            1thorn e0 e0 e101 00

                                                            frac14 1

                                                            1823 0192

                                                            0107frac14 098m2=MN

                                                            k frac14 cvmvw frac14 265 098 98

                                                            60 1440 365 103frac14 81 1010 m=s

                                                            72

                                                            Using Equation 77 (one-dimensional method)

                                                            sc frac14 e0 e11thorn e0 H

                                                            Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                            Figure Q72

                                                            52 Consolidation theory

                                                            Settlement

                                                            Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                            1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                            318

                                                            Notes 5 92y 460thorn 84

                                                            Heave

                                                            Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                            1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                            38

                                                            73

                                                            U frac14 f ethTvTHORN frac14 f cvt

                                                            d2

                                                            Hence if cv is constant

                                                            t1

                                                            t2frac14 d

                                                            21

                                                            d22

                                                            where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                            d1 frac14 95mm and d2 frac14 2500mm

                                                            for U frac14 050 t2 frac14 t1 d22

                                                            d21

                                                            frac14 20

                                                            60 24 365 25002

                                                            952frac14 263 years

                                                            for U lt 060 Tv frac14

                                                            4U2 (Equation 724(a))

                                                            t030 frac14 t050 0302

                                                            0502

                                                            frac14 263 036 frac14 095 years

                                                            Consolidation theory 53

                                                            74

                                                            The layer is open

                                                            d frac14 8

                                                            2frac14 4m

                                                            Tv frac14 cvtd2frac14 24 3

                                                            42frac14 0450

                                                            ui frac14 frac14 84 kN=m2

                                                            The excess pore water pressure is given by Equation 721

                                                            ue frac14Xmfrac141mfrac140

                                                            2ui

                                                            Msin

                                                            Mz

                                                            d

                                                            expethM2TvTHORN

                                                            In this case z frac14 d

                                                            sinMz

                                                            d

                                                            frac14 sinM

                                                            where

                                                            M frac14

                                                            23

                                                            25

                                                            2

                                                            M sin M M2Tv exp (M2Tv)

                                                            2thorn1 1110 0329

                                                            3

                                                            21 9993 457 105

                                                            ue frac14 2 84 2

                                                            1 0329 ethother terms negligibleTHORN

                                                            frac14 352 kN=m2

                                                            75

                                                            The layer is open

                                                            d frac14 6

                                                            2frac14 3m

                                                            Tv frac14 cvtd2frac14 10 3

                                                            32frac14 0333

                                                            The layer thickness will be divided into six equal parts ie m frac14 6

                                                            54 Consolidation theory

                                                            For an open layer

                                                            Tv frac14 4n

                                                            m2

                                                            n frac14 0333 62

                                                            4frac14 300

                                                            The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                            ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                            i j

                                                            0 1 2 3 4 5 6 7 8 9 10 11 12

                                                            0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                            The initial and 3-year isochrones are plotted in Figure Q75

                                                            Area under initial isochrone frac14 180 units

                                                            Area under 3-year isochrone frac14 63 units

                                                            The average degree of consolidation is given by Equation 725Thus

                                                            U frac14 1 63

                                                            180frac14 065

                                                            Figure Q75

                                                            Consolidation theory 55

                                                            76

                                                            At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                            0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                            The final consolidation settlement (one-dimensional method) is

                                                            sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                            Corrected time t frac14 2 1

                                                            2

                                                            40

                                                            52

                                                            frac14 1615 years

                                                            Tv frac14 cvtd2frac14 44 1615

                                                            42frac14 0444

                                                            From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                            77

                                                            The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                            Figure Q77

                                                            56 Consolidation theory

                                                            Point m n Ir (kNm2) sc (mm)

                                                            13020frac14 15 20

                                                            20frac14 10 0194 (4) 113 124

                                                            260

                                                            20frac14 30

                                                            20

                                                            20frac14 10 0204 (2) 59 65

                                                            360

                                                            20frac14 30

                                                            40

                                                            20frac14 20 0238 (1) 35 38

                                                            430

                                                            20frac14 15

                                                            40

                                                            20frac14 20 0224 (2) 65 72

                                                            Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                            78

                                                            Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                            (a) Immediate settlement

                                                            H

                                                            Bfrac14 30

                                                            35frac14 086

                                                            D

                                                            Bfrac14 2

                                                            35frac14 006

                                                            Figure Q78

                                                            Consolidation theory 57

                                                            From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                            si frac14 130131qB

                                                            Eufrac14 10 032 105 35

                                                            40frac14 30mm

                                                            (b) Consolidation settlement

                                                            Layer z (m) Dz Ic (kNm2) syod (mm)

                                                            1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                            3150

                                                            Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                            Now

                                                            H

                                                            Bfrac14 30

                                                            35frac14 086 and A frac14 065

                                                            from Figure 712 13 frac14 079

                                                            sc frac14 13sod frac14 079 315 frac14 250mm

                                                            Total settlement

                                                            s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                            79

                                                            Without sand drains

                                                            Uv frac14 025

                                                            Tv frac14 0049 ethfrom Figure 718THORN

                                                            t frac14 Tvd2

                                                            cvfrac14 0049 82

                                                            cvWith sand drains

                                                            R frac14 0564S frac14 0564 3 frac14 169m

                                                            n frac14 Rrfrac14 169

                                                            015frac14 113

                                                            Tr frac14 cht

                                                            4R2frac14 ch

                                                            4 1692 0049 82

                                                            cvethand ch frac14 cvTHORN

                                                            frac14 0275

                                                            Ur frac14 073 (from Figure 730)

                                                            58 Consolidation theory

                                                            Using Equation 740

                                                            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                            U frac14 080

                                                            710

                                                            Without sand drains

                                                            Uv frac14 090

                                                            Tv frac14 0848

                                                            t frac14 Tvd2

                                                            cvfrac14 0848 102

                                                            96frac14 88 years

                                                            With sand drains

                                                            R frac14 0564S frac14 0564 4 frac14 226m

                                                            n frac14 Rrfrac14 226

                                                            015frac14 15

                                                            Tr

                                                            Tvfrac14 chcv

                                                            d2

                                                            4R2ethsame tTHORN

                                                            Tr

                                                            Tvfrac14 140

                                                            96 102

                                                            4 2262frac14 714 eth1THORN

                                                            Using Equation 740

                                                            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                            An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                            Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                            040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                            Thus

                                                            Uv frac14 0295 and Ur frac14 086

                                                            t frac14 88 00683

                                                            0848frac14 07 years

                                                            Consolidation theory 59

                                                            Chapter 8

                                                            Bearing capacity

                                                            81

                                                            (a) The ultimate bearing capacity is given by Equation 83

                                                            qf frac14 cNc thorn DNq thorn 1

                                                            2BN

                                                            For u frac14 0

                                                            Nc frac14 514 Nq frac14 1 N frac14 0

                                                            qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                            The net ultimate bearing capacity is

                                                            qnf frac14 qf D frac14 540 kN=m2

                                                            The net foundation pressure is

                                                            qn frac14 q D frac14 425

                                                            2 eth21 1THORN frac14 192 kN=m2

                                                            The factor of safety (Equation 86) is

                                                            F frac14 qnfqnfrac14 540

                                                            192frac14 28

                                                            (b) For 0 frac14 28

                                                            Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                            qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                            2 112 2 13

                                                            frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                            qnf frac14 574 112 frac14 563 kN=m2

                                                            F frac14 563

                                                            192frac14 29

                                                            (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                            82

                                                            For 0 frac14 38

                                                            Nq frac14 49 N frac14 67

                                                            qnf frac14 DethNq 1THORN thorn 1

                                                            2BN ethfrom Equation 83THORN

                                                            frac14 eth18 075 48THORN thorn 1

                                                            2 18 15 67

                                                            frac14 648thorn 905 frac14 1553 kN=m2

                                                            qn frac14 500

                                                            15 eth18 075THORN frac14 320 kN=m2

                                                            F frac14 qnfqnfrac14 1553

                                                            320frac14 48

                                                            0d frac14 tan1tan 38

                                                            125

                                                            frac14 32 therefore Nq frac14 23 and N frac14 25

                                                            Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                            2 18 15 25

                                                            frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                            Design load (action) Vd frac14 500 kN=m

                                                            The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                            83

                                                            D

                                                            Bfrac14 350

                                                            225frac14 155

                                                            From Figure 85 for a square foundation

                                                            Nc frac14 81

                                                            Bearing capacity 61

                                                            For a rectangular foundation (L frac14 450m B frac14 225m)

                                                            Nc frac14 084thorn 016B

                                                            L

                                                            81 frac14 745

                                                            Using Equation 810

                                                            qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                            For F frac14 3

                                                            qn frac14 1006

                                                            3frac14 335 kN=m2

                                                            q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                            Design load frac14 405 450 225 frac14 4100 kN

                                                            Design undrained strength cud frac14 135

                                                            14frac14 96 kN=m2

                                                            Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                            frac14 7241 kN

                                                            Design load Vd frac14 4100 kN

                                                            Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                            84

                                                            For 0 frac14 40

                                                            Nq frac14 64 N frac14 95

                                                            qnf frac14 DethNq 1THORN thorn 04BN

                                                            (a) Water table 5m below ground level

                                                            qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                            qn frac14 400 17 frac14 383 kN=m2

                                                            F frac14 2686

                                                            383frac14 70

                                                            (b) Water table 1m below ground level (ie at foundation level)

                                                            0 frac14 20 98 frac14 102 kN=m3

                                                            62 Bearing capacity

                                                            qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                            F frac14 2040

                                                            383frac14 53

                                                            (c) Water table at ground level with upward hydraulic gradient 02

                                                            eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                            qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                            F frac14 1296

                                                            392frac14 33

                                                            85

                                                            The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                            Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                            Design value of 0 frac14 tan1tan 39

                                                            125

                                                            frac14 33

                                                            For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                            Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                            Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                            86

                                                            (a) Undrained shear for u frac14 0

                                                            Nc frac14 514 Nq frac14 1 N frac14 0

                                                            qnf frac14 12cuNc

                                                            frac14 12 100 514 frac14 617 kN=m2

                                                            qn frac14 qnfFfrac14 617

                                                            3frac14 206 kN=m2

                                                            q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                            Bearing capacity 63

                                                            Drained shear for 0 frac14 32

                                                            Nq frac14 23 N frac14 25

                                                            0 frac14 21 98 frac14 112 kN=m3

                                                            qnf frac14 0DethNq 1THORN thorn 040BN

                                                            frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                            frac14 694 kN=m2

                                                            q frac14 694

                                                            3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                            Design load frac14 42 227 frac14 3632 kN

                                                            (b) Design undrained strength cud frac14 100

                                                            14frac14 71 kNm2

                                                            Design bearing resistance Rd frac14 12cudNe area

                                                            frac14 12 71 514 42

                                                            frac14 7007 kN

                                                            For drained shear 0d frac14 tan1tan 32

                                                            125

                                                            frac14 26

                                                            Nq frac14 12 N frac14 10

                                                            Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                            (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                            Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                            1 2 100 0175 0700qn 0182qn

                                                            2 6 033 0044 0176qn 0046qn

                                                            3 10 020 0017 0068qn 0018qn

                                                            0246qn

                                                            Diameter of equivalent circle B frac14 45m

                                                            H

                                                            Bfrac14 12

                                                            45frac14 27 and A frac14 042

                                                            13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                            64 Bearing capacity

                                                            For sc frac14 30mm

                                                            qn frac14 30

                                                            0147frac14 204 kN=m2

                                                            q frac14 204thorn 21 frac14 225 kN=m2

                                                            Design load frac14 42 225 frac14 3600 kN

                                                            The design load is 3600 kN settlement being the limiting criterion

                                                            87

                                                            D

                                                            Bfrac14 8

                                                            4frac14 20

                                                            From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                            F frac14 cuNc

                                                            Dfrac14 40 71

                                                            20 8frac14 18

                                                            88

                                                            Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                            Design value of 0 frac14 tan1tan 38

                                                            125

                                                            frac14 32

                                                            Figure Q86

                                                            Bearing capacity 65

                                                            For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                            Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                            The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                            Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                            For B frac14 250m qn frac14 3750

                                                            2502 17 frac14 583 kN=m2

                                                            From Figure 510 m frac14 n frac14 126

                                                            6frac14 021

                                                            Ir frac14 0019

                                                            Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                            Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                            The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                            89

                                                            Depth (m) N 0v (kNm2) CN N1

                                                            070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                            Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                            (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                            Cw frac14 05thorn 0530

                                                            47

                                                            frac14 082

                                                            66 Bearing capacity

                                                            Thus

                                                            qa frac14 150 082 frac14 120 kN=m2

                                                            (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                            Thus

                                                            qa frac14 90 15 frac14 135 kN=m2

                                                            (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                            Ic frac14 171

                                                            1014frac14 0068

                                                            From Equation 819(a) with s frac14 25mm

                                                            q frac14 25

                                                            3507 0068frac14 150 kN=m2

                                                            810

                                                            Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                            Izp frac14 05thorn 01130

                                                            16 27

                                                            05

                                                            frac14 067

                                                            Refer to Figure Q810

                                                            E frac14 25qc

                                                            Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                            Ez (mm3MN)

                                                            1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                            0203

                                                            C1 frac14 1 0500qnfrac14 1 05 12 16

                                                            130frac14 093

                                                            C2 frac14 1 ethsayTHORN

                                                            s frac14 C1C2qnX Iz

                                                            Ez frac14 093 1 130 0203 frac14 25mm

                                                            Bearing capacity 67

                                                            811

                                                            At pile base level

                                                            cu frac14 220 kN=m2

                                                            qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                            Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                            00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                            qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                            Then

                                                            Qf frac14 Abqb thorn Asqs

                                                            frac14

                                                            4 32 1980

                                                            thorn eth 105 139 86THORN

                                                            frac14 13 996thorn 3941 frac14 17 937 kN

                                                            0 01 02 03 04 05 06 07

                                                            0 2 4 6 8 10 12 14

                                                            1

                                                            2

                                                            3

                                                            4

                                                            5

                                                            6

                                                            7

                                                            8

                                                            (1)

                                                            (2)

                                                            (3)

                                                            (4)

                                                            (5)

                                                            qc

                                                            qc

                                                            Iz

                                                            Iz

                                                            (MNm2)

                                                            z (m)

                                                            Figure Q810

                                                            68 Bearing capacity

                                                            Allowable load

                                                            ethaTHORN Qf

                                                            2frac14 17 937

                                                            2frac14 8968 kN

                                                            ethbTHORN Abqb

                                                            3thorn Asqs frac14 13 996

                                                            3thorn 3941 frac14 8606 kN

                                                            ie allowable load frac14 8600 kN

                                                            Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                            According to the limit state method

                                                            Characteristic undrained strength at base level cuk frac14 220

                                                            150kN=m2

                                                            Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                            150frac14 1320 kN=m2

                                                            Characteristic shaft resistance qsk frac14 00150

                                                            frac14 86

                                                            150frac14 57 kN=m2

                                                            Characteristic base and shaft resistances

                                                            Rbk frac14

                                                            4 32 1320 frac14 9330 kN

                                                            Rsk frac14 105 139 86

                                                            150frac14 2629 kN

                                                            For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                            Design bearing resistance Rcd frac14 9330

                                                            160thorn 2629

                                                            130

                                                            frac14 5831thorn 2022

                                                            frac14 7850 kN

                                                            Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                            812

                                                            ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                            qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                            For a single pile

                                                            Qf frac14 Abqb thorn Asqs

                                                            frac14

                                                            4 062 1305

                                                            thorn eth 06 15 42THORN

                                                            frac14 369thorn 1187 frac14 1556 kN

                                                            Bearing capacity 69

                                                            Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                            eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                            (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                            (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                            qbkfrac14 9cuk frac14 9 220

                                                            150frac14 1320 kN=m2

                                                            qskfrac14cuk frac14 040 105

                                                            150frac14 28 kN=m2

                                                            Rbkfrac14

                                                            4 0602 1320 frac14 373 kN

                                                            Rskfrac14 060 15 28 frac14 791 kN

                                                            Rcdfrac14 373

                                                            160thorn 791

                                                            130frac14 233thorn 608 frac14 841 kN

                                                            Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                            Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                            (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                            q frac14 21 000

                                                            1762frac14 68 kN=m2

                                                            Immediate settlement

                                                            H

                                                            Bfrac14 15

                                                            176frac14 085

                                                            D

                                                            Bfrac14 13

                                                            176frac14 074

                                                            L

                                                            Bfrac14 1

                                                            Hence from Figure 515

                                                            130 frac14 078 and 131 frac14 041

                                                            70 Bearing capacity

                                                            Thus using Equation 528

                                                            si frac14 078 041 68 176

                                                            65frac14 6mm

                                                            Consolidation settlement

                                                            Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                            1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                            434 (sod)

                                                            Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                            sc frac14 056 434 frac14 24mm

                                                            The total settlement is (6thorn 24) frac14 30mm

                                                            813

                                                            At base level N frac14 26 Then using Equation 830

                                                            qb frac14 40NDb

                                                            Bfrac14 40 26 2

                                                            025frac14 8320 kN=m2

                                                            ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                            Figure Q812

                                                            Bearing capacity 71

                                                            Over the length embedded in sand

                                                            N frac14 21 ie18thorn 24

                                                            2

                                                            Using Equation 831

                                                            qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                            For a single pile

                                                            Qf frac14 Abqb thorn Asqs

                                                            frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                            For the pile group assuming a group efficiency of 12

                                                            XQf frac14 12 9 604 frac14 6523 kN

                                                            Then the load factor is

                                                            F frac14 6523

                                                            2000thorn 1000frac14 21

                                                            (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                            Characteristic base resistance per unit area qbk frac14 8320

                                                            150frac14 5547 kNm2

                                                            Characteristic shaft resistance per unit area qsk frac14 42

                                                            150frac14 28 kNm2

                                                            Characteristic base and shaft resistances for a single pile

                                                            Rbk frac14 0252 5547 frac14 347 kN

                                                            Rsk frac14 4 025 2 28 frac14 56 kN

                                                            For a driven pile the partial factors are b frac14 s frac14 130

                                                            Design bearing resistance Rcd frac14 347

                                                            130thorn 56

                                                            130frac14 310 kN

                                                            For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                            Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                            (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                            From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                            72 Bearing capacity

                                                            N frac14 24thorn 26thorn 34

                                                            3frac14 28

                                                            Ic frac14 171

                                                            2814frac14 0016 ethEquation 818THORN

                                                            s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                            814

                                                            Using Equation 841

                                                            Tf frac14 DLcu thorn

                                                            4ethD2 d2THORNcuNc

                                                            frac14 eth 02 5 06 110THORN thorn

                                                            4eth022 012THORN110 9

                                                            frac14 207thorn 23 frac14 230 kN

                                                            Figure Q813

                                                            Bearing capacity 73

                                                            Chapter 9

                                                            Stability of slopes

                                                            91

                                                            Referring to Figure Q91

                                                            W frac14 417 19 frac14 792 kN=m

                                                            Q frac14 20 28 frac14 56 kN=m

                                                            Arc lengthAB frac14

                                                            180 73 90 frac14 115m

                                                            Arc length BC frac14

                                                            180 28 90 frac14 44m

                                                            The factor of safety is given by

                                                            F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                            90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                            Depth of tension crack z0 frac14 2cu

                                                            frac14 2 20

                                                            19frac14 21m

                                                            Arc length BD frac14

                                                            180 13

                                                            1

                                                            2 90 frac14 21m

                                                            F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                            The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                            Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                            14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                            Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                            The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                            92

                                                            u frac14 0

                                                            Depth factor D frac14 11

                                                            9frac14 122

                                                            Using Equation 92 with F frac14 10

                                                            Ns frac14 cu

                                                            FHfrac14 30

                                                            10 19 9frac14 0175

                                                            Hence from Figure 93

                                                            frac14 50

                                                            For F frac14 12

                                                            Ns frac14 30

                                                            12 19 9frac14 0146

                                                            frac14 27

                                                            93

                                                            Refer to Figure Q93

                                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                            1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                            74 m

                                                            214 1deg

                                                            213 1deg

                                                            39 m

                                                            WB

                                                            D

                                                            C

                                                            28 m

                                                            21 m

                                                            A

                                                            Q

                                                            Soil (1)Soil (2)

                                                            73deg

                                                            Figure Q91

                                                            Stability of slopes 75

                                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                            9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                            599 256 328 1372

                                                            Figure Q93

                                                            76 Stability of slopes

                                                            XW cos frac14 b

                                                            Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                            W sin frac14 bX

                                                            h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                            Arc length La frac14

                                                            180 57

                                                            1

                                                            2 326 frac14 327m

                                                            The factor of safety is given by

                                                            F frac14 c0La thorn tan0ethW cos ulTHORN

                                                            W sin

                                                            frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                            frac14 091

                                                            According to the limit state method

                                                            0d frac14 tan1tan 32

                                                            125

                                                            frac14 265

                                                            c0 frac14 8

                                                            160frac14 5 kN=m2

                                                            Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                            Design disturbing moment frac14 1075 kN=m

                                                            The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                            94

                                                            F frac14 1

                                                            W sin

                                                            Xfc0bthorn ethW ubTHORN tan0g sec

                                                            1thorn ethtan tan0=FTHORN

                                                            c0 frac14 8 kN=m2

                                                            0 frac14 32

                                                            c0b frac14 8 2 frac14 16 kN=m

                                                            W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                            Try F frac14 100

                                                            tan0

                                                            Ffrac14 0625

                                                            Stability of slopes 77

                                                            Values of u are as obtained in Figure Q93

                                                            SliceNo

                                                            h(m)

                                                            W frac14 bh(kNm)

                                                            W sin(kNm)

                                                            ub(kNm)

                                                            c0bthorn (W ub) tan0(kNm)

                                                            sec

                                                            1thorn (tan tan0)FProduct(kNm)

                                                            1 05 21 6 2 8 24 1078 262 13 55 31

                                                            23 33 30 1042 31

                                                            3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                            224 92 72 0931 67

                                                            6 50 210 11 40 100 85 0907 777 55 231 14

                                                            12 58 112 90 0889 80

                                                            8 60 252 1812

                                                            80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                            10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                            2154 88 116 0853 99

                                                            14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                            1074 1091

                                                            F frac14 1091

                                                            1074frac14 102 (assumed value 100)

                                                            Thus

                                                            F frac14 101

                                                            95

                                                            F frac14 1

                                                            W sin

                                                            XfWeth1 ruTHORN tan0g sec

                                                            1thorn ethtan tan0THORN=F

                                                            0 frac14 33

                                                            ru frac14 020

                                                            W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                            eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                            Try F frac14 110

                                                            tan 0

                                                            Ffrac14 tan 33

                                                            110frac14 0590

                                                            78 Stability of slopes

                                                            Referring to Figure Q95

                                                            SliceNo

                                                            h(m)

                                                            W frac14 bh(kNm)

                                                            W sin(kNm)

                                                            W(1 ru) tan0(kNm)

                                                            sec

                                                            1thorn ( tan tan0)FProduct(kNm)

                                                            1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                            2120 234 0892 209

                                                            4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                            1185 1271

                                                            Figure Q95

                                                            Stability of slopes 79

                                                            F frac14 1271

                                                            1185frac14 107

                                                            The trial value was 110 therefore take F to be 108

                                                            96

                                                            (a) Water table at surface the factor of safety is given by Equation 912

                                                            F frac14 0

                                                            sat

                                                            tan0

                                                            tan

                                                            ptie 15 frac14 92

                                                            19

                                                            tan 36

                                                            tan

                                                            tan frac14 0234

                                                            frac14 13

                                                            Water table well below surface the factor of safety is given by Equation 911

                                                            F frac14 tan0

                                                            tan

                                                            frac14 tan 36

                                                            tan 13

                                                            frac14 31

                                                            (b) 0d frac14 tan1tan 36

                                                            125

                                                            frac14 30

                                                            Depth of potential failure surface frac14 z

                                                            Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                            frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                            frac14 504z kN

                                                            Design disturbing moment per unit area Sd frac14 sat sin cos

                                                            frac14 19 z sin 13 cos 13

                                                            frac14 416z kN

                                                            Rd gtSd therefore the limit state for overall stability is satisfied

                                                            80 Stability of slopes

                                                            • Book Cover
                                                            • Title
                                                            • Contents
                                                            • Basic characteristics of soils
                                                            • Seepage
                                                            • Effective stress
                                                            • Shear strength
                                                            • Stresses and displacements
                                                            • Lateral earth pressure
                                                            • Consolidation theory
                                                            • Bearing capacity
                                                            • Stability of slopes

                                                              47

                                                              The torque required to produce shear failure is given by

                                                              T frac14 dh cud

                                                              2thorn 2

                                                              Z d=2

                                                              0

                                                              2r drcur

                                                              frac14 cud2h

                                                              2thorn 4cu

                                                              Z d=2

                                                              0

                                                              r2dr

                                                              frac14 cud2h

                                                              2thorn d

                                                              3

                                                              6

                                                              Then

                                                              35 frac14 cu52 10

                                                              2thorn 53

                                                              6

                                                              103

                                                              cu frac14 76 kN=m3

                                                              400

                                                              0 400 800 1200 1600

                                                              τ (k

                                                              Nm

                                                              2 )

                                                              σprime (kNm2)

                                                              34deg

                                                              315deg29deg

                                                              (a)

                                                              (b)

                                                              0 400

                                                              400

                                                              800 1200 1600

                                                              Failure envelope

                                                              300 500

                                                              σprime (kNm2)

                                                              τ (k

                                                              Nm

                                                              2 )

                                                              20 (kNm2)

                                                              31deg

                                                              Figure Q46

                                                              Shear strength 25

                                                              48

                                                              The relevant stress values are calculated as follows

                                                              3 frac14 600 kN=m2

                                                              1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                                              2(1 3) 0 40 79 107 139 159

                                                              1

                                                              2(01 thorn 03) 400 411 402 389 351 326

                                                              1

                                                              2(1 thorn 3) 600 640 679 707 739 759

                                                              The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                                              Af frac14 433 200

                                                              319frac14 073

                                                              49

                                                              B frac14 u33

                                                              frac14 144

                                                              350 200frac14 096

                                                              a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                                              0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                                              10 283 65 023

                                                              Figure Q48

                                                              26 Shear strength

                                                              The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                                              Figure Q49

                                                              Shear strength 27

                                                              Chapter 5

                                                              Stresses and displacements

                                                              51

                                                              Vertical stress is given by

                                                              z frac14 Qz2Ip frac14 5000

                                                              52Ip

                                                              Values of Ip are obtained from Table 51

                                                              r (m) rz Ip z (kNm2)

                                                              0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                                              10 20 0009 2

                                                              The variation of z with radial distance (r) is plotted in Figure Q51

                                                              Figure Q51

                                                              52

                                                              Below the centre load (Figure Q52)

                                                              r

                                                              zfrac14 0 for the 7500-kN load

                                                              Ip frac14 0478

                                                              r

                                                              zfrac14 5

                                                              4frac14 125 for the 10 000- and 9000-kN loads

                                                              Ip frac14 0045

                                                              Then

                                                              z frac14X Q

                                                              z2Ip

                                                              frac14 7500 0478

                                                              42thorn 10 000 0045

                                                              42thorn 9000 0045

                                                              42

                                                              frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                                              53

                                                              The vertical stress under a corner of a rectangular area is given by

                                                              z frac14 qIr

                                                              where values of Ir are obtained from Figure 510 In this case

                                                              z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                                              z

                                                              Figure Q52

                                                              Stresses and displacements 29

                                                              z (m) m n Ir z (kNm2)

                                                              0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                                              10 010 0005 5

                                                              z is plotted against z in Figure Q53

                                                              54

                                                              (a)

                                                              m frac14 125

                                                              12frac14 104

                                                              n frac14 18

                                                              12frac14 150

                                                              From Figure 510 Irfrac14 0196

                                                              z frac14 2 175 0196 frac14 68 kN=m2

                                                              Figure Q53

                                                              30 Stresses and displacements

                                                              (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                                              z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                                              55

                                                              Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                                              Px frac14 2Q

                                                              1

                                                              m2 thorn 1frac14 2 150

                                                              125frac14 76 kN=m

                                                              Equation 517 is used to obtain the pressure distribution

                                                              px frac14 4Q

                                                              h

                                                              m2n

                                                              ethm2 thorn n2THORN2 frac14150

                                                              m2n

                                                              ethm2 thorn n2THORN2 ethkN=m2THORN

                                                              Figure Q54

                                                              Stresses and displacements 31

                                                              n m2n

                                                              (m2 thorn n2)2

                                                              px(kNm2)

                                                              0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                              The pressure distribution is plotted in Figure Q55

                                                              56

                                                              H

                                                              Bfrac14 10

                                                              2frac14 5

                                                              L

                                                              Bfrac14 4

                                                              2frac14 2

                                                              D

                                                              Bfrac14 1

                                                              2frac14 05

                                                              Hence from Figure 515

                                                              131 frac14 082

                                                              130 frac14 094

                                                              Figure Q55

                                                              32 Stresses and displacements

                                                              The immediate settlement is given by Equation 528

                                                              si frac14 130131qB

                                                              Eu

                                                              frac14 094 082 200 2

                                                              45frac14 7mm

                                                              Stresses and displacements 33

                                                              Chapter 6

                                                              Lateral earth pressure

                                                              61

                                                              For 0 frac14 37 the active pressure coefficient is given by

                                                              Ka frac14 1 sin 37

                                                              1thorn sin 37frac14 025

                                                              The total active thrust (Equation 66a with c0 frac14 0) is

                                                              Pa frac14 1

                                                              2KaH

                                                              2 frac14 1

                                                              2 025 17 62 frac14 765 kN=m

                                                              If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                              K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                              and the thrust on the wall is

                                                              P0 frac14 1

                                                              2K0H

                                                              2 frac14 1

                                                              2 040 17 62 frac14 122 kN=m

                                                              62

                                                              The active pressure coefficients for the three soil types are as follows

                                                              Ka1 frac141 sin 35

                                                              1thorn sin 35frac14 0271

                                                              Ka2 frac141 sin 27

                                                              1thorn sin 27frac14 0375

                                                              ffiffiffiffiffiffiffiKa2

                                                              p frac14 0613

                                                              Ka3 frac141 sin 42

                                                              1thorn sin 42frac14 0198

                                                              Distribution of active pressure (plotted in Figure Q62)

                                                              Depth (m) Soil Active pressure (kNm2)

                                                              3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                              12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                              At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                              Total thrust frac14 571 kNm

                                                              Point of application is (4893571) m from the top of the wall ie 857m

                                                              Force (kN) Arm (m) Moment (kN m)

                                                              (1)1

                                                              2 0271 16 32 frac14 195 20 390

                                                              (2) 0271 16 3 2 frac14 260 40 1040

                                                              (3)1

                                                              2 0271 92 22 frac14 50 433 217

                                                              (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                              (5)1

                                                              2 0375 102 32 frac14 172 70 1204

                                                              (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                              (7)1

                                                              2 0198 112 42 frac14 177 1067 1889

                                                              (8)1

                                                              2 98 92 frac14 3969 90 35721

                                                              5713 48934

                                                              Figure Q62

                                                              Lateral earth pressure 35

                                                              63

                                                              (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                              Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                              frac14 245

                                                              At the lower end of the piling

                                                              pa frac14 Kaqthorn Kasatz Kaccu

                                                              frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                              frac14 115 kN=m2

                                                              pp frac14 Kpsatzthorn Kpccu

                                                              frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                              frac14 202 kN=m2

                                                              (b) For 0 frac14 26 and frac14 1

                                                              20

                                                              Ka frac14 035

                                                              Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                              pfrac14 145 ethEquation 619THORN

                                                              Kp frac14 37

                                                              Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                              pfrac14 47 ethEquation 624THORN

                                                              At the lower end of the piling

                                                              pa frac14 Kaqthorn Ka0z Kacc

                                                              0

                                                              frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                              frac14 187 kN=m2

                                                              pp frac14 Kp0zthorn Kpcc

                                                              0

                                                              frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                              frac14 198 kN=m2

                                                              36 Lateral earth pressure

                                                              64

                                                              (a) For 0 frac14 38 Ka frac14 024

                                                              0 frac14 20 98 frac14 102 kN=m3

                                                              The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                              Force (kN) Arm (m) Moment (kN m)

                                                              (1) 024 10 66 frac14 159 33 525

                                                              (2)1

                                                              2 024 17 392 frac14 310 400 1240

                                                              (3) 024 17 39 27 frac14 430 135 580

                                                              (4)1

                                                              2 024 102 272 frac14 89 090 80

                                                              (5)1

                                                              2 98 272 frac14 357 090 321

                                                              Hfrac14 1345 MH frac14 2746

                                                              (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                              (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                              XM frac14MV MH frac14 7790 kNm

                                                              Lever arm of base resultant

                                                              M

                                                              Vfrac14 779

                                                              488frac14 160

                                                              Eccentricity of base resultant

                                                              e frac14 200 160 frac14 040m

                                                              39 m

                                                              27 m

                                                              40 m

                                                              04 m

                                                              04 m

                                                              26 m

                                                              (7)

                                                              (9)

                                                              (1)(2)

                                                              (3)

                                                              (4)

                                                              (5)

                                                              (8)(6)

                                                              (10)

                                                              WT

                                                              10 kNm2

                                                              Hydrostatic

                                                              Figure Q64

                                                              Lateral earth pressure 37

                                                              Base pressures (Equation 627)

                                                              p frac14 VB

                                                              1 6e

                                                              B

                                                              frac14 488

                                                              4eth1 060THORN

                                                              frac14 195 kN=m2 and 49 kN=m2

                                                              Factor of safety against sliding (Equation 628)

                                                              F frac14 V tan

                                                              Hfrac14 488 tan 25

                                                              1345frac14 17

                                                              (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                              Hfrac14 1633 kN

                                                              V frac14 4879 kN

                                                              MH frac14 3453 kNm

                                                              MV frac14 10536 kNm

                                                              The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                              65

                                                              For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                              Kp

                                                              Ffrac14 385

                                                              2

                                                              0 frac14 20 98 frac14 102 kN=m3

                                                              The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                              Force (kN) Arm (m) Moment (kN m)

                                                              (1)1

                                                              2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                              (2) 026 17 45 d frac14 199d d2 995d2

                                                              (3)1

                                                              2 026 102 d2 frac14 133d2 d3 044d3

                                                              (4)1

                                                              2 385

                                                              2 17 152 frac14 368 dthorn 05 368d 184

                                                              (5)385

                                                              2 17 15 d frac14 491d d2 2455d2

                                                              (6)1

                                                              2 385

                                                              2 102 d2 frac14 982d2 d3 327d3

                                                              38 Lateral earth pressure

                                                              XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                              d3 thorn 516d2 283d 1724 frac14 0

                                                              d frac14 179m

                                                              Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                              Over additional 20 embedded depth

                                                              pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                              Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                              66

                                                              The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                              Ka frac14 sin 69=sin 105

                                                              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                              pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                              26664

                                                              37775

                                                              2

                                                              frac14 050

                                                              The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                              Pa frac14 1

                                                              2 050 19 7502 frac14 267 kN=m

                                                              Figure Q65

                                                              Lateral earth pressure 39

                                                              Horizontal component

                                                              Ph frac14 267 cos 40 frac14 205 kN=m

                                                              Vertical component

                                                              Pv frac14 267 sin 40 frac14 172 kN=m

                                                              Consider moments about the toe of the wall (Figure Q66) (per m)

                                                              Force (kN) Arm (m) Moment (kN m)

                                                              (1)1

                                                              2 175 650 235 frac14 1337 258 345

                                                              (2) 050 650 235 frac14 764 175 134

                                                              (3)1

                                                              2 070 650 235 frac14 535 127 68

                                                              (4) 100 400 235 frac14 940 200 188

                                                              (5) 1

                                                              2 080 050 235 frac14 47 027 1

                                                              Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                              Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                              Lever arm of base resultant

                                                              M

                                                              Vfrac14 795

                                                              525frac14 151m

                                                              Eccentricity of base resultant

                                                              e frac14 200 151 frac14 049m

                                                              Figure Q66

                                                              40 Lateral earth pressure

                                                              Base pressures (Equation 627)

                                                              p frac14 525

                                                              41 6 049

                                                              4

                                                              frac14 228 kN=m2 and 35 kN=m2

                                                              The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                              The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                              The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                              67

                                                              For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                              Force (kN) Arm (m) Moment (kNm)

                                                              (1)1

                                                              2 027 17 52 frac14 574 183 1050

                                                              (2) 027 17 5 3 frac14 689 500 3445

                                                              (3)1

                                                              2 027 102 32 frac14 124 550 682

                                                              (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                              (5)1

                                                              2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                              (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                              (7) 1

                                                              2 267

                                                              2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                              (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                              2 d frac14 163d d2thorn 650 82d2 1060d

                                                              Tie rod force per m frac14 T 0 0

                                                              XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                              d3 thorn 77d2 269d 1438 frac14 0

                                                              d frac14 467m

                                                              Depth of penetration frac14 12d frac14 560m

                                                              Lateral earth pressure 41

                                                              Algebraic sum of forces for d frac14 467m isX

                                                              F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                              T frac14 905 kN=m

                                                              Force in each tie rod frac14 25T frac14 226 kN

                                                              68

                                                              (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                              0 frac14 21 98 frac14 112 kN=m3

                                                              The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                              uC frac14 150

                                                              165 15 98 frac14 134 kN=m2

                                                              The average seepage pressure is

                                                              j frac14 15

                                                              165 98 frac14 09 kN=m3

                                                              Hence

                                                              0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                              0 j frac14 112 09 frac14 103 kN=m3

                                                              Figure Q67

                                                              42 Lateral earth pressure

                                                              Consider moments about the anchor point A (per m)

                                                              Force (kN) Arm (m) Moment (kN m)

                                                              (1) 10 026 150 frac14 390 60 2340

                                                              (2)1

                                                              2 026 18 452 frac14 474 15 711

                                                              (3) 026 18 45 105 frac14 2211 825 18240

                                                              (4)1

                                                              2 026 121 1052 frac14 1734 100 17340

                                                              (5)1

                                                              2 134 15 frac14 101 40 404

                                                              (6) 134 30 frac14 402 60 2412

                                                              (7)1

                                                              2 134 60 frac14 402 95 3819

                                                              571 4527(8) Ppm

                                                              115 115PPm

                                                              XM frac14 0

                                                              Ppm frac144527

                                                              115frac14 394 kN=m

                                                              Available passive resistance

                                                              Pp frac14 1

                                                              2 385 103 62 frac14 714 kN=m

                                                              Factor of safety

                                                              Fp frac14 Pp

                                                              Ppm

                                                              frac14 714

                                                              394frac14 18

                                                              Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                              Figure Q68

                                                              Lateral earth pressure 43

                                                              (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                              Consider moments (per m) about the tie point A

                                                              Force (kN) Arm (m)

                                                              (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                              (2)1

                                                              2 033 18 452 frac14 601 15

                                                              (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                              (4)1

                                                              2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                              (5)1

                                                              2 134 15 frac14 101 40

                                                              (6) 134 30 frac14 402 60

                                                              (7)1

                                                              2 134 d frac14 67d d3thorn 75

                                                              (8) 1

                                                              2 30 103 d2 frac141545d2 2d3thorn 75

                                                              Moment (kN m)

                                                              (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                              XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                              d3 thorn 827d2 466d 1518 frac14 0

                                                              By trial

                                                              d frac14 544m

                                                              The minimum depth of embedment required is 544m

                                                              69

                                                              For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                              0 frac14 20 98 frac14 102 kN=m3

                                                              The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                              44 Lateral earth pressure

                                                              uC frac14 147

                                                              173 26 98 frac14 216 kN=m2

                                                              and the average seepage pressure around the wall is

                                                              j frac14 26

                                                              173 98 frac14 15 kN=m3

                                                              Consider moments about the prop (A) (per m)

                                                              Force (kN) Arm (m) Moment (kN m)

                                                              (1)1

                                                              2 03 17 272 frac14 186 020 37

                                                              (2) 03 17 27 53 frac14 730 335 2445

                                                              (3)1

                                                              2 03 (102thorn 15) 532 frac14 493 423 2085

                                                              (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                              (5)1

                                                              2 216 26 frac14 281 243 684

                                                              (6) 216 27 frac14 583 465 2712

                                                              (7)1

                                                              2 216 60 frac14 648 800 5184

                                                              3055(8)

                                                              1

                                                              2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                              Factor of safety

                                                              Fr frac14 6885

                                                              3055frac14 225

                                                              Figure Q69

                                                              Lateral earth pressure 45

                                                              610

                                                              For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                              p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                              Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                              Using the recommendations of Twine and Roscoe

                                                              p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                              Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                              611

                                                              frac14 18 kN=m3 0 frac14 34

                                                              H frac14 350m nH frac14 335m mH frac14 185m

                                                              Consider a trial value of F frac14 20 Refer to Figure 635

                                                              0m frac14 tan1tan 34

                                                              20

                                                              frac14 186

                                                              Then

                                                              frac14 45 thorn 0m2frac14 543

                                                              W frac14 1

                                                              2 18 3502 cot 543 frac14 792 kN=m

                                                              Figure Q610

                                                              46 Lateral earth pressure

                                                              P frac14 1

                                                              2 s 3352 frac14 561s kN=m

                                                              U frac14 1

                                                              2 98 1852 cosec 543 frac14 206 kN=m

                                                              Equations 630 and 631 then become

                                                              561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                              792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                              ie

                                                              561s 0616N 405 frac14 0

                                                              792 0857N thorn 563 frac14 0

                                                              N frac14 848

                                                              0857frac14 989 kN=m

                                                              Then

                                                              561s 609 405 frac14 0

                                                              s frac14 649

                                                              561frac14 116 kN=m3

                                                              The calculations for trial values of F of 20 15 and 10 are summarized below

                                                              F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                              20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                              s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                              Figure Q611

                                                              Lateral earth pressure 47

                                                              612

                                                              For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                              45 thorn 0

                                                              2frac14 63

                                                              For the retained material between the surface and a depth of 36m

                                                              Pa frac14 1

                                                              2 030 18 362 frac14 350 kN=m

                                                              Weight of reinforced fill between the surface and a depth of 36m is

                                                              Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                              eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                              Eccentricity of Rv

                                                              e frac14 263 250 frac14 013m

                                                              The average vertical stress at a depth of 36m is

                                                              z frac14 Rv

                                                              L 2efrac14 324

                                                              474frac14 68 kN=m2

                                                              (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                              Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                              Tensile stress in the element frac14 138 103

                                                              65 3frac14 71N=mm2

                                                              Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                              Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                              Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                              The weight of ABC is

                                                              W frac14 1

                                                              2 18 52 265 frac14 124 kN=m

                                                              From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                              48 Lateral earth pressure

                                                              (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                              Tp frac14 032 68 120 065 frac14 170 kN

                                                              Tr frac14 213 420

                                                              418frac14 214 kN

                                                              Again the tensile failure and slipping limit states are satisfied for this element

                                                              Figure Q612

                                                              Lateral earth pressure 49

                                                              Chapter 7

                                                              Consolidation theory

                                                              71

                                                              Total change in thickness

                                                              H frac14 782 602 frac14 180mm

                                                              Average thickness frac14 1530thorn 180

                                                              2frac14 1620mm

                                                              Length of drainage path d frac14 1620

                                                              2frac14 810mm

                                                              Root time plot (Figure Q71a)

                                                              ffiffiffiffiffiffit90p frac14 33

                                                              t90 frac14 109min

                                                              cv frac14 0848d2

                                                              t90frac14 0848 8102

                                                              109 1440 365

                                                              106frac14 27m2=year

                                                              r0 frac14 782 764

                                                              782 602frac14 018

                                                              180frac14 0100

                                                              rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                              10 119

                                                              9 180frac14 0735

                                                              rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                              Log time plot (Figure Q71b)

                                                              t50 frac14 26min

                                                              cv frac14 0196d2

                                                              t50frac14 0196 8102

                                                              26 1440 365

                                                              106frac14 26m2=year

                                                              r0 frac14 782 763

                                                              782 602frac14 019

                                                              180frac14 0106

                                                              rp frac14 763 623

                                                              782 602frac14 140

                                                              180frac14 0778

                                                              rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                              Figure Q71(a)

                                                              Figure Q71(b)

                                                              Final void ratio

                                                              e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                              e

                                                              Hfrac14 1thorn e0

                                                              H0frac14 1thorn e1 thorne

                                                              H0

                                                              ie

                                                              e

                                                              180frac14 1631thorne

                                                              1710

                                                              e frac14 2936

                                                              1530frac14 0192

                                                              Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                              mv frac14 1

                                                              1thorn e0 e0 e101 00

                                                              frac14 1

                                                              1823 0192

                                                              0107frac14 098m2=MN

                                                              k frac14 cvmvw frac14 265 098 98

                                                              60 1440 365 103frac14 81 1010 m=s

                                                              72

                                                              Using Equation 77 (one-dimensional method)

                                                              sc frac14 e0 e11thorn e0 H

                                                              Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                              Figure Q72

                                                              52 Consolidation theory

                                                              Settlement

                                                              Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                              1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                              318

                                                              Notes 5 92y 460thorn 84

                                                              Heave

                                                              Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                              1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                              38

                                                              73

                                                              U frac14 f ethTvTHORN frac14 f cvt

                                                              d2

                                                              Hence if cv is constant

                                                              t1

                                                              t2frac14 d

                                                              21

                                                              d22

                                                              where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                              d1 frac14 95mm and d2 frac14 2500mm

                                                              for U frac14 050 t2 frac14 t1 d22

                                                              d21

                                                              frac14 20

                                                              60 24 365 25002

                                                              952frac14 263 years

                                                              for U lt 060 Tv frac14

                                                              4U2 (Equation 724(a))

                                                              t030 frac14 t050 0302

                                                              0502

                                                              frac14 263 036 frac14 095 years

                                                              Consolidation theory 53

                                                              74

                                                              The layer is open

                                                              d frac14 8

                                                              2frac14 4m

                                                              Tv frac14 cvtd2frac14 24 3

                                                              42frac14 0450

                                                              ui frac14 frac14 84 kN=m2

                                                              The excess pore water pressure is given by Equation 721

                                                              ue frac14Xmfrac141mfrac140

                                                              2ui

                                                              Msin

                                                              Mz

                                                              d

                                                              expethM2TvTHORN

                                                              In this case z frac14 d

                                                              sinMz

                                                              d

                                                              frac14 sinM

                                                              where

                                                              M frac14

                                                              23

                                                              25

                                                              2

                                                              M sin M M2Tv exp (M2Tv)

                                                              2thorn1 1110 0329

                                                              3

                                                              21 9993 457 105

                                                              ue frac14 2 84 2

                                                              1 0329 ethother terms negligibleTHORN

                                                              frac14 352 kN=m2

                                                              75

                                                              The layer is open

                                                              d frac14 6

                                                              2frac14 3m

                                                              Tv frac14 cvtd2frac14 10 3

                                                              32frac14 0333

                                                              The layer thickness will be divided into six equal parts ie m frac14 6

                                                              54 Consolidation theory

                                                              For an open layer

                                                              Tv frac14 4n

                                                              m2

                                                              n frac14 0333 62

                                                              4frac14 300

                                                              The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                              ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                              i j

                                                              0 1 2 3 4 5 6 7 8 9 10 11 12

                                                              0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                              The initial and 3-year isochrones are plotted in Figure Q75

                                                              Area under initial isochrone frac14 180 units

                                                              Area under 3-year isochrone frac14 63 units

                                                              The average degree of consolidation is given by Equation 725Thus

                                                              U frac14 1 63

                                                              180frac14 065

                                                              Figure Q75

                                                              Consolidation theory 55

                                                              76

                                                              At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                              0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                              The final consolidation settlement (one-dimensional method) is

                                                              sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                              Corrected time t frac14 2 1

                                                              2

                                                              40

                                                              52

                                                              frac14 1615 years

                                                              Tv frac14 cvtd2frac14 44 1615

                                                              42frac14 0444

                                                              From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                              77

                                                              The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                              Figure Q77

                                                              56 Consolidation theory

                                                              Point m n Ir (kNm2) sc (mm)

                                                              13020frac14 15 20

                                                              20frac14 10 0194 (4) 113 124

                                                              260

                                                              20frac14 30

                                                              20

                                                              20frac14 10 0204 (2) 59 65

                                                              360

                                                              20frac14 30

                                                              40

                                                              20frac14 20 0238 (1) 35 38

                                                              430

                                                              20frac14 15

                                                              40

                                                              20frac14 20 0224 (2) 65 72

                                                              Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                              78

                                                              Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                              (a) Immediate settlement

                                                              H

                                                              Bfrac14 30

                                                              35frac14 086

                                                              D

                                                              Bfrac14 2

                                                              35frac14 006

                                                              Figure Q78

                                                              Consolidation theory 57

                                                              From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                              si frac14 130131qB

                                                              Eufrac14 10 032 105 35

                                                              40frac14 30mm

                                                              (b) Consolidation settlement

                                                              Layer z (m) Dz Ic (kNm2) syod (mm)

                                                              1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                              3150

                                                              Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                              Now

                                                              H

                                                              Bfrac14 30

                                                              35frac14 086 and A frac14 065

                                                              from Figure 712 13 frac14 079

                                                              sc frac14 13sod frac14 079 315 frac14 250mm

                                                              Total settlement

                                                              s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                              79

                                                              Without sand drains

                                                              Uv frac14 025

                                                              Tv frac14 0049 ethfrom Figure 718THORN

                                                              t frac14 Tvd2

                                                              cvfrac14 0049 82

                                                              cvWith sand drains

                                                              R frac14 0564S frac14 0564 3 frac14 169m

                                                              n frac14 Rrfrac14 169

                                                              015frac14 113

                                                              Tr frac14 cht

                                                              4R2frac14 ch

                                                              4 1692 0049 82

                                                              cvethand ch frac14 cvTHORN

                                                              frac14 0275

                                                              Ur frac14 073 (from Figure 730)

                                                              58 Consolidation theory

                                                              Using Equation 740

                                                              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                              U frac14 080

                                                              710

                                                              Without sand drains

                                                              Uv frac14 090

                                                              Tv frac14 0848

                                                              t frac14 Tvd2

                                                              cvfrac14 0848 102

                                                              96frac14 88 years

                                                              With sand drains

                                                              R frac14 0564S frac14 0564 4 frac14 226m

                                                              n frac14 Rrfrac14 226

                                                              015frac14 15

                                                              Tr

                                                              Tvfrac14 chcv

                                                              d2

                                                              4R2ethsame tTHORN

                                                              Tr

                                                              Tvfrac14 140

                                                              96 102

                                                              4 2262frac14 714 eth1THORN

                                                              Using Equation 740

                                                              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                              An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                              Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                              040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                              Thus

                                                              Uv frac14 0295 and Ur frac14 086

                                                              t frac14 88 00683

                                                              0848frac14 07 years

                                                              Consolidation theory 59

                                                              Chapter 8

                                                              Bearing capacity

                                                              81

                                                              (a) The ultimate bearing capacity is given by Equation 83

                                                              qf frac14 cNc thorn DNq thorn 1

                                                              2BN

                                                              For u frac14 0

                                                              Nc frac14 514 Nq frac14 1 N frac14 0

                                                              qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                              The net ultimate bearing capacity is

                                                              qnf frac14 qf D frac14 540 kN=m2

                                                              The net foundation pressure is

                                                              qn frac14 q D frac14 425

                                                              2 eth21 1THORN frac14 192 kN=m2

                                                              The factor of safety (Equation 86) is

                                                              F frac14 qnfqnfrac14 540

                                                              192frac14 28

                                                              (b) For 0 frac14 28

                                                              Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                              qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                              2 112 2 13

                                                              frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                              qnf frac14 574 112 frac14 563 kN=m2

                                                              F frac14 563

                                                              192frac14 29

                                                              (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                              82

                                                              For 0 frac14 38

                                                              Nq frac14 49 N frac14 67

                                                              qnf frac14 DethNq 1THORN thorn 1

                                                              2BN ethfrom Equation 83THORN

                                                              frac14 eth18 075 48THORN thorn 1

                                                              2 18 15 67

                                                              frac14 648thorn 905 frac14 1553 kN=m2

                                                              qn frac14 500

                                                              15 eth18 075THORN frac14 320 kN=m2

                                                              F frac14 qnfqnfrac14 1553

                                                              320frac14 48

                                                              0d frac14 tan1tan 38

                                                              125

                                                              frac14 32 therefore Nq frac14 23 and N frac14 25

                                                              Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                              2 18 15 25

                                                              frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                              Design load (action) Vd frac14 500 kN=m

                                                              The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                              83

                                                              D

                                                              Bfrac14 350

                                                              225frac14 155

                                                              From Figure 85 for a square foundation

                                                              Nc frac14 81

                                                              Bearing capacity 61

                                                              For a rectangular foundation (L frac14 450m B frac14 225m)

                                                              Nc frac14 084thorn 016B

                                                              L

                                                              81 frac14 745

                                                              Using Equation 810

                                                              qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                              For F frac14 3

                                                              qn frac14 1006

                                                              3frac14 335 kN=m2

                                                              q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                              Design load frac14 405 450 225 frac14 4100 kN

                                                              Design undrained strength cud frac14 135

                                                              14frac14 96 kN=m2

                                                              Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                              frac14 7241 kN

                                                              Design load Vd frac14 4100 kN

                                                              Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                              84

                                                              For 0 frac14 40

                                                              Nq frac14 64 N frac14 95

                                                              qnf frac14 DethNq 1THORN thorn 04BN

                                                              (a) Water table 5m below ground level

                                                              qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                              qn frac14 400 17 frac14 383 kN=m2

                                                              F frac14 2686

                                                              383frac14 70

                                                              (b) Water table 1m below ground level (ie at foundation level)

                                                              0 frac14 20 98 frac14 102 kN=m3

                                                              62 Bearing capacity

                                                              qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                              F frac14 2040

                                                              383frac14 53

                                                              (c) Water table at ground level with upward hydraulic gradient 02

                                                              eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                              qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                              F frac14 1296

                                                              392frac14 33

                                                              85

                                                              The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                              Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                              Design value of 0 frac14 tan1tan 39

                                                              125

                                                              frac14 33

                                                              For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                              Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                              Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                              86

                                                              (a) Undrained shear for u frac14 0

                                                              Nc frac14 514 Nq frac14 1 N frac14 0

                                                              qnf frac14 12cuNc

                                                              frac14 12 100 514 frac14 617 kN=m2

                                                              qn frac14 qnfFfrac14 617

                                                              3frac14 206 kN=m2

                                                              q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                              Bearing capacity 63

                                                              Drained shear for 0 frac14 32

                                                              Nq frac14 23 N frac14 25

                                                              0 frac14 21 98 frac14 112 kN=m3

                                                              qnf frac14 0DethNq 1THORN thorn 040BN

                                                              frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                              frac14 694 kN=m2

                                                              q frac14 694

                                                              3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                              Design load frac14 42 227 frac14 3632 kN

                                                              (b) Design undrained strength cud frac14 100

                                                              14frac14 71 kNm2

                                                              Design bearing resistance Rd frac14 12cudNe area

                                                              frac14 12 71 514 42

                                                              frac14 7007 kN

                                                              For drained shear 0d frac14 tan1tan 32

                                                              125

                                                              frac14 26

                                                              Nq frac14 12 N frac14 10

                                                              Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                              (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                              Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                              1 2 100 0175 0700qn 0182qn

                                                              2 6 033 0044 0176qn 0046qn

                                                              3 10 020 0017 0068qn 0018qn

                                                              0246qn

                                                              Diameter of equivalent circle B frac14 45m

                                                              H

                                                              Bfrac14 12

                                                              45frac14 27 and A frac14 042

                                                              13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                              64 Bearing capacity

                                                              For sc frac14 30mm

                                                              qn frac14 30

                                                              0147frac14 204 kN=m2

                                                              q frac14 204thorn 21 frac14 225 kN=m2

                                                              Design load frac14 42 225 frac14 3600 kN

                                                              The design load is 3600 kN settlement being the limiting criterion

                                                              87

                                                              D

                                                              Bfrac14 8

                                                              4frac14 20

                                                              From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                              F frac14 cuNc

                                                              Dfrac14 40 71

                                                              20 8frac14 18

                                                              88

                                                              Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                              Design value of 0 frac14 tan1tan 38

                                                              125

                                                              frac14 32

                                                              Figure Q86

                                                              Bearing capacity 65

                                                              For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                              Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                              The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                              Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                              For B frac14 250m qn frac14 3750

                                                              2502 17 frac14 583 kN=m2

                                                              From Figure 510 m frac14 n frac14 126

                                                              6frac14 021

                                                              Ir frac14 0019

                                                              Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                              Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                              The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                              89

                                                              Depth (m) N 0v (kNm2) CN N1

                                                              070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                              Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                              (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                              Cw frac14 05thorn 0530

                                                              47

                                                              frac14 082

                                                              66 Bearing capacity

                                                              Thus

                                                              qa frac14 150 082 frac14 120 kN=m2

                                                              (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                              Thus

                                                              qa frac14 90 15 frac14 135 kN=m2

                                                              (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                              Ic frac14 171

                                                              1014frac14 0068

                                                              From Equation 819(a) with s frac14 25mm

                                                              q frac14 25

                                                              3507 0068frac14 150 kN=m2

                                                              810

                                                              Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                              Izp frac14 05thorn 01130

                                                              16 27

                                                              05

                                                              frac14 067

                                                              Refer to Figure Q810

                                                              E frac14 25qc

                                                              Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                              Ez (mm3MN)

                                                              1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                              0203

                                                              C1 frac14 1 0500qnfrac14 1 05 12 16

                                                              130frac14 093

                                                              C2 frac14 1 ethsayTHORN

                                                              s frac14 C1C2qnX Iz

                                                              Ez frac14 093 1 130 0203 frac14 25mm

                                                              Bearing capacity 67

                                                              811

                                                              At pile base level

                                                              cu frac14 220 kN=m2

                                                              qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                              Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                              00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                              qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                              Then

                                                              Qf frac14 Abqb thorn Asqs

                                                              frac14

                                                              4 32 1980

                                                              thorn eth 105 139 86THORN

                                                              frac14 13 996thorn 3941 frac14 17 937 kN

                                                              0 01 02 03 04 05 06 07

                                                              0 2 4 6 8 10 12 14

                                                              1

                                                              2

                                                              3

                                                              4

                                                              5

                                                              6

                                                              7

                                                              8

                                                              (1)

                                                              (2)

                                                              (3)

                                                              (4)

                                                              (5)

                                                              qc

                                                              qc

                                                              Iz

                                                              Iz

                                                              (MNm2)

                                                              z (m)

                                                              Figure Q810

                                                              68 Bearing capacity

                                                              Allowable load

                                                              ethaTHORN Qf

                                                              2frac14 17 937

                                                              2frac14 8968 kN

                                                              ethbTHORN Abqb

                                                              3thorn Asqs frac14 13 996

                                                              3thorn 3941 frac14 8606 kN

                                                              ie allowable load frac14 8600 kN

                                                              Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                              According to the limit state method

                                                              Characteristic undrained strength at base level cuk frac14 220

                                                              150kN=m2

                                                              Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                              150frac14 1320 kN=m2

                                                              Characteristic shaft resistance qsk frac14 00150

                                                              frac14 86

                                                              150frac14 57 kN=m2

                                                              Characteristic base and shaft resistances

                                                              Rbk frac14

                                                              4 32 1320 frac14 9330 kN

                                                              Rsk frac14 105 139 86

                                                              150frac14 2629 kN

                                                              For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                              Design bearing resistance Rcd frac14 9330

                                                              160thorn 2629

                                                              130

                                                              frac14 5831thorn 2022

                                                              frac14 7850 kN

                                                              Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                              812

                                                              ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                              qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                              For a single pile

                                                              Qf frac14 Abqb thorn Asqs

                                                              frac14

                                                              4 062 1305

                                                              thorn eth 06 15 42THORN

                                                              frac14 369thorn 1187 frac14 1556 kN

                                                              Bearing capacity 69

                                                              Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                              eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                              (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                              (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                              qbkfrac14 9cuk frac14 9 220

                                                              150frac14 1320 kN=m2

                                                              qskfrac14cuk frac14 040 105

                                                              150frac14 28 kN=m2

                                                              Rbkfrac14

                                                              4 0602 1320 frac14 373 kN

                                                              Rskfrac14 060 15 28 frac14 791 kN

                                                              Rcdfrac14 373

                                                              160thorn 791

                                                              130frac14 233thorn 608 frac14 841 kN

                                                              Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                              Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                              (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                              q frac14 21 000

                                                              1762frac14 68 kN=m2

                                                              Immediate settlement

                                                              H

                                                              Bfrac14 15

                                                              176frac14 085

                                                              D

                                                              Bfrac14 13

                                                              176frac14 074

                                                              L

                                                              Bfrac14 1

                                                              Hence from Figure 515

                                                              130 frac14 078 and 131 frac14 041

                                                              70 Bearing capacity

                                                              Thus using Equation 528

                                                              si frac14 078 041 68 176

                                                              65frac14 6mm

                                                              Consolidation settlement

                                                              Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                              1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                              434 (sod)

                                                              Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                              sc frac14 056 434 frac14 24mm

                                                              The total settlement is (6thorn 24) frac14 30mm

                                                              813

                                                              At base level N frac14 26 Then using Equation 830

                                                              qb frac14 40NDb

                                                              Bfrac14 40 26 2

                                                              025frac14 8320 kN=m2

                                                              ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                              Figure Q812

                                                              Bearing capacity 71

                                                              Over the length embedded in sand

                                                              N frac14 21 ie18thorn 24

                                                              2

                                                              Using Equation 831

                                                              qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                              For a single pile

                                                              Qf frac14 Abqb thorn Asqs

                                                              frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                              For the pile group assuming a group efficiency of 12

                                                              XQf frac14 12 9 604 frac14 6523 kN

                                                              Then the load factor is

                                                              F frac14 6523

                                                              2000thorn 1000frac14 21

                                                              (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                              Characteristic base resistance per unit area qbk frac14 8320

                                                              150frac14 5547 kNm2

                                                              Characteristic shaft resistance per unit area qsk frac14 42

                                                              150frac14 28 kNm2

                                                              Characteristic base and shaft resistances for a single pile

                                                              Rbk frac14 0252 5547 frac14 347 kN

                                                              Rsk frac14 4 025 2 28 frac14 56 kN

                                                              For a driven pile the partial factors are b frac14 s frac14 130

                                                              Design bearing resistance Rcd frac14 347

                                                              130thorn 56

                                                              130frac14 310 kN

                                                              For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                              Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                              (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                              From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                              72 Bearing capacity

                                                              N frac14 24thorn 26thorn 34

                                                              3frac14 28

                                                              Ic frac14 171

                                                              2814frac14 0016 ethEquation 818THORN

                                                              s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                              814

                                                              Using Equation 841

                                                              Tf frac14 DLcu thorn

                                                              4ethD2 d2THORNcuNc

                                                              frac14 eth 02 5 06 110THORN thorn

                                                              4eth022 012THORN110 9

                                                              frac14 207thorn 23 frac14 230 kN

                                                              Figure Q813

                                                              Bearing capacity 73

                                                              Chapter 9

                                                              Stability of slopes

                                                              91

                                                              Referring to Figure Q91

                                                              W frac14 417 19 frac14 792 kN=m

                                                              Q frac14 20 28 frac14 56 kN=m

                                                              Arc lengthAB frac14

                                                              180 73 90 frac14 115m

                                                              Arc length BC frac14

                                                              180 28 90 frac14 44m

                                                              The factor of safety is given by

                                                              F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                              90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                              Depth of tension crack z0 frac14 2cu

                                                              frac14 2 20

                                                              19frac14 21m

                                                              Arc length BD frac14

                                                              180 13

                                                              1

                                                              2 90 frac14 21m

                                                              F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                              The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                              Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                              14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                              Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                              The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                              92

                                                              u frac14 0

                                                              Depth factor D frac14 11

                                                              9frac14 122

                                                              Using Equation 92 with F frac14 10

                                                              Ns frac14 cu

                                                              FHfrac14 30

                                                              10 19 9frac14 0175

                                                              Hence from Figure 93

                                                              frac14 50

                                                              For F frac14 12

                                                              Ns frac14 30

                                                              12 19 9frac14 0146

                                                              frac14 27

                                                              93

                                                              Refer to Figure Q93

                                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                              1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                              74 m

                                                              214 1deg

                                                              213 1deg

                                                              39 m

                                                              WB

                                                              D

                                                              C

                                                              28 m

                                                              21 m

                                                              A

                                                              Q

                                                              Soil (1)Soil (2)

                                                              73deg

                                                              Figure Q91

                                                              Stability of slopes 75

                                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                              9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                              599 256 328 1372

                                                              Figure Q93

                                                              76 Stability of slopes

                                                              XW cos frac14 b

                                                              Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                              W sin frac14 bX

                                                              h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                              Arc length La frac14

                                                              180 57

                                                              1

                                                              2 326 frac14 327m

                                                              The factor of safety is given by

                                                              F frac14 c0La thorn tan0ethW cos ulTHORN

                                                              W sin

                                                              frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                              frac14 091

                                                              According to the limit state method

                                                              0d frac14 tan1tan 32

                                                              125

                                                              frac14 265

                                                              c0 frac14 8

                                                              160frac14 5 kN=m2

                                                              Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                              Design disturbing moment frac14 1075 kN=m

                                                              The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                              94

                                                              F frac14 1

                                                              W sin

                                                              Xfc0bthorn ethW ubTHORN tan0g sec

                                                              1thorn ethtan tan0=FTHORN

                                                              c0 frac14 8 kN=m2

                                                              0 frac14 32

                                                              c0b frac14 8 2 frac14 16 kN=m

                                                              W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                              Try F frac14 100

                                                              tan0

                                                              Ffrac14 0625

                                                              Stability of slopes 77

                                                              Values of u are as obtained in Figure Q93

                                                              SliceNo

                                                              h(m)

                                                              W frac14 bh(kNm)

                                                              W sin(kNm)

                                                              ub(kNm)

                                                              c0bthorn (W ub) tan0(kNm)

                                                              sec

                                                              1thorn (tan tan0)FProduct(kNm)

                                                              1 05 21 6 2 8 24 1078 262 13 55 31

                                                              23 33 30 1042 31

                                                              3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                              224 92 72 0931 67

                                                              6 50 210 11 40 100 85 0907 777 55 231 14

                                                              12 58 112 90 0889 80

                                                              8 60 252 1812

                                                              80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                              10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                              2154 88 116 0853 99

                                                              14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                              1074 1091

                                                              F frac14 1091

                                                              1074frac14 102 (assumed value 100)

                                                              Thus

                                                              F frac14 101

                                                              95

                                                              F frac14 1

                                                              W sin

                                                              XfWeth1 ruTHORN tan0g sec

                                                              1thorn ethtan tan0THORN=F

                                                              0 frac14 33

                                                              ru frac14 020

                                                              W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                              eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                              Try F frac14 110

                                                              tan 0

                                                              Ffrac14 tan 33

                                                              110frac14 0590

                                                              78 Stability of slopes

                                                              Referring to Figure Q95

                                                              SliceNo

                                                              h(m)

                                                              W frac14 bh(kNm)

                                                              W sin(kNm)

                                                              W(1 ru) tan0(kNm)

                                                              sec

                                                              1thorn ( tan tan0)FProduct(kNm)

                                                              1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                              2120 234 0892 209

                                                              4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                              1185 1271

                                                              Figure Q95

                                                              Stability of slopes 79

                                                              F frac14 1271

                                                              1185frac14 107

                                                              The trial value was 110 therefore take F to be 108

                                                              96

                                                              (a) Water table at surface the factor of safety is given by Equation 912

                                                              F frac14 0

                                                              sat

                                                              tan0

                                                              tan

                                                              ptie 15 frac14 92

                                                              19

                                                              tan 36

                                                              tan

                                                              tan frac14 0234

                                                              frac14 13

                                                              Water table well below surface the factor of safety is given by Equation 911

                                                              F frac14 tan0

                                                              tan

                                                              frac14 tan 36

                                                              tan 13

                                                              frac14 31

                                                              (b) 0d frac14 tan1tan 36

                                                              125

                                                              frac14 30

                                                              Depth of potential failure surface frac14 z

                                                              Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                              frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                              frac14 504z kN

                                                              Design disturbing moment per unit area Sd frac14 sat sin cos

                                                              frac14 19 z sin 13 cos 13

                                                              frac14 416z kN

                                                              Rd gtSd therefore the limit state for overall stability is satisfied

                                                              80 Stability of slopes

                                                              • Book Cover
                                                              • Title
                                                              • Contents
                                                              • Basic characteristics of soils
                                                              • Seepage
                                                              • Effective stress
                                                              • Shear strength
                                                              • Stresses and displacements
                                                              • Lateral earth pressure
                                                              • Consolidation theory
                                                              • Bearing capacity
                                                              • Stability of slopes

                                                                48

                                                                The relevant stress values are calculated as follows

                                                                3 frac14 600 kN=m2

                                                                1 3 0 80 158 214 279 3191 600 680 758 814 879 919u 200 229 277 318 388 43301 400 451 481 496 491 48603 400 371 323 282 212 1671

                                                                2(1 3) 0 40 79 107 139 159

                                                                1

                                                                2(01 thorn 03) 400 411 402 389 351 326

                                                                1

                                                                2(1 thorn 3) 600 640 679 707 739 759

                                                                The stress paths are plotted in Figure Q48 The initial points on the effective and totalstress paths are separated by the value of the back pressure (usfrac14 200 kNm2)

                                                                Af frac14 433 200

                                                                319frac14 073

                                                                49

                                                                B frac14 u33

                                                                frac14 144

                                                                350 200frac14 096

                                                                a () 1 frac14 1 3 (kNm2) u1 (kNm2) A frac14 u1=1

                                                                0 0 0 ndash2 201 100 0504 252 96 0386 275 78 0288 282 68 024

                                                                10 283 65 023

                                                                Figure Q48

                                                                26 Shear strength

                                                                The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                                                Figure Q49

                                                                Shear strength 27

                                                                Chapter 5

                                                                Stresses and displacements

                                                                51

                                                                Vertical stress is given by

                                                                z frac14 Qz2Ip frac14 5000

                                                                52Ip

                                                                Values of Ip are obtained from Table 51

                                                                r (m) rz Ip z (kNm2)

                                                                0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                                                10 20 0009 2

                                                                The variation of z with radial distance (r) is plotted in Figure Q51

                                                                Figure Q51

                                                                52

                                                                Below the centre load (Figure Q52)

                                                                r

                                                                zfrac14 0 for the 7500-kN load

                                                                Ip frac14 0478

                                                                r

                                                                zfrac14 5

                                                                4frac14 125 for the 10 000- and 9000-kN loads

                                                                Ip frac14 0045

                                                                Then

                                                                z frac14X Q

                                                                z2Ip

                                                                frac14 7500 0478

                                                                42thorn 10 000 0045

                                                                42thorn 9000 0045

                                                                42

                                                                frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                                                53

                                                                The vertical stress under a corner of a rectangular area is given by

                                                                z frac14 qIr

                                                                where values of Ir are obtained from Figure 510 In this case

                                                                z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                                                z

                                                                Figure Q52

                                                                Stresses and displacements 29

                                                                z (m) m n Ir z (kNm2)

                                                                0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                                                10 010 0005 5

                                                                z is plotted against z in Figure Q53

                                                                54

                                                                (a)

                                                                m frac14 125

                                                                12frac14 104

                                                                n frac14 18

                                                                12frac14 150

                                                                From Figure 510 Irfrac14 0196

                                                                z frac14 2 175 0196 frac14 68 kN=m2

                                                                Figure Q53

                                                                30 Stresses and displacements

                                                                (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                                                z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                                                55

                                                                Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                                                Px frac14 2Q

                                                                1

                                                                m2 thorn 1frac14 2 150

                                                                125frac14 76 kN=m

                                                                Equation 517 is used to obtain the pressure distribution

                                                                px frac14 4Q

                                                                h

                                                                m2n

                                                                ethm2 thorn n2THORN2 frac14150

                                                                m2n

                                                                ethm2 thorn n2THORN2 ethkN=m2THORN

                                                                Figure Q54

                                                                Stresses and displacements 31

                                                                n m2n

                                                                (m2 thorn n2)2

                                                                px(kNm2)

                                                                0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                                The pressure distribution is plotted in Figure Q55

                                                                56

                                                                H

                                                                Bfrac14 10

                                                                2frac14 5

                                                                L

                                                                Bfrac14 4

                                                                2frac14 2

                                                                D

                                                                Bfrac14 1

                                                                2frac14 05

                                                                Hence from Figure 515

                                                                131 frac14 082

                                                                130 frac14 094

                                                                Figure Q55

                                                                32 Stresses and displacements

                                                                The immediate settlement is given by Equation 528

                                                                si frac14 130131qB

                                                                Eu

                                                                frac14 094 082 200 2

                                                                45frac14 7mm

                                                                Stresses and displacements 33

                                                                Chapter 6

                                                                Lateral earth pressure

                                                                61

                                                                For 0 frac14 37 the active pressure coefficient is given by

                                                                Ka frac14 1 sin 37

                                                                1thorn sin 37frac14 025

                                                                The total active thrust (Equation 66a with c0 frac14 0) is

                                                                Pa frac14 1

                                                                2KaH

                                                                2 frac14 1

                                                                2 025 17 62 frac14 765 kN=m

                                                                If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                                K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                                and the thrust on the wall is

                                                                P0 frac14 1

                                                                2K0H

                                                                2 frac14 1

                                                                2 040 17 62 frac14 122 kN=m

                                                                62

                                                                The active pressure coefficients for the three soil types are as follows

                                                                Ka1 frac141 sin 35

                                                                1thorn sin 35frac14 0271

                                                                Ka2 frac141 sin 27

                                                                1thorn sin 27frac14 0375

                                                                ffiffiffiffiffiffiffiKa2

                                                                p frac14 0613

                                                                Ka3 frac141 sin 42

                                                                1thorn sin 42frac14 0198

                                                                Distribution of active pressure (plotted in Figure Q62)

                                                                Depth (m) Soil Active pressure (kNm2)

                                                                3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                                12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                                At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                                Total thrust frac14 571 kNm

                                                                Point of application is (4893571) m from the top of the wall ie 857m

                                                                Force (kN) Arm (m) Moment (kN m)

                                                                (1)1

                                                                2 0271 16 32 frac14 195 20 390

                                                                (2) 0271 16 3 2 frac14 260 40 1040

                                                                (3)1

                                                                2 0271 92 22 frac14 50 433 217

                                                                (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                                (5)1

                                                                2 0375 102 32 frac14 172 70 1204

                                                                (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                                (7)1

                                                                2 0198 112 42 frac14 177 1067 1889

                                                                (8)1

                                                                2 98 92 frac14 3969 90 35721

                                                                5713 48934

                                                                Figure Q62

                                                                Lateral earth pressure 35

                                                                63

                                                                (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                                Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                                frac14 245

                                                                At the lower end of the piling

                                                                pa frac14 Kaqthorn Kasatz Kaccu

                                                                frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                                frac14 115 kN=m2

                                                                pp frac14 Kpsatzthorn Kpccu

                                                                frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                                frac14 202 kN=m2

                                                                (b) For 0 frac14 26 and frac14 1

                                                                20

                                                                Ka frac14 035

                                                                Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                                pfrac14 145 ethEquation 619THORN

                                                                Kp frac14 37

                                                                Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                                pfrac14 47 ethEquation 624THORN

                                                                At the lower end of the piling

                                                                pa frac14 Kaqthorn Ka0z Kacc

                                                                0

                                                                frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                                frac14 187 kN=m2

                                                                pp frac14 Kp0zthorn Kpcc

                                                                0

                                                                frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                                frac14 198 kN=m2

                                                                36 Lateral earth pressure

                                                                64

                                                                (a) For 0 frac14 38 Ka frac14 024

                                                                0 frac14 20 98 frac14 102 kN=m3

                                                                The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                                Force (kN) Arm (m) Moment (kN m)

                                                                (1) 024 10 66 frac14 159 33 525

                                                                (2)1

                                                                2 024 17 392 frac14 310 400 1240

                                                                (3) 024 17 39 27 frac14 430 135 580

                                                                (4)1

                                                                2 024 102 272 frac14 89 090 80

                                                                (5)1

                                                                2 98 272 frac14 357 090 321

                                                                Hfrac14 1345 MH frac14 2746

                                                                (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                                (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                                XM frac14MV MH frac14 7790 kNm

                                                                Lever arm of base resultant

                                                                M

                                                                Vfrac14 779

                                                                488frac14 160

                                                                Eccentricity of base resultant

                                                                e frac14 200 160 frac14 040m

                                                                39 m

                                                                27 m

                                                                40 m

                                                                04 m

                                                                04 m

                                                                26 m

                                                                (7)

                                                                (9)

                                                                (1)(2)

                                                                (3)

                                                                (4)

                                                                (5)

                                                                (8)(6)

                                                                (10)

                                                                WT

                                                                10 kNm2

                                                                Hydrostatic

                                                                Figure Q64

                                                                Lateral earth pressure 37

                                                                Base pressures (Equation 627)

                                                                p frac14 VB

                                                                1 6e

                                                                B

                                                                frac14 488

                                                                4eth1 060THORN

                                                                frac14 195 kN=m2 and 49 kN=m2

                                                                Factor of safety against sliding (Equation 628)

                                                                F frac14 V tan

                                                                Hfrac14 488 tan 25

                                                                1345frac14 17

                                                                (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                                Hfrac14 1633 kN

                                                                V frac14 4879 kN

                                                                MH frac14 3453 kNm

                                                                MV frac14 10536 kNm

                                                                The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                                65

                                                                For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                Kp

                                                                Ffrac14 385

                                                                2

                                                                0 frac14 20 98 frac14 102 kN=m3

                                                                The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                                Force (kN) Arm (m) Moment (kN m)

                                                                (1)1

                                                                2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                                (2) 026 17 45 d frac14 199d d2 995d2

                                                                (3)1

                                                                2 026 102 d2 frac14 133d2 d3 044d3

                                                                (4)1

                                                                2 385

                                                                2 17 152 frac14 368 dthorn 05 368d 184

                                                                (5)385

                                                                2 17 15 d frac14 491d d2 2455d2

                                                                (6)1

                                                                2 385

                                                                2 102 d2 frac14 982d2 d3 327d3

                                                                38 Lateral earth pressure

                                                                XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                                d3 thorn 516d2 283d 1724 frac14 0

                                                                d frac14 179m

                                                                Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                                Over additional 20 embedded depth

                                                                pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                                Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                                66

                                                                The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                                Ka frac14 sin 69=sin 105

                                                                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                                pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                                26664

                                                                37775

                                                                2

                                                                frac14 050

                                                                The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                                Pa frac14 1

                                                                2 050 19 7502 frac14 267 kN=m

                                                                Figure Q65

                                                                Lateral earth pressure 39

                                                                Horizontal component

                                                                Ph frac14 267 cos 40 frac14 205 kN=m

                                                                Vertical component

                                                                Pv frac14 267 sin 40 frac14 172 kN=m

                                                                Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                Force (kN) Arm (m) Moment (kN m)

                                                                (1)1

                                                                2 175 650 235 frac14 1337 258 345

                                                                (2) 050 650 235 frac14 764 175 134

                                                                (3)1

                                                                2 070 650 235 frac14 535 127 68

                                                                (4) 100 400 235 frac14 940 200 188

                                                                (5) 1

                                                                2 080 050 235 frac14 47 027 1

                                                                Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                Lever arm of base resultant

                                                                M

                                                                Vfrac14 795

                                                                525frac14 151m

                                                                Eccentricity of base resultant

                                                                e frac14 200 151 frac14 049m

                                                                Figure Q66

                                                                40 Lateral earth pressure

                                                                Base pressures (Equation 627)

                                                                p frac14 525

                                                                41 6 049

                                                                4

                                                                frac14 228 kN=m2 and 35 kN=m2

                                                                The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                67

                                                                For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                Force (kN) Arm (m) Moment (kNm)

                                                                (1)1

                                                                2 027 17 52 frac14 574 183 1050

                                                                (2) 027 17 5 3 frac14 689 500 3445

                                                                (3)1

                                                                2 027 102 32 frac14 124 550 682

                                                                (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                (5)1

                                                                2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                (7) 1

                                                                2 267

                                                                2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                2 d frac14 163d d2thorn 650 82d2 1060d

                                                                Tie rod force per m frac14 T 0 0

                                                                XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                d3 thorn 77d2 269d 1438 frac14 0

                                                                d frac14 467m

                                                                Depth of penetration frac14 12d frac14 560m

                                                                Lateral earth pressure 41

                                                                Algebraic sum of forces for d frac14 467m isX

                                                                F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                T frac14 905 kN=m

                                                                Force in each tie rod frac14 25T frac14 226 kN

                                                                68

                                                                (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                0 frac14 21 98 frac14 112 kN=m3

                                                                The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                uC frac14 150

                                                                165 15 98 frac14 134 kN=m2

                                                                The average seepage pressure is

                                                                j frac14 15

                                                                165 98 frac14 09 kN=m3

                                                                Hence

                                                                0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                0 j frac14 112 09 frac14 103 kN=m3

                                                                Figure Q67

                                                                42 Lateral earth pressure

                                                                Consider moments about the anchor point A (per m)

                                                                Force (kN) Arm (m) Moment (kN m)

                                                                (1) 10 026 150 frac14 390 60 2340

                                                                (2)1

                                                                2 026 18 452 frac14 474 15 711

                                                                (3) 026 18 45 105 frac14 2211 825 18240

                                                                (4)1

                                                                2 026 121 1052 frac14 1734 100 17340

                                                                (5)1

                                                                2 134 15 frac14 101 40 404

                                                                (6) 134 30 frac14 402 60 2412

                                                                (7)1

                                                                2 134 60 frac14 402 95 3819

                                                                571 4527(8) Ppm

                                                                115 115PPm

                                                                XM frac14 0

                                                                Ppm frac144527

                                                                115frac14 394 kN=m

                                                                Available passive resistance

                                                                Pp frac14 1

                                                                2 385 103 62 frac14 714 kN=m

                                                                Factor of safety

                                                                Fp frac14 Pp

                                                                Ppm

                                                                frac14 714

                                                                394frac14 18

                                                                Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                Figure Q68

                                                                Lateral earth pressure 43

                                                                (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                Consider moments (per m) about the tie point A

                                                                Force (kN) Arm (m)

                                                                (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                (2)1

                                                                2 033 18 452 frac14 601 15

                                                                (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                (4)1

                                                                2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                (5)1

                                                                2 134 15 frac14 101 40

                                                                (6) 134 30 frac14 402 60

                                                                (7)1

                                                                2 134 d frac14 67d d3thorn 75

                                                                (8) 1

                                                                2 30 103 d2 frac141545d2 2d3thorn 75

                                                                Moment (kN m)

                                                                (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                d3 thorn 827d2 466d 1518 frac14 0

                                                                By trial

                                                                d frac14 544m

                                                                The minimum depth of embedment required is 544m

                                                                69

                                                                For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                0 frac14 20 98 frac14 102 kN=m3

                                                                The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                44 Lateral earth pressure

                                                                uC frac14 147

                                                                173 26 98 frac14 216 kN=m2

                                                                and the average seepage pressure around the wall is

                                                                j frac14 26

                                                                173 98 frac14 15 kN=m3

                                                                Consider moments about the prop (A) (per m)

                                                                Force (kN) Arm (m) Moment (kN m)

                                                                (1)1

                                                                2 03 17 272 frac14 186 020 37

                                                                (2) 03 17 27 53 frac14 730 335 2445

                                                                (3)1

                                                                2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                (5)1

                                                                2 216 26 frac14 281 243 684

                                                                (6) 216 27 frac14 583 465 2712

                                                                (7)1

                                                                2 216 60 frac14 648 800 5184

                                                                3055(8)

                                                                1

                                                                2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                Factor of safety

                                                                Fr frac14 6885

                                                                3055frac14 225

                                                                Figure Q69

                                                                Lateral earth pressure 45

                                                                610

                                                                For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                Using the recommendations of Twine and Roscoe

                                                                p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                611

                                                                frac14 18 kN=m3 0 frac14 34

                                                                H frac14 350m nH frac14 335m mH frac14 185m

                                                                Consider a trial value of F frac14 20 Refer to Figure 635

                                                                0m frac14 tan1tan 34

                                                                20

                                                                frac14 186

                                                                Then

                                                                frac14 45 thorn 0m2frac14 543

                                                                W frac14 1

                                                                2 18 3502 cot 543 frac14 792 kN=m

                                                                Figure Q610

                                                                46 Lateral earth pressure

                                                                P frac14 1

                                                                2 s 3352 frac14 561s kN=m

                                                                U frac14 1

                                                                2 98 1852 cosec 543 frac14 206 kN=m

                                                                Equations 630 and 631 then become

                                                                561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                ie

                                                                561s 0616N 405 frac14 0

                                                                792 0857N thorn 563 frac14 0

                                                                N frac14 848

                                                                0857frac14 989 kN=m

                                                                Then

                                                                561s 609 405 frac14 0

                                                                s frac14 649

                                                                561frac14 116 kN=m3

                                                                The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                Figure Q611

                                                                Lateral earth pressure 47

                                                                612

                                                                For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                45 thorn 0

                                                                2frac14 63

                                                                For the retained material between the surface and a depth of 36m

                                                                Pa frac14 1

                                                                2 030 18 362 frac14 350 kN=m

                                                                Weight of reinforced fill between the surface and a depth of 36m is

                                                                Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                Eccentricity of Rv

                                                                e frac14 263 250 frac14 013m

                                                                The average vertical stress at a depth of 36m is

                                                                z frac14 Rv

                                                                L 2efrac14 324

                                                                474frac14 68 kN=m2

                                                                (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                Tensile stress in the element frac14 138 103

                                                                65 3frac14 71N=mm2

                                                                Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                The weight of ABC is

                                                                W frac14 1

                                                                2 18 52 265 frac14 124 kN=m

                                                                From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                48 Lateral earth pressure

                                                                (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                Tp frac14 032 68 120 065 frac14 170 kN

                                                                Tr frac14 213 420

                                                                418frac14 214 kN

                                                                Again the tensile failure and slipping limit states are satisfied for this element

                                                                Figure Q612

                                                                Lateral earth pressure 49

                                                                Chapter 7

                                                                Consolidation theory

                                                                71

                                                                Total change in thickness

                                                                H frac14 782 602 frac14 180mm

                                                                Average thickness frac14 1530thorn 180

                                                                2frac14 1620mm

                                                                Length of drainage path d frac14 1620

                                                                2frac14 810mm

                                                                Root time plot (Figure Q71a)

                                                                ffiffiffiffiffiffit90p frac14 33

                                                                t90 frac14 109min

                                                                cv frac14 0848d2

                                                                t90frac14 0848 8102

                                                                109 1440 365

                                                                106frac14 27m2=year

                                                                r0 frac14 782 764

                                                                782 602frac14 018

                                                                180frac14 0100

                                                                rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                10 119

                                                                9 180frac14 0735

                                                                rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                Log time plot (Figure Q71b)

                                                                t50 frac14 26min

                                                                cv frac14 0196d2

                                                                t50frac14 0196 8102

                                                                26 1440 365

                                                                106frac14 26m2=year

                                                                r0 frac14 782 763

                                                                782 602frac14 019

                                                                180frac14 0106

                                                                rp frac14 763 623

                                                                782 602frac14 140

                                                                180frac14 0778

                                                                rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                Figure Q71(a)

                                                                Figure Q71(b)

                                                                Final void ratio

                                                                e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                e

                                                                Hfrac14 1thorn e0

                                                                H0frac14 1thorn e1 thorne

                                                                H0

                                                                ie

                                                                e

                                                                180frac14 1631thorne

                                                                1710

                                                                e frac14 2936

                                                                1530frac14 0192

                                                                Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                mv frac14 1

                                                                1thorn e0 e0 e101 00

                                                                frac14 1

                                                                1823 0192

                                                                0107frac14 098m2=MN

                                                                k frac14 cvmvw frac14 265 098 98

                                                                60 1440 365 103frac14 81 1010 m=s

                                                                72

                                                                Using Equation 77 (one-dimensional method)

                                                                sc frac14 e0 e11thorn e0 H

                                                                Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                Figure Q72

                                                                52 Consolidation theory

                                                                Settlement

                                                                Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                318

                                                                Notes 5 92y 460thorn 84

                                                                Heave

                                                                Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                38

                                                                73

                                                                U frac14 f ethTvTHORN frac14 f cvt

                                                                d2

                                                                Hence if cv is constant

                                                                t1

                                                                t2frac14 d

                                                                21

                                                                d22

                                                                where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                d1 frac14 95mm and d2 frac14 2500mm

                                                                for U frac14 050 t2 frac14 t1 d22

                                                                d21

                                                                frac14 20

                                                                60 24 365 25002

                                                                952frac14 263 years

                                                                for U lt 060 Tv frac14

                                                                4U2 (Equation 724(a))

                                                                t030 frac14 t050 0302

                                                                0502

                                                                frac14 263 036 frac14 095 years

                                                                Consolidation theory 53

                                                                74

                                                                The layer is open

                                                                d frac14 8

                                                                2frac14 4m

                                                                Tv frac14 cvtd2frac14 24 3

                                                                42frac14 0450

                                                                ui frac14 frac14 84 kN=m2

                                                                The excess pore water pressure is given by Equation 721

                                                                ue frac14Xmfrac141mfrac140

                                                                2ui

                                                                Msin

                                                                Mz

                                                                d

                                                                expethM2TvTHORN

                                                                In this case z frac14 d

                                                                sinMz

                                                                d

                                                                frac14 sinM

                                                                where

                                                                M frac14

                                                                23

                                                                25

                                                                2

                                                                M sin M M2Tv exp (M2Tv)

                                                                2thorn1 1110 0329

                                                                3

                                                                21 9993 457 105

                                                                ue frac14 2 84 2

                                                                1 0329 ethother terms negligibleTHORN

                                                                frac14 352 kN=m2

                                                                75

                                                                The layer is open

                                                                d frac14 6

                                                                2frac14 3m

                                                                Tv frac14 cvtd2frac14 10 3

                                                                32frac14 0333

                                                                The layer thickness will be divided into six equal parts ie m frac14 6

                                                                54 Consolidation theory

                                                                For an open layer

                                                                Tv frac14 4n

                                                                m2

                                                                n frac14 0333 62

                                                                4frac14 300

                                                                The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                i j

                                                                0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                The initial and 3-year isochrones are plotted in Figure Q75

                                                                Area under initial isochrone frac14 180 units

                                                                Area under 3-year isochrone frac14 63 units

                                                                The average degree of consolidation is given by Equation 725Thus

                                                                U frac14 1 63

                                                                180frac14 065

                                                                Figure Q75

                                                                Consolidation theory 55

                                                                76

                                                                At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                The final consolidation settlement (one-dimensional method) is

                                                                sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                Corrected time t frac14 2 1

                                                                2

                                                                40

                                                                52

                                                                frac14 1615 years

                                                                Tv frac14 cvtd2frac14 44 1615

                                                                42frac14 0444

                                                                From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                77

                                                                The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                Figure Q77

                                                                56 Consolidation theory

                                                                Point m n Ir (kNm2) sc (mm)

                                                                13020frac14 15 20

                                                                20frac14 10 0194 (4) 113 124

                                                                260

                                                                20frac14 30

                                                                20

                                                                20frac14 10 0204 (2) 59 65

                                                                360

                                                                20frac14 30

                                                                40

                                                                20frac14 20 0238 (1) 35 38

                                                                430

                                                                20frac14 15

                                                                40

                                                                20frac14 20 0224 (2) 65 72

                                                                Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                78

                                                                Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                (a) Immediate settlement

                                                                H

                                                                Bfrac14 30

                                                                35frac14 086

                                                                D

                                                                Bfrac14 2

                                                                35frac14 006

                                                                Figure Q78

                                                                Consolidation theory 57

                                                                From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                si frac14 130131qB

                                                                Eufrac14 10 032 105 35

                                                                40frac14 30mm

                                                                (b) Consolidation settlement

                                                                Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                3150

                                                                Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                Now

                                                                H

                                                                Bfrac14 30

                                                                35frac14 086 and A frac14 065

                                                                from Figure 712 13 frac14 079

                                                                sc frac14 13sod frac14 079 315 frac14 250mm

                                                                Total settlement

                                                                s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                79

                                                                Without sand drains

                                                                Uv frac14 025

                                                                Tv frac14 0049 ethfrom Figure 718THORN

                                                                t frac14 Tvd2

                                                                cvfrac14 0049 82

                                                                cvWith sand drains

                                                                R frac14 0564S frac14 0564 3 frac14 169m

                                                                n frac14 Rrfrac14 169

                                                                015frac14 113

                                                                Tr frac14 cht

                                                                4R2frac14 ch

                                                                4 1692 0049 82

                                                                cvethand ch frac14 cvTHORN

                                                                frac14 0275

                                                                Ur frac14 073 (from Figure 730)

                                                                58 Consolidation theory

                                                                Using Equation 740

                                                                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                U frac14 080

                                                                710

                                                                Without sand drains

                                                                Uv frac14 090

                                                                Tv frac14 0848

                                                                t frac14 Tvd2

                                                                cvfrac14 0848 102

                                                                96frac14 88 years

                                                                With sand drains

                                                                R frac14 0564S frac14 0564 4 frac14 226m

                                                                n frac14 Rrfrac14 226

                                                                015frac14 15

                                                                Tr

                                                                Tvfrac14 chcv

                                                                d2

                                                                4R2ethsame tTHORN

                                                                Tr

                                                                Tvfrac14 140

                                                                96 102

                                                                4 2262frac14 714 eth1THORN

                                                                Using Equation 740

                                                                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                Thus

                                                                Uv frac14 0295 and Ur frac14 086

                                                                t frac14 88 00683

                                                                0848frac14 07 years

                                                                Consolidation theory 59

                                                                Chapter 8

                                                                Bearing capacity

                                                                81

                                                                (a) The ultimate bearing capacity is given by Equation 83

                                                                qf frac14 cNc thorn DNq thorn 1

                                                                2BN

                                                                For u frac14 0

                                                                Nc frac14 514 Nq frac14 1 N frac14 0

                                                                qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                The net ultimate bearing capacity is

                                                                qnf frac14 qf D frac14 540 kN=m2

                                                                The net foundation pressure is

                                                                qn frac14 q D frac14 425

                                                                2 eth21 1THORN frac14 192 kN=m2

                                                                The factor of safety (Equation 86) is

                                                                F frac14 qnfqnfrac14 540

                                                                192frac14 28

                                                                (b) For 0 frac14 28

                                                                Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                2 112 2 13

                                                                frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                qnf frac14 574 112 frac14 563 kN=m2

                                                                F frac14 563

                                                                192frac14 29

                                                                (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                82

                                                                For 0 frac14 38

                                                                Nq frac14 49 N frac14 67

                                                                qnf frac14 DethNq 1THORN thorn 1

                                                                2BN ethfrom Equation 83THORN

                                                                frac14 eth18 075 48THORN thorn 1

                                                                2 18 15 67

                                                                frac14 648thorn 905 frac14 1553 kN=m2

                                                                qn frac14 500

                                                                15 eth18 075THORN frac14 320 kN=m2

                                                                F frac14 qnfqnfrac14 1553

                                                                320frac14 48

                                                                0d frac14 tan1tan 38

                                                                125

                                                                frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                2 18 15 25

                                                                frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                Design load (action) Vd frac14 500 kN=m

                                                                The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                83

                                                                D

                                                                Bfrac14 350

                                                                225frac14 155

                                                                From Figure 85 for a square foundation

                                                                Nc frac14 81

                                                                Bearing capacity 61

                                                                For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                Nc frac14 084thorn 016B

                                                                L

                                                                81 frac14 745

                                                                Using Equation 810

                                                                qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                For F frac14 3

                                                                qn frac14 1006

                                                                3frac14 335 kN=m2

                                                                q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                Design load frac14 405 450 225 frac14 4100 kN

                                                                Design undrained strength cud frac14 135

                                                                14frac14 96 kN=m2

                                                                Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                frac14 7241 kN

                                                                Design load Vd frac14 4100 kN

                                                                Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                84

                                                                For 0 frac14 40

                                                                Nq frac14 64 N frac14 95

                                                                qnf frac14 DethNq 1THORN thorn 04BN

                                                                (a) Water table 5m below ground level

                                                                qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                qn frac14 400 17 frac14 383 kN=m2

                                                                F frac14 2686

                                                                383frac14 70

                                                                (b) Water table 1m below ground level (ie at foundation level)

                                                                0 frac14 20 98 frac14 102 kN=m3

                                                                62 Bearing capacity

                                                                qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                F frac14 2040

                                                                383frac14 53

                                                                (c) Water table at ground level with upward hydraulic gradient 02

                                                                eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                F frac14 1296

                                                                392frac14 33

                                                                85

                                                                The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                Design value of 0 frac14 tan1tan 39

                                                                125

                                                                frac14 33

                                                                For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                86

                                                                (a) Undrained shear for u frac14 0

                                                                Nc frac14 514 Nq frac14 1 N frac14 0

                                                                qnf frac14 12cuNc

                                                                frac14 12 100 514 frac14 617 kN=m2

                                                                qn frac14 qnfFfrac14 617

                                                                3frac14 206 kN=m2

                                                                q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                Bearing capacity 63

                                                                Drained shear for 0 frac14 32

                                                                Nq frac14 23 N frac14 25

                                                                0 frac14 21 98 frac14 112 kN=m3

                                                                qnf frac14 0DethNq 1THORN thorn 040BN

                                                                frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                frac14 694 kN=m2

                                                                q frac14 694

                                                                3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                Design load frac14 42 227 frac14 3632 kN

                                                                (b) Design undrained strength cud frac14 100

                                                                14frac14 71 kNm2

                                                                Design bearing resistance Rd frac14 12cudNe area

                                                                frac14 12 71 514 42

                                                                frac14 7007 kN

                                                                For drained shear 0d frac14 tan1tan 32

                                                                125

                                                                frac14 26

                                                                Nq frac14 12 N frac14 10

                                                                Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                1 2 100 0175 0700qn 0182qn

                                                                2 6 033 0044 0176qn 0046qn

                                                                3 10 020 0017 0068qn 0018qn

                                                                0246qn

                                                                Diameter of equivalent circle B frac14 45m

                                                                H

                                                                Bfrac14 12

                                                                45frac14 27 and A frac14 042

                                                                13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                64 Bearing capacity

                                                                For sc frac14 30mm

                                                                qn frac14 30

                                                                0147frac14 204 kN=m2

                                                                q frac14 204thorn 21 frac14 225 kN=m2

                                                                Design load frac14 42 225 frac14 3600 kN

                                                                The design load is 3600 kN settlement being the limiting criterion

                                                                87

                                                                D

                                                                Bfrac14 8

                                                                4frac14 20

                                                                From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                F frac14 cuNc

                                                                Dfrac14 40 71

                                                                20 8frac14 18

                                                                88

                                                                Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                Design value of 0 frac14 tan1tan 38

                                                                125

                                                                frac14 32

                                                                Figure Q86

                                                                Bearing capacity 65

                                                                For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                For B frac14 250m qn frac14 3750

                                                                2502 17 frac14 583 kN=m2

                                                                From Figure 510 m frac14 n frac14 126

                                                                6frac14 021

                                                                Ir frac14 0019

                                                                Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                89

                                                                Depth (m) N 0v (kNm2) CN N1

                                                                070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                Cw frac14 05thorn 0530

                                                                47

                                                                frac14 082

                                                                66 Bearing capacity

                                                                Thus

                                                                qa frac14 150 082 frac14 120 kN=m2

                                                                (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                Thus

                                                                qa frac14 90 15 frac14 135 kN=m2

                                                                (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                Ic frac14 171

                                                                1014frac14 0068

                                                                From Equation 819(a) with s frac14 25mm

                                                                q frac14 25

                                                                3507 0068frac14 150 kN=m2

                                                                810

                                                                Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                Izp frac14 05thorn 01130

                                                                16 27

                                                                05

                                                                frac14 067

                                                                Refer to Figure Q810

                                                                E frac14 25qc

                                                                Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                Ez (mm3MN)

                                                                1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                0203

                                                                C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                130frac14 093

                                                                C2 frac14 1 ethsayTHORN

                                                                s frac14 C1C2qnX Iz

                                                                Ez frac14 093 1 130 0203 frac14 25mm

                                                                Bearing capacity 67

                                                                811

                                                                At pile base level

                                                                cu frac14 220 kN=m2

                                                                qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                Then

                                                                Qf frac14 Abqb thorn Asqs

                                                                frac14

                                                                4 32 1980

                                                                thorn eth 105 139 86THORN

                                                                frac14 13 996thorn 3941 frac14 17 937 kN

                                                                0 01 02 03 04 05 06 07

                                                                0 2 4 6 8 10 12 14

                                                                1

                                                                2

                                                                3

                                                                4

                                                                5

                                                                6

                                                                7

                                                                8

                                                                (1)

                                                                (2)

                                                                (3)

                                                                (4)

                                                                (5)

                                                                qc

                                                                qc

                                                                Iz

                                                                Iz

                                                                (MNm2)

                                                                z (m)

                                                                Figure Q810

                                                                68 Bearing capacity

                                                                Allowable load

                                                                ethaTHORN Qf

                                                                2frac14 17 937

                                                                2frac14 8968 kN

                                                                ethbTHORN Abqb

                                                                3thorn Asqs frac14 13 996

                                                                3thorn 3941 frac14 8606 kN

                                                                ie allowable load frac14 8600 kN

                                                                Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                According to the limit state method

                                                                Characteristic undrained strength at base level cuk frac14 220

                                                                150kN=m2

                                                                Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                150frac14 1320 kN=m2

                                                                Characteristic shaft resistance qsk frac14 00150

                                                                frac14 86

                                                                150frac14 57 kN=m2

                                                                Characteristic base and shaft resistances

                                                                Rbk frac14

                                                                4 32 1320 frac14 9330 kN

                                                                Rsk frac14 105 139 86

                                                                150frac14 2629 kN

                                                                For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                Design bearing resistance Rcd frac14 9330

                                                                160thorn 2629

                                                                130

                                                                frac14 5831thorn 2022

                                                                frac14 7850 kN

                                                                Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                812

                                                                ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                For a single pile

                                                                Qf frac14 Abqb thorn Asqs

                                                                frac14

                                                                4 062 1305

                                                                thorn eth 06 15 42THORN

                                                                frac14 369thorn 1187 frac14 1556 kN

                                                                Bearing capacity 69

                                                                Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                qbkfrac14 9cuk frac14 9 220

                                                                150frac14 1320 kN=m2

                                                                qskfrac14cuk frac14 040 105

                                                                150frac14 28 kN=m2

                                                                Rbkfrac14

                                                                4 0602 1320 frac14 373 kN

                                                                Rskfrac14 060 15 28 frac14 791 kN

                                                                Rcdfrac14 373

                                                                160thorn 791

                                                                130frac14 233thorn 608 frac14 841 kN

                                                                Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                q frac14 21 000

                                                                1762frac14 68 kN=m2

                                                                Immediate settlement

                                                                H

                                                                Bfrac14 15

                                                                176frac14 085

                                                                D

                                                                Bfrac14 13

                                                                176frac14 074

                                                                L

                                                                Bfrac14 1

                                                                Hence from Figure 515

                                                                130 frac14 078 and 131 frac14 041

                                                                70 Bearing capacity

                                                                Thus using Equation 528

                                                                si frac14 078 041 68 176

                                                                65frac14 6mm

                                                                Consolidation settlement

                                                                Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                434 (sod)

                                                                Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                sc frac14 056 434 frac14 24mm

                                                                The total settlement is (6thorn 24) frac14 30mm

                                                                813

                                                                At base level N frac14 26 Then using Equation 830

                                                                qb frac14 40NDb

                                                                Bfrac14 40 26 2

                                                                025frac14 8320 kN=m2

                                                                ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                Figure Q812

                                                                Bearing capacity 71

                                                                Over the length embedded in sand

                                                                N frac14 21 ie18thorn 24

                                                                2

                                                                Using Equation 831

                                                                qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                For a single pile

                                                                Qf frac14 Abqb thorn Asqs

                                                                frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                For the pile group assuming a group efficiency of 12

                                                                XQf frac14 12 9 604 frac14 6523 kN

                                                                Then the load factor is

                                                                F frac14 6523

                                                                2000thorn 1000frac14 21

                                                                (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                Characteristic base resistance per unit area qbk frac14 8320

                                                                150frac14 5547 kNm2

                                                                Characteristic shaft resistance per unit area qsk frac14 42

                                                                150frac14 28 kNm2

                                                                Characteristic base and shaft resistances for a single pile

                                                                Rbk frac14 0252 5547 frac14 347 kN

                                                                Rsk frac14 4 025 2 28 frac14 56 kN

                                                                For a driven pile the partial factors are b frac14 s frac14 130

                                                                Design bearing resistance Rcd frac14 347

                                                                130thorn 56

                                                                130frac14 310 kN

                                                                For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                72 Bearing capacity

                                                                N frac14 24thorn 26thorn 34

                                                                3frac14 28

                                                                Ic frac14 171

                                                                2814frac14 0016 ethEquation 818THORN

                                                                s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                814

                                                                Using Equation 841

                                                                Tf frac14 DLcu thorn

                                                                4ethD2 d2THORNcuNc

                                                                frac14 eth 02 5 06 110THORN thorn

                                                                4eth022 012THORN110 9

                                                                frac14 207thorn 23 frac14 230 kN

                                                                Figure Q813

                                                                Bearing capacity 73

                                                                Chapter 9

                                                                Stability of slopes

                                                                91

                                                                Referring to Figure Q91

                                                                W frac14 417 19 frac14 792 kN=m

                                                                Q frac14 20 28 frac14 56 kN=m

                                                                Arc lengthAB frac14

                                                                180 73 90 frac14 115m

                                                                Arc length BC frac14

                                                                180 28 90 frac14 44m

                                                                The factor of safety is given by

                                                                F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                Depth of tension crack z0 frac14 2cu

                                                                frac14 2 20

                                                                19frac14 21m

                                                                Arc length BD frac14

                                                                180 13

                                                                1

                                                                2 90 frac14 21m

                                                                F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                92

                                                                u frac14 0

                                                                Depth factor D frac14 11

                                                                9frac14 122

                                                                Using Equation 92 with F frac14 10

                                                                Ns frac14 cu

                                                                FHfrac14 30

                                                                10 19 9frac14 0175

                                                                Hence from Figure 93

                                                                frac14 50

                                                                For F frac14 12

                                                                Ns frac14 30

                                                                12 19 9frac14 0146

                                                                frac14 27

                                                                93

                                                                Refer to Figure Q93

                                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                74 m

                                                                214 1deg

                                                                213 1deg

                                                                39 m

                                                                WB

                                                                D

                                                                C

                                                                28 m

                                                                21 m

                                                                A

                                                                Q

                                                                Soil (1)Soil (2)

                                                                73deg

                                                                Figure Q91

                                                                Stability of slopes 75

                                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                599 256 328 1372

                                                                Figure Q93

                                                                76 Stability of slopes

                                                                XW cos frac14 b

                                                                Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                W sin frac14 bX

                                                                h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                Arc length La frac14

                                                                180 57

                                                                1

                                                                2 326 frac14 327m

                                                                The factor of safety is given by

                                                                F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                W sin

                                                                frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                frac14 091

                                                                According to the limit state method

                                                                0d frac14 tan1tan 32

                                                                125

                                                                frac14 265

                                                                c0 frac14 8

                                                                160frac14 5 kN=m2

                                                                Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                Design disturbing moment frac14 1075 kN=m

                                                                The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                94

                                                                F frac14 1

                                                                W sin

                                                                Xfc0bthorn ethW ubTHORN tan0g sec

                                                                1thorn ethtan tan0=FTHORN

                                                                c0 frac14 8 kN=m2

                                                                0 frac14 32

                                                                c0b frac14 8 2 frac14 16 kN=m

                                                                W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                Try F frac14 100

                                                                tan0

                                                                Ffrac14 0625

                                                                Stability of slopes 77

                                                                Values of u are as obtained in Figure Q93

                                                                SliceNo

                                                                h(m)

                                                                W frac14 bh(kNm)

                                                                W sin(kNm)

                                                                ub(kNm)

                                                                c0bthorn (W ub) tan0(kNm)

                                                                sec

                                                                1thorn (tan tan0)FProduct(kNm)

                                                                1 05 21 6 2 8 24 1078 262 13 55 31

                                                                23 33 30 1042 31

                                                                3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                224 92 72 0931 67

                                                                6 50 210 11 40 100 85 0907 777 55 231 14

                                                                12 58 112 90 0889 80

                                                                8 60 252 1812

                                                                80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                2154 88 116 0853 99

                                                                14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                1074 1091

                                                                F frac14 1091

                                                                1074frac14 102 (assumed value 100)

                                                                Thus

                                                                F frac14 101

                                                                95

                                                                F frac14 1

                                                                W sin

                                                                XfWeth1 ruTHORN tan0g sec

                                                                1thorn ethtan tan0THORN=F

                                                                0 frac14 33

                                                                ru frac14 020

                                                                W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                Try F frac14 110

                                                                tan 0

                                                                Ffrac14 tan 33

                                                                110frac14 0590

                                                                78 Stability of slopes

                                                                Referring to Figure Q95

                                                                SliceNo

                                                                h(m)

                                                                W frac14 bh(kNm)

                                                                W sin(kNm)

                                                                W(1 ru) tan0(kNm)

                                                                sec

                                                                1thorn ( tan tan0)FProduct(kNm)

                                                                1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                2120 234 0892 209

                                                                4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                1185 1271

                                                                Figure Q95

                                                                Stability of slopes 79

                                                                F frac14 1271

                                                                1185frac14 107

                                                                The trial value was 110 therefore take F to be 108

                                                                96

                                                                (a) Water table at surface the factor of safety is given by Equation 912

                                                                F frac14 0

                                                                sat

                                                                tan0

                                                                tan

                                                                ptie 15 frac14 92

                                                                19

                                                                tan 36

                                                                tan

                                                                tan frac14 0234

                                                                frac14 13

                                                                Water table well below surface the factor of safety is given by Equation 911

                                                                F frac14 tan0

                                                                tan

                                                                frac14 tan 36

                                                                tan 13

                                                                frac14 31

                                                                (b) 0d frac14 tan1tan 36

                                                                125

                                                                frac14 30

                                                                Depth of potential failure surface frac14 z

                                                                Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                frac14 504z kN

                                                                Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                frac14 19 z sin 13 cos 13

                                                                frac14 416z kN

                                                                Rd gtSd therefore the limit state for overall stability is satisfied

                                                                80 Stability of slopes

                                                                • Book Cover
                                                                • Title
                                                                • Contents
                                                                • Basic characteristics of soils
                                                                • Seepage
                                                                • Effective stress
                                                                • Shear strength
                                                                • Stresses and displacements
                                                                • Lateral earth pressure
                                                                • Consolidation theory
                                                                • Bearing capacity
                                                                • Stability of slopes

                                                                  The variation of A with axial strain is plotted in Figure Q49 At failure Afrac14 023

                                                                  Figure Q49

                                                                  Shear strength 27

                                                                  Chapter 5

                                                                  Stresses and displacements

                                                                  51

                                                                  Vertical stress is given by

                                                                  z frac14 Qz2Ip frac14 5000

                                                                  52Ip

                                                                  Values of Ip are obtained from Table 51

                                                                  r (m) rz Ip z (kNm2)

                                                                  0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                                                  10 20 0009 2

                                                                  The variation of z with radial distance (r) is plotted in Figure Q51

                                                                  Figure Q51

                                                                  52

                                                                  Below the centre load (Figure Q52)

                                                                  r

                                                                  zfrac14 0 for the 7500-kN load

                                                                  Ip frac14 0478

                                                                  r

                                                                  zfrac14 5

                                                                  4frac14 125 for the 10 000- and 9000-kN loads

                                                                  Ip frac14 0045

                                                                  Then

                                                                  z frac14X Q

                                                                  z2Ip

                                                                  frac14 7500 0478

                                                                  42thorn 10 000 0045

                                                                  42thorn 9000 0045

                                                                  42

                                                                  frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                                                  53

                                                                  The vertical stress under a corner of a rectangular area is given by

                                                                  z frac14 qIr

                                                                  where values of Ir are obtained from Figure 510 In this case

                                                                  z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                                                  z

                                                                  Figure Q52

                                                                  Stresses and displacements 29

                                                                  z (m) m n Ir z (kNm2)

                                                                  0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                                                  10 010 0005 5

                                                                  z is plotted against z in Figure Q53

                                                                  54

                                                                  (a)

                                                                  m frac14 125

                                                                  12frac14 104

                                                                  n frac14 18

                                                                  12frac14 150

                                                                  From Figure 510 Irfrac14 0196

                                                                  z frac14 2 175 0196 frac14 68 kN=m2

                                                                  Figure Q53

                                                                  30 Stresses and displacements

                                                                  (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                                                  z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                                                  55

                                                                  Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                                                  Px frac14 2Q

                                                                  1

                                                                  m2 thorn 1frac14 2 150

                                                                  125frac14 76 kN=m

                                                                  Equation 517 is used to obtain the pressure distribution

                                                                  px frac14 4Q

                                                                  h

                                                                  m2n

                                                                  ethm2 thorn n2THORN2 frac14150

                                                                  m2n

                                                                  ethm2 thorn n2THORN2 ethkN=m2THORN

                                                                  Figure Q54

                                                                  Stresses and displacements 31

                                                                  n m2n

                                                                  (m2 thorn n2)2

                                                                  px(kNm2)

                                                                  0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                                  The pressure distribution is plotted in Figure Q55

                                                                  56

                                                                  H

                                                                  Bfrac14 10

                                                                  2frac14 5

                                                                  L

                                                                  Bfrac14 4

                                                                  2frac14 2

                                                                  D

                                                                  Bfrac14 1

                                                                  2frac14 05

                                                                  Hence from Figure 515

                                                                  131 frac14 082

                                                                  130 frac14 094

                                                                  Figure Q55

                                                                  32 Stresses and displacements

                                                                  The immediate settlement is given by Equation 528

                                                                  si frac14 130131qB

                                                                  Eu

                                                                  frac14 094 082 200 2

                                                                  45frac14 7mm

                                                                  Stresses and displacements 33

                                                                  Chapter 6

                                                                  Lateral earth pressure

                                                                  61

                                                                  For 0 frac14 37 the active pressure coefficient is given by

                                                                  Ka frac14 1 sin 37

                                                                  1thorn sin 37frac14 025

                                                                  The total active thrust (Equation 66a with c0 frac14 0) is

                                                                  Pa frac14 1

                                                                  2KaH

                                                                  2 frac14 1

                                                                  2 025 17 62 frac14 765 kN=m

                                                                  If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                                  K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                                  and the thrust on the wall is

                                                                  P0 frac14 1

                                                                  2K0H

                                                                  2 frac14 1

                                                                  2 040 17 62 frac14 122 kN=m

                                                                  62

                                                                  The active pressure coefficients for the three soil types are as follows

                                                                  Ka1 frac141 sin 35

                                                                  1thorn sin 35frac14 0271

                                                                  Ka2 frac141 sin 27

                                                                  1thorn sin 27frac14 0375

                                                                  ffiffiffiffiffiffiffiKa2

                                                                  p frac14 0613

                                                                  Ka3 frac141 sin 42

                                                                  1thorn sin 42frac14 0198

                                                                  Distribution of active pressure (plotted in Figure Q62)

                                                                  Depth (m) Soil Active pressure (kNm2)

                                                                  3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                                  12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                                  At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                                  Total thrust frac14 571 kNm

                                                                  Point of application is (4893571) m from the top of the wall ie 857m

                                                                  Force (kN) Arm (m) Moment (kN m)

                                                                  (1)1

                                                                  2 0271 16 32 frac14 195 20 390

                                                                  (2) 0271 16 3 2 frac14 260 40 1040

                                                                  (3)1

                                                                  2 0271 92 22 frac14 50 433 217

                                                                  (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                                  (5)1

                                                                  2 0375 102 32 frac14 172 70 1204

                                                                  (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                                  (7)1

                                                                  2 0198 112 42 frac14 177 1067 1889

                                                                  (8)1

                                                                  2 98 92 frac14 3969 90 35721

                                                                  5713 48934

                                                                  Figure Q62

                                                                  Lateral earth pressure 35

                                                                  63

                                                                  (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                                  Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                                  frac14 245

                                                                  At the lower end of the piling

                                                                  pa frac14 Kaqthorn Kasatz Kaccu

                                                                  frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                                  frac14 115 kN=m2

                                                                  pp frac14 Kpsatzthorn Kpccu

                                                                  frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                                  frac14 202 kN=m2

                                                                  (b) For 0 frac14 26 and frac14 1

                                                                  20

                                                                  Ka frac14 035

                                                                  Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                                  pfrac14 145 ethEquation 619THORN

                                                                  Kp frac14 37

                                                                  Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                                  pfrac14 47 ethEquation 624THORN

                                                                  At the lower end of the piling

                                                                  pa frac14 Kaqthorn Ka0z Kacc

                                                                  0

                                                                  frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                                  frac14 187 kN=m2

                                                                  pp frac14 Kp0zthorn Kpcc

                                                                  0

                                                                  frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                                  frac14 198 kN=m2

                                                                  36 Lateral earth pressure

                                                                  64

                                                                  (a) For 0 frac14 38 Ka frac14 024

                                                                  0 frac14 20 98 frac14 102 kN=m3

                                                                  The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                                  Force (kN) Arm (m) Moment (kN m)

                                                                  (1) 024 10 66 frac14 159 33 525

                                                                  (2)1

                                                                  2 024 17 392 frac14 310 400 1240

                                                                  (3) 024 17 39 27 frac14 430 135 580

                                                                  (4)1

                                                                  2 024 102 272 frac14 89 090 80

                                                                  (5)1

                                                                  2 98 272 frac14 357 090 321

                                                                  Hfrac14 1345 MH frac14 2746

                                                                  (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                                  (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                                  XM frac14MV MH frac14 7790 kNm

                                                                  Lever arm of base resultant

                                                                  M

                                                                  Vfrac14 779

                                                                  488frac14 160

                                                                  Eccentricity of base resultant

                                                                  e frac14 200 160 frac14 040m

                                                                  39 m

                                                                  27 m

                                                                  40 m

                                                                  04 m

                                                                  04 m

                                                                  26 m

                                                                  (7)

                                                                  (9)

                                                                  (1)(2)

                                                                  (3)

                                                                  (4)

                                                                  (5)

                                                                  (8)(6)

                                                                  (10)

                                                                  WT

                                                                  10 kNm2

                                                                  Hydrostatic

                                                                  Figure Q64

                                                                  Lateral earth pressure 37

                                                                  Base pressures (Equation 627)

                                                                  p frac14 VB

                                                                  1 6e

                                                                  B

                                                                  frac14 488

                                                                  4eth1 060THORN

                                                                  frac14 195 kN=m2 and 49 kN=m2

                                                                  Factor of safety against sliding (Equation 628)

                                                                  F frac14 V tan

                                                                  Hfrac14 488 tan 25

                                                                  1345frac14 17

                                                                  (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                                  Hfrac14 1633 kN

                                                                  V frac14 4879 kN

                                                                  MH frac14 3453 kNm

                                                                  MV frac14 10536 kNm

                                                                  The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                                  65

                                                                  For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                  Kp

                                                                  Ffrac14 385

                                                                  2

                                                                  0 frac14 20 98 frac14 102 kN=m3

                                                                  The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                                  Force (kN) Arm (m) Moment (kN m)

                                                                  (1)1

                                                                  2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                                  (2) 026 17 45 d frac14 199d d2 995d2

                                                                  (3)1

                                                                  2 026 102 d2 frac14 133d2 d3 044d3

                                                                  (4)1

                                                                  2 385

                                                                  2 17 152 frac14 368 dthorn 05 368d 184

                                                                  (5)385

                                                                  2 17 15 d frac14 491d d2 2455d2

                                                                  (6)1

                                                                  2 385

                                                                  2 102 d2 frac14 982d2 d3 327d3

                                                                  38 Lateral earth pressure

                                                                  XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                                  d3 thorn 516d2 283d 1724 frac14 0

                                                                  d frac14 179m

                                                                  Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                                  Over additional 20 embedded depth

                                                                  pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                                  Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                                  66

                                                                  The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                                  Ka frac14 sin 69=sin 105

                                                                  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                                  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                                  pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                                  26664

                                                                  37775

                                                                  2

                                                                  frac14 050

                                                                  The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                                  Pa frac14 1

                                                                  2 050 19 7502 frac14 267 kN=m

                                                                  Figure Q65

                                                                  Lateral earth pressure 39

                                                                  Horizontal component

                                                                  Ph frac14 267 cos 40 frac14 205 kN=m

                                                                  Vertical component

                                                                  Pv frac14 267 sin 40 frac14 172 kN=m

                                                                  Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                  Force (kN) Arm (m) Moment (kN m)

                                                                  (1)1

                                                                  2 175 650 235 frac14 1337 258 345

                                                                  (2) 050 650 235 frac14 764 175 134

                                                                  (3)1

                                                                  2 070 650 235 frac14 535 127 68

                                                                  (4) 100 400 235 frac14 940 200 188

                                                                  (5) 1

                                                                  2 080 050 235 frac14 47 027 1

                                                                  Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                  Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                  Lever arm of base resultant

                                                                  M

                                                                  Vfrac14 795

                                                                  525frac14 151m

                                                                  Eccentricity of base resultant

                                                                  e frac14 200 151 frac14 049m

                                                                  Figure Q66

                                                                  40 Lateral earth pressure

                                                                  Base pressures (Equation 627)

                                                                  p frac14 525

                                                                  41 6 049

                                                                  4

                                                                  frac14 228 kN=m2 and 35 kN=m2

                                                                  The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                  The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                  The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                  67

                                                                  For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                  Force (kN) Arm (m) Moment (kNm)

                                                                  (1)1

                                                                  2 027 17 52 frac14 574 183 1050

                                                                  (2) 027 17 5 3 frac14 689 500 3445

                                                                  (3)1

                                                                  2 027 102 32 frac14 124 550 682

                                                                  (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                  (5)1

                                                                  2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                  (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                  (7) 1

                                                                  2 267

                                                                  2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                  (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                  2 d frac14 163d d2thorn 650 82d2 1060d

                                                                  Tie rod force per m frac14 T 0 0

                                                                  XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                  d3 thorn 77d2 269d 1438 frac14 0

                                                                  d frac14 467m

                                                                  Depth of penetration frac14 12d frac14 560m

                                                                  Lateral earth pressure 41

                                                                  Algebraic sum of forces for d frac14 467m isX

                                                                  F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                  T frac14 905 kN=m

                                                                  Force in each tie rod frac14 25T frac14 226 kN

                                                                  68

                                                                  (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                  0 frac14 21 98 frac14 112 kN=m3

                                                                  The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                  uC frac14 150

                                                                  165 15 98 frac14 134 kN=m2

                                                                  The average seepage pressure is

                                                                  j frac14 15

                                                                  165 98 frac14 09 kN=m3

                                                                  Hence

                                                                  0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                  0 j frac14 112 09 frac14 103 kN=m3

                                                                  Figure Q67

                                                                  42 Lateral earth pressure

                                                                  Consider moments about the anchor point A (per m)

                                                                  Force (kN) Arm (m) Moment (kN m)

                                                                  (1) 10 026 150 frac14 390 60 2340

                                                                  (2)1

                                                                  2 026 18 452 frac14 474 15 711

                                                                  (3) 026 18 45 105 frac14 2211 825 18240

                                                                  (4)1

                                                                  2 026 121 1052 frac14 1734 100 17340

                                                                  (5)1

                                                                  2 134 15 frac14 101 40 404

                                                                  (6) 134 30 frac14 402 60 2412

                                                                  (7)1

                                                                  2 134 60 frac14 402 95 3819

                                                                  571 4527(8) Ppm

                                                                  115 115PPm

                                                                  XM frac14 0

                                                                  Ppm frac144527

                                                                  115frac14 394 kN=m

                                                                  Available passive resistance

                                                                  Pp frac14 1

                                                                  2 385 103 62 frac14 714 kN=m

                                                                  Factor of safety

                                                                  Fp frac14 Pp

                                                                  Ppm

                                                                  frac14 714

                                                                  394frac14 18

                                                                  Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                  Figure Q68

                                                                  Lateral earth pressure 43

                                                                  (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                  Consider moments (per m) about the tie point A

                                                                  Force (kN) Arm (m)

                                                                  (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                  (2)1

                                                                  2 033 18 452 frac14 601 15

                                                                  (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                  (4)1

                                                                  2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                  (5)1

                                                                  2 134 15 frac14 101 40

                                                                  (6) 134 30 frac14 402 60

                                                                  (7)1

                                                                  2 134 d frac14 67d d3thorn 75

                                                                  (8) 1

                                                                  2 30 103 d2 frac141545d2 2d3thorn 75

                                                                  Moment (kN m)

                                                                  (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                  XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                  d3 thorn 827d2 466d 1518 frac14 0

                                                                  By trial

                                                                  d frac14 544m

                                                                  The minimum depth of embedment required is 544m

                                                                  69

                                                                  For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                  0 frac14 20 98 frac14 102 kN=m3

                                                                  The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                  44 Lateral earth pressure

                                                                  uC frac14 147

                                                                  173 26 98 frac14 216 kN=m2

                                                                  and the average seepage pressure around the wall is

                                                                  j frac14 26

                                                                  173 98 frac14 15 kN=m3

                                                                  Consider moments about the prop (A) (per m)

                                                                  Force (kN) Arm (m) Moment (kN m)

                                                                  (1)1

                                                                  2 03 17 272 frac14 186 020 37

                                                                  (2) 03 17 27 53 frac14 730 335 2445

                                                                  (3)1

                                                                  2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                  (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                  (5)1

                                                                  2 216 26 frac14 281 243 684

                                                                  (6) 216 27 frac14 583 465 2712

                                                                  (7)1

                                                                  2 216 60 frac14 648 800 5184

                                                                  3055(8)

                                                                  1

                                                                  2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                  Factor of safety

                                                                  Fr frac14 6885

                                                                  3055frac14 225

                                                                  Figure Q69

                                                                  Lateral earth pressure 45

                                                                  610

                                                                  For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                  p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                  Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                  Using the recommendations of Twine and Roscoe

                                                                  p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                  Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                  611

                                                                  frac14 18 kN=m3 0 frac14 34

                                                                  H frac14 350m nH frac14 335m mH frac14 185m

                                                                  Consider a trial value of F frac14 20 Refer to Figure 635

                                                                  0m frac14 tan1tan 34

                                                                  20

                                                                  frac14 186

                                                                  Then

                                                                  frac14 45 thorn 0m2frac14 543

                                                                  W frac14 1

                                                                  2 18 3502 cot 543 frac14 792 kN=m

                                                                  Figure Q610

                                                                  46 Lateral earth pressure

                                                                  P frac14 1

                                                                  2 s 3352 frac14 561s kN=m

                                                                  U frac14 1

                                                                  2 98 1852 cosec 543 frac14 206 kN=m

                                                                  Equations 630 and 631 then become

                                                                  561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                  792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                  ie

                                                                  561s 0616N 405 frac14 0

                                                                  792 0857N thorn 563 frac14 0

                                                                  N frac14 848

                                                                  0857frac14 989 kN=m

                                                                  Then

                                                                  561s 609 405 frac14 0

                                                                  s frac14 649

                                                                  561frac14 116 kN=m3

                                                                  The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                  F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                  20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                  s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                  Figure Q611

                                                                  Lateral earth pressure 47

                                                                  612

                                                                  For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                  45 thorn 0

                                                                  2frac14 63

                                                                  For the retained material between the surface and a depth of 36m

                                                                  Pa frac14 1

                                                                  2 030 18 362 frac14 350 kN=m

                                                                  Weight of reinforced fill between the surface and a depth of 36m is

                                                                  Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                  eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                  Eccentricity of Rv

                                                                  e frac14 263 250 frac14 013m

                                                                  The average vertical stress at a depth of 36m is

                                                                  z frac14 Rv

                                                                  L 2efrac14 324

                                                                  474frac14 68 kN=m2

                                                                  (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                  Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                  Tensile stress in the element frac14 138 103

                                                                  65 3frac14 71N=mm2

                                                                  Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                  Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                  Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                  The weight of ABC is

                                                                  W frac14 1

                                                                  2 18 52 265 frac14 124 kN=m

                                                                  From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                  48 Lateral earth pressure

                                                                  (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                  Tp frac14 032 68 120 065 frac14 170 kN

                                                                  Tr frac14 213 420

                                                                  418frac14 214 kN

                                                                  Again the tensile failure and slipping limit states are satisfied for this element

                                                                  Figure Q612

                                                                  Lateral earth pressure 49

                                                                  Chapter 7

                                                                  Consolidation theory

                                                                  71

                                                                  Total change in thickness

                                                                  H frac14 782 602 frac14 180mm

                                                                  Average thickness frac14 1530thorn 180

                                                                  2frac14 1620mm

                                                                  Length of drainage path d frac14 1620

                                                                  2frac14 810mm

                                                                  Root time plot (Figure Q71a)

                                                                  ffiffiffiffiffiffit90p frac14 33

                                                                  t90 frac14 109min

                                                                  cv frac14 0848d2

                                                                  t90frac14 0848 8102

                                                                  109 1440 365

                                                                  106frac14 27m2=year

                                                                  r0 frac14 782 764

                                                                  782 602frac14 018

                                                                  180frac14 0100

                                                                  rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                  10 119

                                                                  9 180frac14 0735

                                                                  rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                  Log time plot (Figure Q71b)

                                                                  t50 frac14 26min

                                                                  cv frac14 0196d2

                                                                  t50frac14 0196 8102

                                                                  26 1440 365

                                                                  106frac14 26m2=year

                                                                  r0 frac14 782 763

                                                                  782 602frac14 019

                                                                  180frac14 0106

                                                                  rp frac14 763 623

                                                                  782 602frac14 140

                                                                  180frac14 0778

                                                                  rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                  Figure Q71(a)

                                                                  Figure Q71(b)

                                                                  Final void ratio

                                                                  e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                  e

                                                                  Hfrac14 1thorn e0

                                                                  H0frac14 1thorn e1 thorne

                                                                  H0

                                                                  ie

                                                                  e

                                                                  180frac14 1631thorne

                                                                  1710

                                                                  e frac14 2936

                                                                  1530frac14 0192

                                                                  Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                  mv frac14 1

                                                                  1thorn e0 e0 e101 00

                                                                  frac14 1

                                                                  1823 0192

                                                                  0107frac14 098m2=MN

                                                                  k frac14 cvmvw frac14 265 098 98

                                                                  60 1440 365 103frac14 81 1010 m=s

                                                                  72

                                                                  Using Equation 77 (one-dimensional method)

                                                                  sc frac14 e0 e11thorn e0 H

                                                                  Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                  Figure Q72

                                                                  52 Consolidation theory

                                                                  Settlement

                                                                  Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                  1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                  318

                                                                  Notes 5 92y 460thorn 84

                                                                  Heave

                                                                  Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                  1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                  38

                                                                  73

                                                                  U frac14 f ethTvTHORN frac14 f cvt

                                                                  d2

                                                                  Hence if cv is constant

                                                                  t1

                                                                  t2frac14 d

                                                                  21

                                                                  d22

                                                                  where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                  d1 frac14 95mm and d2 frac14 2500mm

                                                                  for U frac14 050 t2 frac14 t1 d22

                                                                  d21

                                                                  frac14 20

                                                                  60 24 365 25002

                                                                  952frac14 263 years

                                                                  for U lt 060 Tv frac14

                                                                  4U2 (Equation 724(a))

                                                                  t030 frac14 t050 0302

                                                                  0502

                                                                  frac14 263 036 frac14 095 years

                                                                  Consolidation theory 53

                                                                  74

                                                                  The layer is open

                                                                  d frac14 8

                                                                  2frac14 4m

                                                                  Tv frac14 cvtd2frac14 24 3

                                                                  42frac14 0450

                                                                  ui frac14 frac14 84 kN=m2

                                                                  The excess pore water pressure is given by Equation 721

                                                                  ue frac14Xmfrac141mfrac140

                                                                  2ui

                                                                  Msin

                                                                  Mz

                                                                  d

                                                                  expethM2TvTHORN

                                                                  In this case z frac14 d

                                                                  sinMz

                                                                  d

                                                                  frac14 sinM

                                                                  where

                                                                  M frac14

                                                                  23

                                                                  25

                                                                  2

                                                                  M sin M M2Tv exp (M2Tv)

                                                                  2thorn1 1110 0329

                                                                  3

                                                                  21 9993 457 105

                                                                  ue frac14 2 84 2

                                                                  1 0329 ethother terms negligibleTHORN

                                                                  frac14 352 kN=m2

                                                                  75

                                                                  The layer is open

                                                                  d frac14 6

                                                                  2frac14 3m

                                                                  Tv frac14 cvtd2frac14 10 3

                                                                  32frac14 0333

                                                                  The layer thickness will be divided into six equal parts ie m frac14 6

                                                                  54 Consolidation theory

                                                                  For an open layer

                                                                  Tv frac14 4n

                                                                  m2

                                                                  n frac14 0333 62

                                                                  4frac14 300

                                                                  The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                  ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                  i j

                                                                  0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                  0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                  The initial and 3-year isochrones are plotted in Figure Q75

                                                                  Area under initial isochrone frac14 180 units

                                                                  Area under 3-year isochrone frac14 63 units

                                                                  The average degree of consolidation is given by Equation 725Thus

                                                                  U frac14 1 63

                                                                  180frac14 065

                                                                  Figure Q75

                                                                  Consolidation theory 55

                                                                  76

                                                                  At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                  0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                  The final consolidation settlement (one-dimensional method) is

                                                                  sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                  Corrected time t frac14 2 1

                                                                  2

                                                                  40

                                                                  52

                                                                  frac14 1615 years

                                                                  Tv frac14 cvtd2frac14 44 1615

                                                                  42frac14 0444

                                                                  From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                  77

                                                                  The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                  Figure Q77

                                                                  56 Consolidation theory

                                                                  Point m n Ir (kNm2) sc (mm)

                                                                  13020frac14 15 20

                                                                  20frac14 10 0194 (4) 113 124

                                                                  260

                                                                  20frac14 30

                                                                  20

                                                                  20frac14 10 0204 (2) 59 65

                                                                  360

                                                                  20frac14 30

                                                                  40

                                                                  20frac14 20 0238 (1) 35 38

                                                                  430

                                                                  20frac14 15

                                                                  40

                                                                  20frac14 20 0224 (2) 65 72

                                                                  Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                  78

                                                                  Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                  (a) Immediate settlement

                                                                  H

                                                                  Bfrac14 30

                                                                  35frac14 086

                                                                  D

                                                                  Bfrac14 2

                                                                  35frac14 006

                                                                  Figure Q78

                                                                  Consolidation theory 57

                                                                  From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                  si frac14 130131qB

                                                                  Eufrac14 10 032 105 35

                                                                  40frac14 30mm

                                                                  (b) Consolidation settlement

                                                                  Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                  1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                  3150

                                                                  Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                  Now

                                                                  H

                                                                  Bfrac14 30

                                                                  35frac14 086 and A frac14 065

                                                                  from Figure 712 13 frac14 079

                                                                  sc frac14 13sod frac14 079 315 frac14 250mm

                                                                  Total settlement

                                                                  s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                  79

                                                                  Without sand drains

                                                                  Uv frac14 025

                                                                  Tv frac14 0049 ethfrom Figure 718THORN

                                                                  t frac14 Tvd2

                                                                  cvfrac14 0049 82

                                                                  cvWith sand drains

                                                                  R frac14 0564S frac14 0564 3 frac14 169m

                                                                  n frac14 Rrfrac14 169

                                                                  015frac14 113

                                                                  Tr frac14 cht

                                                                  4R2frac14 ch

                                                                  4 1692 0049 82

                                                                  cvethand ch frac14 cvTHORN

                                                                  frac14 0275

                                                                  Ur frac14 073 (from Figure 730)

                                                                  58 Consolidation theory

                                                                  Using Equation 740

                                                                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                  U frac14 080

                                                                  710

                                                                  Without sand drains

                                                                  Uv frac14 090

                                                                  Tv frac14 0848

                                                                  t frac14 Tvd2

                                                                  cvfrac14 0848 102

                                                                  96frac14 88 years

                                                                  With sand drains

                                                                  R frac14 0564S frac14 0564 4 frac14 226m

                                                                  n frac14 Rrfrac14 226

                                                                  015frac14 15

                                                                  Tr

                                                                  Tvfrac14 chcv

                                                                  d2

                                                                  4R2ethsame tTHORN

                                                                  Tr

                                                                  Tvfrac14 140

                                                                  96 102

                                                                  4 2262frac14 714 eth1THORN

                                                                  Using Equation 740

                                                                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                  An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                  Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                  040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                  Thus

                                                                  Uv frac14 0295 and Ur frac14 086

                                                                  t frac14 88 00683

                                                                  0848frac14 07 years

                                                                  Consolidation theory 59

                                                                  Chapter 8

                                                                  Bearing capacity

                                                                  81

                                                                  (a) The ultimate bearing capacity is given by Equation 83

                                                                  qf frac14 cNc thorn DNq thorn 1

                                                                  2BN

                                                                  For u frac14 0

                                                                  Nc frac14 514 Nq frac14 1 N frac14 0

                                                                  qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                  The net ultimate bearing capacity is

                                                                  qnf frac14 qf D frac14 540 kN=m2

                                                                  The net foundation pressure is

                                                                  qn frac14 q D frac14 425

                                                                  2 eth21 1THORN frac14 192 kN=m2

                                                                  The factor of safety (Equation 86) is

                                                                  F frac14 qnfqnfrac14 540

                                                                  192frac14 28

                                                                  (b) For 0 frac14 28

                                                                  Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                  qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                  2 112 2 13

                                                                  frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                  qnf frac14 574 112 frac14 563 kN=m2

                                                                  F frac14 563

                                                                  192frac14 29

                                                                  (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                  82

                                                                  For 0 frac14 38

                                                                  Nq frac14 49 N frac14 67

                                                                  qnf frac14 DethNq 1THORN thorn 1

                                                                  2BN ethfrom Equation 83THORN

                                                                  frac14 eth18 075 48THORN thorn 1

                                                                  2 18 15 67

                                                                  frac14 648thorn 905 frac14 1553 kN=m2

                                                                  qn frac14 500

                                                                  15 eth18 075THORN frac14 320 kN=m2

                                                                  F frac14 qnfqnfrac14 1553

                                                                  320frac14 48

                                                                  0d frac14 tan1tan 38

                                                                  125

                                                                  frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                  Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                  2 18 15 25

                                                                  frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                  Design load (action) Vd frac14 500 kN=m

                                                                  The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                  83

                                                                  D

                                                                  Bfrac14 350

                                                                  225frac14 155

                                                                  From Figure 85 for a square foundation

                                                                  Nc frac14 81

                                                                  Bearing capacity 61

                                                                  For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                  Nc frac14 084thorn 016B

                                                                  L

                                                                  81 frac14 745

                                                                  Using Equation 810

                                                                  qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                  For F frac14 3

                                                                  qn frac14 1006

                                                                  3frac14 335 kN=m2

                                                                  q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                  Design load frac14 405 450 225 frac14 4100 kN

                                                                  Design undrained strength cud frac14 135

                                                                  14frac14 96 kN=m2

                                                                  Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                  frac14 7241 kN

                                                                  Design load Vd frac14 4100 kN

                                                                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                  84

                                                                  For 0 frac14 40

                                                                  Nq frac14 64 N frac14 95

                                                                  qnf frac14 DethNq 1THORN thorn 04BN

                                                                  (a) Water table 5m below ground level

                                                                  qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                  qn frac14 400 17 frac14 383 kN=m2

                                                                  F frac14 2686

                                                                  383frac14 70

                                                                  (b) Water table 1m below ground level (ie at foundation level)

                                                                  0 frac14 20 98 frac14 102 kN=m3

                                                                  62 Bearing capacity

                                                                  qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                  F frac14 2040

                                                                  383frac14 53

                                                                  (c) Water table at ground level with upward hydraulic gradient 02

                                                                  eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                  qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                  F frac14 1296

                                                                  392frac14 33

                                                                  85

                                                                  The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                  Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                  Design value of 0 frac14 tan1tan 39

                                                                  125

                                                                  frac14 33

                                                                  For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                  Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                  86

                                                                  (a) Undrained shear for u frac14 0

                                                                  Nc frac14 514 Nq frac14 1 N frac14 0

                                                                  qnf frac14 12cuNc

                                                                  frac14 12 100 514 frac14 617 kN=m2

                                                                  qn frac14 qnfFfrac14 617

                                                                  3frac14 206 kN=m2

                                                                  q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                  Bearing capacity 63

                                                                  Drained shear for 0 frac14 32

                                                                  Nq frac14 23 N frac14 25

                                                                  0 frac14 21 98 frac14 112 kN=m3

                                                                  qnf frac14 0DethNq 1THORN thorn 040BN

                                                                  frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                  frac14 694 kN=m2

                                                                  q frac14 694

                                                                  3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                  Design load frac14 42 227 frac14 3632 kN

                                                                  (b) Design undrained strength cud frac14 100

                                                                  14frac14 71 kNm2

                                                                  Design bearing resistance Rd frac14 12cudNe area

                                                                  frac14 12 71 514 42

                                                                  frac14 7007 kN

                                                                  For drained shear 0d frac14 tan1tan 32

                                                                  125

                                                                  frac14 26

                                                                  Nq frac14 12 N frac14 10

                                                                  Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                  (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                  Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                  1 2 100 0175 0700qn 0182qn

                                                                  2 6 033 0044 0176qn 0046qn

                                                                  3 10 020 0017 0068qn 0018qn

                                                                  0246qn

                                                                  Diameter of equivalent circle B frac14 45m

                                                                  H

                                                                  Bfrac14 12

                                                                  45frac14 27 and A frac14 042

                                                                  13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                  64 Bearing capacity

                                                                  For sc frac14 30mm

                                                                  qn frac14 30

                                                                  0147frac14 204 kN=m2

                                                                  q frac14 204thorn 21 frac14 225 kN=m2

                                                                  Design load frac14 42 225 frac14 3600 kN

                                                                  The design load is 3600 kN settlement being the limiting criterion

                                                                  87

                                                                  D

                                                                  Bfrac14 8

                                                                  4frac14 20

                                                                  From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                  F frac14 cuNc

                                                                  Dfrac14 40 71

                                                                  20 8frac14 18

                                                                  88

                                                                  Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                  Design value of 0 frac14 tan1tan 38

                                                                  125

                                                                  frac14 32

                                                                  Figure Q86

                                                                  Bearing capacity 65

                                                                  For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                  Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                  The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                  Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                  For B frac14 250m qn frac14 3750

                                                                  2502 17 frac14 583 kN=m2

                                                                  From Figure 510 m frac14 n frac14 126

                                                                  6frac14 021

                                                                  Ir frac14 0019

                                                                  Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                  Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                  The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                  89

                                                                  Depth (m) N 0v (kNm2) CN N1

                                                                  070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                  Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                  (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                  Cw frac14 05thorn 0530

                                                                  47

                                                                  frac14 082

                                                                  66 Bearing capacity

                                                                  Thus

                                                                  qa frac14 150 082 frac14 120 kN=m2

                                                                  (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                  Thus

                                                                  qa frac14 90 15 frac14 135 kN=m2

                                                                  (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                  Ic frac14 171

                                                                  1014frac14 0068

                                                                  From Equation 819(a) with s frac14 25mm

                                                                  q frac14 25

                                                                  3507 0068frac14 150 kN=m2

                                                                  810

                                                                  Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                  Izp frac14 05thorn 01130

                                                                  16 27

                                                                  05

                                                                  frac14 067

                                                                  Refer to Figure Q810

                                                                  E frac14 25qc

                                                                  Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                  Ez (mm3MN)

                                                                  1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                  0203

                                                                  C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                  130frac14 093

                                                                  C2 frac14 1 ethsayTHORN

                                                                  s frac14 C1C2qnX Iz

                                                                  Ez frac14 093 1 130 0203 frac14 25mm

                                                                  Bearing capacity 67

                                                                  811

                                                                  At pile base level

                                                                  cu frac14 220 kN=m2

                                                                  qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                  Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                  00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                  qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                  Then

                                                                  Qf frac14 Abqb thorn Asqs

                                                                  frac14

                                                                  4 32 1980

                                                                  thorn eth 105 139 86THORN

                                                                  frac14 13 996thorn 3941 frac14 17 937 kN

                                                                  0 01 02 03 04 05 06 07

                                                                  0 2 4 6 8 10 12 14

                                                                  1

                                                                  2

                                                                  3

                                                                  4

                                                                  5

                                                                  6

                                                                  7

                                                                  8

                                                                  (1)

                                                                  (2)

                                                                  (3)

                                                                  (4)

                                                                  (5)

                                                                  qc

                                                                  qc

                                                                  Iz

                                                                  Iz

                                                                  (MNm2)

                                                                  z (m)

                                                                  Figure Q810

                                                                  68 Bearing capacity

                                                                  Allowable load

                                                                  ethaTHORN Qf

                                                                  2frac14 17 937

                                                                  2frac14 8968 kN

                                                                  ethbTHORN Abqb

                                                                  3thorn Asqs frac14 13 996

                                                                  3thorn 3941 frac14 8606 kN

                                                                  ie allowable load frac14 8600 kN

                                                                  Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                  According to the limit state method

                                                                  Characteristic undrained strength at base level cuk frac14 220

                                                                  150kN=m2

                                                                  Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                  150frac14 1320 kN=m2

                                                                  Characteristic shaft resistance qsk frac14 00150

                                                                  frac14 86

                                                                  150frac14 57 kN=m2

                                                                  Characteristic base and shaft resistances

                                                                  Rbk frac14

                                                                  4 32 1320 frac14 9330 kN

                                                                  Rsk frac14 105 139 86

                                                                  150frac14 2629 kN

                                                                  For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                  Design bearing resistance Rcd frac14 9330

                                                                  160thorn 2629

                                                                  130

                                                                  frac14 5831thorn 2022

                                                                  frac14 7850 kN

                                                                  Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                  812

                                                                  ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                  qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                  For a single pile

                                                                  Qf frac14 Abqb thorn Asqs

                                                                  frac14

                                                                  4 062 1305

                                                                  thorn eth 06 15 42THORN

                                                                  frac14 369thorn 1187 frac14 1556 kN

                                                                  Bearing capacity 69

                                                                  Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                  eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                  (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                  (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                  qbkfrac14 9cuk frac14 9 220

                                                                  150frac14 1320 kN=m2

                                                                  qskfrac14cuk frac14 040 105

                                                                  150frac14 28 kN=m2

                                                                  Rbkfrac14

                                                                  4 0602 1320 frac14 373 kN

                                                                  Rskfrac14 060 15 28 frac14 791 kN

                                                                  Rcdfrac14 373

                                                                  160thorn 791

                                                                  130frac14 233thorn 608 frac14 841 kN

                                                                  Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                  Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                  (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                  q frac14 21 000

                                                                  1762frac14 68 kN=m2

                                                                  Immediate settlement

                                                                  H

                                                                  Bfrac14 15

                                                                  176frac14 085

                                                                  D

                                                                  Bfrac14 13

                                                                  176frac14 074

                                                                  L

                                                                  Bfrac14 1

                                                                  Hence from Figure 515

                                                                  130 frac14 078 and 131 frac14 041

                                                                  70 Bearing capacity

                                                                  Thus using Equation 528

                                                                  si frac14 078 041 68 176

                                                                  65frac14 6mm

                                                                  Consolidation settlement

                                                                  Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                  1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                  434 (sod)

                                                                  Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                  sc frac14 056 434 frac14 24mm

                                                                  The total settlement is (6thorn 24) frac14 30mm

                                                                  813

                                                                  At base level N frac14 26 Then using Equation 830

                                                                  qb frac14 40NDb

                                                                  Bfrac14 40 26 2

                                                                  025frac14 8320 kN=m2

                                                                  ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                  Figure Q812

                                                                  Bearing capacity 71

                                                                  Over the length embedded in sand

                                                                  N frac14 21 ie18thorn 24

                                                                  2

                                                                  Using Equation 831

                                                                  qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                  For a single pile

                                                                  Qf frac14 Abqb thorn Asqs

                                                                  frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                  For the pile group assuming a group efficiency of 12

                                                                  XQf frac14 12 9 604 frac14 6523 kN

                                                                  Then the load factor is

                                                                  F frac14 6523

                                                                  2000thorn 1000frac14 21

                                                                  (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                  Characteristic base resistance per unit area qbk frac14 8320

                                                                  150frac14 5547 kNm2

                                                                  Characteristic shaft resistance per unit area qsk frac14 42

                                                                  150frac14 28 kNm2

                                                                  Characteristic base and shaft resistances for a single pile

                                                                  Rbk frac14 0252 5547 frac14 347 kN

                                                                  Rsk frac14 4 025 2 28 frac14 56 kN

                                                                  For a driven pile the partial factors are b frac14 s frac14 130

                                                                  Design bearing resistance Rcd frac14 347

                                                                  130thorn 56

                                                                  130frac14 310 kN

                                                                  For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                  Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                  (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                  From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                  72 Bearing capacity

                                                                  N frac14 24thorn 26thorn 34

                                                                  3frac14 28

                                                                  Ic frac14 171

                                                                  2814frac14 0016 ethEquation 818THORN

                                                                  s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                  814

                                                                  Using Equation 841

                                                                  Tf frac14 DLcu thorn

                                                                  4ethD2 d2THORNcuNc

                                                                  frac14 eth 02 5 06 110THORN thorn

                                                                  4eth022 012THORN110 9

                                                                  frac14 207thorn 23 frac14 230 kN

                                                                  Figure Q813

                                                                  Bearing capacity 73

                                                                  Chapter 9

                                                                  Stability of slopes

                                                                  91

                                                                  Referring to Figure Q91

                                                                  W frac14 417 19 frac14 792 kN=m

                                                                  Q frac14 20 28 frac14 56 kN=m

                                                                  Arc lengthAB frac14

                                                                  180 73 90 frac14 115m

                                                                  Arc length BC frac14

                                                                  180 28 90 frac14 44m

                                                                  The factor of safety is given by

                                                                  F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                  90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                  Depth of tension crack z0 frac14 2cu

                                                                  frac14 2 20

                                                                  19frac14 21m

                                                                  Arc length BD frac14

                                                                  180 13

                                                                  1

                                                                  2 90 frac14 21m

                                                                  F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                  The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                  Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                  14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                  Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                  The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                  92

                                                                  u frac14 0

                                                                  Depth factor D frac14 11

                                                                  9frac14 122

                                                                  Using Equation 92 with F frac14 10

                                                                  Ns frac14 cu

                                                                  FHfrac14 30

                                                                  10 19 9frac14 0175

                                                                  Hence from Figure 93

                                                                  frac14 50

                                                                  For F frac14 12

                                                                  Ns frac14 30

                                                                  12 19 9frac14 0146

                                                                  frac14 27

                                                                  93

                                                                  Refer to Figure Q93

                                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                  1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                  74 m

                                                                  214 1deg

                                                                  213 1deg

                                                                  39 m

                                                                  WB

                                                                  D

                                                                  C

                                                                  28 m

                                                                  21 m

                                                                  A

                                                                  Q

                                                                  Soil (1)Soil (2)

                                                                  73deg

                                                                  Figure Q91

                                                                  Stability of slopes 75

                                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                  9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                  599 256 328 1372

                                                                  Figure Q93

                                                                  76 Stability of slopes

                                                                  XW cos frac14 b

                                                                  Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                  W sin frac14 bX

                                                                  h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                  Arc length La frac14

                                                                  180 57

                                                                  1

                                                                  2 326 frac14 327m

                                                                  The factor of safety is given by

                                                                  F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                  W sin

                                                                  frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                  frac14 091

                                                                  According to the limit state method

                                                                  0d frac14 tan1tan 32

                                                                  125

                                                                  frac14 265

                                                                  c0 frac14 8

                                                                  160frac14 5 kN=m2

                                                                  Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                  Design disturbing moment frac14 1075 kN=m

                                                                  The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                  94

                                                                  F frac14 1

                                                                  W sin

                                                                  Xfc0bthorn ethW ubTHORN tan0g sec

                                                                  1thorn ethtan tan0=FTHORN

                                                                  c0 frac14 8 kN=m2

                                                                  0 frac14 32

                                                                  c0b frac14 8 2 frac14 16 kN=m

                                                                  W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                  Try F frac14 100

                                                                  tan0

                                                                  Ffrac14 0625

                                                                  Stability of slopes 77

                                                                  Values of u are as obtained in Figure Q93

                                                                  SliceNo

                                                                  h(m)

                                                                  W frac14 bh(kNm)

                                                                  W sin(kNm)

                                                                  ub(kNm)

                                                                  c0bthorn (W ub) tan0(kNm)

                                                                  sec

                                                                  1thorn (tan tan0)FProduct(kNm)

                                                                  1 05 21 6 2 8 24 1078 262 13 55 31

                                                                  23 33 30 1042 31

                                                                  3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                  224 92 72 0931 67

                                                                  6 50 210 11 40 100 85 0907 777 55 231 14

                                                                  12 58 112 90 0889 80

                                                                  8 60 252 1812

                                                                  80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                  10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                  2154 88 116 0853 99

                                                                  14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                  1074 1091

                                                                  F frac14 1091

                                                                  1074frac14 102 (assumed value 100)

                                                                  Thus

                                                                  F frac14 101

                                                                  95

                                                                  F frac14 1

                                                                  W sin

                                                                  XfWeth1 ruTHORN tan0g sec

                                                                  1thorn ethtan tan0THORN=F

                                                                  0 frac14 33

                                                                  ru frac14 020

                                                                  W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                  eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                  Try F frac14 110

                                                                  tan 0

                                                                  Ffrac14 tan 33

                                                                  110frac14 0590

                                                                  78 Stability of slopes

                                                                  Referring to Figure Q95

                                                                  SliceNo

                                                                  h(m)

                                                                  W frac14 bh(kNm)

                                                                  W sin(kNm)

                                                                  W(1 ru) tan0(kNm)

                                                                  sec

                                                                  1thorn ( tan tan0)FProduct(kNm)

                                                                  1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                  2120 234 0892 209

                                                                  4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                  1185 1271

                                                                  Figure Q95

                                                                  Stability of slopes 79

                                                                  F frac14 1271

                                                                  1185frac14 107

                                                                  The trial value was 110 therefore take F to be 108

                                                                  96

                                                                  (a) Water table at surface the factor of safety is given by Equation 912

                                                                  F frac14 0

                                                                  sat

                                                                  tan0

                                                                  tan

                                                                  ptie 15 frac14 92

                                                                  19

                                                                  tan 36

                                                                  tan

                                                                  tan frac14 0234

                                                                  frac14 13

                                                                  Water table well below surface the factor of safety is given by Equation 911

                                                                  F frac14 tan0

                                                                  tan

                                                                  frac14 tan 36

                                                                  tan 13

                                                                  frac14 31

                                                                  (b) 0d frac14 tan1tan 36

                                                                  125

                                                                  frac14 30

                                                                  Depth of potential failure surface frac14 z

                                                                  Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                  frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                  frac14 504z kN

                                                                  Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                  frac14 19 z sin 13 cos 13

                                                                  frac14 416z kN

                                                                  Rd gtSd therefore the limit state for overall stability is satisfied

                                                                  80 Stability of slopes

                                                                  • Book Cover
                                                                  • Title
                                                                  • Contents
                                                                  • Basic characteristics of soils
                                                                  • Seepage
                                                                  • Effective stress
                                                                  • Shear strength
                                                                  • Stresses and displacements
                                                                  • Lateral earth pressure
                                                                  • Consolidation theory
                                                                  • Bearing capacity
                                                                  • Stability of slopes

                                                                    Chapter 5

                                                                    Stresses and displacements

                                                                    51

                                                                    Vertical stress is given by

                                                                    z frac14 Qz2Ip frac14 5000

                                                                    52Ip

                                                                    Values of Ip are obtained from Table 51

                                                                    r (m) rz Ip z (kNm2)

                                                                    0 0 0478 961 02 0433 872 04 0329 663 06 0221 444 08 0139 285 10 0084 177 14 0032 6

                                                                    10 20 0009 2

                                                                    The variation of z with radial distance (r) is plotted in Figure Q51

                                                                    Figure Q51

                                                                    52

                                                                    Below the centre load (Figure Q52)

                                                                    r

                                                                    zfrac14 0 for the 7500-kN load

                                                                    Ip frac14 0478

                                                                    r

                                                                    zfrac14 5

                                                                    4frac14 125 for the 10 000- and 9000-kN loads

                                                                    Ip frac14 0045

                                                                    Then

                                                                    z frac14X Q

                                                                    z2Ip

                                                                    frac14 7500 0478

                                                                    42thorn 10 000 0045

                                                                    42thorn 9000 0045

                                                                    42

                                                                    frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                                                    53

                                                                    The vertical stress under a corner of a rectangular area is given by

                                                                    z frac14 qIr

                                                                    where values of Ir are obtained from Figure 510 In this case

                                                                    z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                                                    z

                                                                    Figure Q52

                                                                    Stresses and displacements 29

                                                                    z (m) m n Ir z (kNm2)

                                                                    0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                                                    10 010 0005 5

                                                                    z is plotted against z in Figure Q53

                                                                    54

                                                                    (a)

                                                                    m frac14 125

                                                                    12frac14 104

                                                                    n frac14 18

                                                                    12frac14 150

                                                                    From Figure 510 Irfrac14 0196

                                                                    z frac14 2 175 0196 frac14 68 kN=m2

                                                                    Figure Q53

                                                                    30 Stresses and displacements

                                                                    (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                                                    z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                                                    55

                                                                    Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                                                    Px frac14 2Q

                                                                    1

                                                                    m2 thorn 1frac14 2 150

                                                                    125frac14 76 kN=m

                                                                    Equation 517 is used to obtain the pressure distribution

                                                                    px frac14 4Q

                                                                    h

                                                                    m2n

                                                                    ethm2 thorn n2THORN2 frac14150

                                                                    m2n

                                                                    ethm2 thorn n2THORN2 ethkN=m2THORN

                                                                    Figure Q54

                                                                    Stresses and displacements 31

                                                                    n m2n

                                                                    (m2 thorn n2)2

                                                                    px(kNm2)

                                                                    0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                                    The pressure distribution is plotted in Figure Q55

                                                                    56

                                                                    H

                                                                    Bfrac14 10

                                                                    2frac14 5

                                                                    L

                                                                    Bfrac14 4

                                                                    2frac14 2

                                                                    D

                                                                    Bfrac14 1

                                                                    2frac14 05

                                                                    Hence from Figure 515

                                                                    131 frac14 082

                                                                    130 frac14 094

                                                                    Figure Q55

                                                                    32 Stresses and displacements

                                                                    The immediate settlement is given by Equation 528

                                                                    si frac14 130131qB

                                                                    Eu

                                                                    frac14 094 082 200 2

                                                                    45frac14 7mm

                                                                    Stresses and displacements 33

                                                                    Chapter 6

                                                                    Lateral earth pressure

                                                                    61

                                                                    For 0 frac14 37 the active pressure coefficient is given by

                                                                    Ka frac14 1 sin 37

                                                                    1thorn sin 37frac14 025

                                                                    The total active thrust (Equation 66a with c0 frac14 0) is

                                                                    Pa frac14 1

                                                                    2KaH

                                                                    2 frac14 1

                                                                    2 025 17 62 frac14 765 kN=m

                                                                    If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                                    K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                                    and the thrust on the wall is

                                                                    P0 frac14 1

                                                                    2K0H

                                                                    2 frac14 1

                                                                    2 040 17 62 frac14 122 kN=m

                                                                    62

                                                                    The active pressure coefficients for the three soil types are as follows

                                                                    Ka1 frac141 sin 35

                                                                    1thorn sin 35frac14 0271

                                                                    Ka2 frac141 sin 27

                                                                    1thorn sin 27frac14 0375

                                                                    ffiffiffiffiffiffiffiKa2

                                                                    p frac14 0613

                                                                    Ka3 frac141 sin 42

                                                                    1thorn sin 42frac14 0198

                                                                    Distribution of active pressure (plotted in Figure Q62)

                                                                    Depth (m) Soil Active pressure (kNm2)

                                                                    3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                                    12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                                    At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                                    Total thrust frac14 571 kNm

                                                                    Point of application is (4893571) m from the top of the wall ie 857m

                                                                    Force (kN) Arm (m) Moment (kN m)

                                                                    (1)1

                                                                    2 0271 16 32 frac14 195 20 390

                                                                    (2) 0271 16 3 2 frac14 260 40 1040

                                                                    (3)1

                                                                    2 0271 92 22 frac14 50 433 217

                                                                    (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                                    (5)1

                                                                    2 0375 102 32 frac14 172 70 1204

                                                                    (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                                    (7)1

                                                                    2 0198 112 42 frac14 177 1067 1889

                                                                    (8)1

                                                                    2 98 92 frac14 3969 90 35721

                                                                    5713 48934

                                                                    Figure Q62

                                                                    Lateral earth pressure 35

                                                                    63

                                                                    (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                                    Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                                    frac14 245

                                                                    At the lower end of the piling

                                                                    pa frac14 Kaqthorn Kasatz Kaccu

                                                                    frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                                    frac14 115 kN=m2

                                                                    pp frac14 Kpsatzthorn Kpccu

                                                                    frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                                    frac14 202 kN=m2

                                                                    (b) For 0 frac14 26 and frac14 1

                                                                    20

                                                                    Ka frac14 035

                                                                    Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                                    pfrac14 145 ethEquation 619THORN

                                                                    Kp frac14 37

                                                                    Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                                    pfrac14 47 ethEquation 624THORN

                                                                    At the lower end of the piling

                                                                    pa frac14 Kaqthorn Ka0z Kacc

                                                                    0

                                                                    frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                                    frac14 187 kN=m2

                                                                    pp frac14 Kp0zthorn Kpcc

                                                                    0

                                                                    frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                                    frac14 198 kN=m2

                                                                    36 Lateral earth pressure

                                                                    64

                                                                    (a) For 0 frac14 38 Ka frac14 024

                                                                    0 frac14 20 98 frac14 102 kN=m3

                                                                    The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                                    Force (kN) Arm (m) Moment (kN m)

                                                                    (1) 024 10 66 frac14 159 33 525

                                                                    (2)1

                                                                    2 024 17 392 frac14 310 400 1240

                                                                    (3) 024 17 39 27 frac14 430 135 580

                                                                    (4)1

                                                                    2 024 102 272 frac14 89 090 80

                                                                    (5)1

                                                                    2 98 272 frac14 357 090 321

                                                                    Hfrac14 1345 MH frac14 2746

                                                                    (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                                    (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                                    XM frac14MV MH frac14 7790 kNm

                                                                    Lever arm of base resultant

                                                                    M

                                                                    Vfrac14 779

                                                                    488frac14 160

                                                                    Eccentricity of base resultant

                                                                    e frac14 200 160 frac14 040m

                                                                    39 m

                                                                    27 m

                                                                    40 m

                                                                    04 m

                                                                    04 m

                                                                    26 m

                                                                    (7)

                                                                    (9)

                                                                    (1)(2)

                                                                    (3)

                                                                    (4)

                                                                    (5)

                                                                    (8)(6)

                                                                    (10)

                                                                    WT

                                                                    10 kNm2

                                                                    Hydrostatic

                                                                    Figure Q64

                                                                    Lateral earth pressure 37

                                                                    Base pressures (Equation 627)

                                                                    p frac14 VB

                                                                    1 6e

                                                                    B

                                                                    frac14 488

                                                                    4eth1 060THORN

                                                                    frac14 195 kN=m2 and 49 kN=m2

                                                                    Factor of safety against sliding (Equation 628)

                                                                    F frac14 V tan

                                                                    Hfrac14 488 tan 25

                                                                    1345frac14 17

                                                                    (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                                    Hfrac14 1633 kN

                                                                    V frac14 4879 kN

                                                                    MH frac14 3453 kNm

                                                                    MV frac14 10536 kNm

                                                                    The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                                    65

                                                                    For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                    Kp

                                                                    Ffrac14 385

                                                                    2

                                                                    0 frac14 20 98 frac14 102 kN=m3

                                                                    The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                                    Force (kN) Arm (m) Moment (kN m)

                                                                    (1)1

                                                                    2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                                    (2) 026 17 45 d frac14 199d d2 995d2

                                                                    (3)1

                                                                    2 026 102 d2 frac14 133d2 d3 044d3

                                                                    (4)1

                                                                    2 385

                                                                    2 17 152 frac14 368 dthorn 05 368d 184

                                                                    (5)385

                                                                    2 17 15 d frac14 491d d2 2455d2

                                                                    (6)1

                                                                    2 385

                                                                    2 102 d2 frac14 982d2 d3 327d3

                                                                    38 Lateral earth pressure

                                                                    XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                                    d3 thorn 516d2 283d 1724 frac14 0

                                                                    d frac14 179m

                                                                    Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                                    Over additional 20 embedded depth

                                                                    pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                                    Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                                    66

                                                                    The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                                    Ka frac14 sin 69=sin 105

                                                                    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                                    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                                    pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                                    26664

                                                                    37775

                                                                    2

                                                                    frac14 050

                                                                    The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                                    Pa frac14 1

                                                                    2 050 19 7502 frac14 267 kN=m

                                                                    Figure Q65

                                                                    Lateral earth pressure 39

                                                                    Horizontal component

                                                                    Ph frac14 267 cos 40 frac14 205 kN=m

                                                                    Vertical component

                                                                    Pv frac14 267 sin 40 frac14 172 kN=m

                                                                    Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                    Force (kN) Arm (m) Moment (kN m)

                                                                    (1)1

                                                                    2 175 650 235 frac14 1337 258 345

                                                                    (2) 050 650 235 frac14 764 175 134

                                                                    (3)1

                                                                    2 070 650 235 frac14 535 127 68

                                                                    (4) 100 400 235 frac14 940 200 188

                                                                    (5) 1

                                                                    2 080 050 235 frac14 47 027 1

                                                                    Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                    Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                    Lever arm of base resultant

                                                                    M

                                                                    Vfrac14 795

                                                                    525frac14 151m

                                                                    Eccentricity of base resultant

                                                                    e frac14 200 151 frac14 049m

                                                                    Figure Q66

                                                                    40 Lateral earth pressure

                                                                    Base pressures (Equation 627)

                                                                    p frac14 525

                                                                    41 6 049

                                                                    4

                                                                    frac14 228 kN=m2 and 35 kN=m2

                                                                    The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                    The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                    The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                    67

                                                                    For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                    Force (kN) Arm (m) Moment (kNm)

                                                                    (1)1

                                                                    2 027 17 52 frac14 574 183 1050

                                                                    (2) 027 17 5 3 frac14 689 500 3445

                                                                    (3)1

                                                                    2 027 102 32 frac14 124 550 682

                                                                    (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                    (5)1

                                                                    2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                    (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                    (7) 1

                                                                    2 267

                                                                    2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                    (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                    2 d frac14 163d d2thorn 650 82d2 1060d

                                                                    Tie rod force per m frac14 T 0 0

                                                                    XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                    d3 thorn 77d2 269d 1438 frac14 0

                                                                    d frac14 467m

                                                                    Depth of penetration frac14 12d frac14 560m

                                                                    Lateral earth pressure 41

                                                                    Algebraic sum of forces for d frac14 467m isX

                                                                    F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                    T frac14 905 kN=m

                                                                    Force in each tie rod frac14 25T frac14 226 kN

                                                                    68

                                                                    (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                    0 frac14 21 98 frac14 112 kN=m3

                                                                    The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                    uC frac14 150

                                                                    165 15 98 frac14 134 kN=m2

                                                                    The average seepage pressure is

                                                                    j frac14 15

                                                                    165 98 frac14 09 kN=m3

                                                                    Hence

                                                                    0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                    0 j frac14 112 09 frac14 103 kN=m3

                                                                    Figure Q67

                                                                    42 Lateral earth pressure

                                                                    Consider moments about the anchor point A (per m)

                                                                    Force (kN) Arm (m) Moment (kN m)

                                                                    (1) 10 026 150 frac14 390 60 2340

                                                                    (2)1

                                                                    2 026 18 452 frac14 474 15 711

                                                                    (3) 026 18 45 105 frac14 2211 825 18240

                                                                    (4)1

                                                                    2 026 121 1052 frac14 1734 100 17340

                                                                    (5)1

                                                                    2 134 15 frac14 101 40 404

                                                                    (6) 134 30 frac14 402 60 2412

                                                                    (7)1

                                                                    2 134 60 frac14 402 95 3819

                                                                    571 4527(8) Ppm

                                                                    115 115PPm

                                                                    XM frac14 0

                                                                    Ppm frac144527

                                                                    115frac14 394 kN=m

                                                                    Available passive resistance

                                                                    Pp frac14 1

                                                                    2 385 103 62 frac14 714 kN=m

                                                                    Factor of safety

                                                                    Fp frac14 Pp

                                                                    Ppm

                                                                    frac14 714

                                                                    394frac14 18

                                                                    Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                    Figure Q68

                                                                    Lateral earth pressure 43

                                                                    (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                    Consider moments (per m) about the tie point A

                                                                    Force (kN) Arm (m)

                                                                    (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                    (2)1

                                                                    2 033 18 452 frac14 601 15

                                                                    (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                    (4)1

                                                                    2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                    (5)1

                                                                    2 134 15 frac14 101 40

                                                                    (6) 134 30 frac14 402 60

                                                                    (7)1

                                                                    2 134 d frac14 67d d3thorn 75

                                                                    (8) 1

                                                                    2 30 103 d2 frac141545d2 2d3thorn 75

                                                                    Moment (kN m)

                                                                    (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                    XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                    d3 thorn 827d2 466d 1518 frac14 0

                                                                    By trial

                                                                    d frac14 544m

                                                                    The minimum depth of embedment required is 544m

                                                                    69

                                                                    For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                    0 frac14 20 98 frac14 102 kN=m3

                                                                    The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                    44 Lateral earth pressure

                                                                    uC frac14 147

                                                                    173 26 98 frac14 216 kN=m2

                                                                    and the average seepage pressure around the wall is

                                                                    j frac14 26

                                                                    173 98 frac14 15 kN=m3

                                                                    Consider moments about the prop (A) (per m)

                                                                    Force (kN) Arm (m) Moment (kN m)

                                                                    (1)1

                                                                    2 03 17 272 frac14 186 020 37

                                                                    (2) 03 17 27 53 frac14 730 335 2445

                                                                    (3)1

                                                                    2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                    (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                    (5)1

                                                                    2 216 26 frac14 281 243 684

                                                                    (6) 216 27 frac14 583 465 2712

                                                                    (7)1

                                                                    2 216 60 frac14 648 800 5184

                                                                    3055(8)

                                                                    1

                                                                    2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                    Factor of safety

                                                                    Fr frac14 6885

                                                                    3055frac14 225

                                                                    Figure Q69

                                                                    Lateral earth pressure 45

                                                                    610

                                                                    For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                    p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                    Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                    Using the recommendations of Twine and Roscoe

                                                                    p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                    Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                    611

                                                                    frac14 18 kN=m3 0 frac14 34

                                                                    H frac14 350m nH frac14 335m mH frac14 185m

                                                                    Consider a trial value of F frac14 20 Refer to Figure 635

                                                                    0m frac14 tan1tan 34

                                                                    20

                                                                    frac14 186

                                                                    Then

                                                                    frac14 45 thorn 0m2frac14 543

                                                                    W frac14 1

                                                                    2 18 3502 cot 543 frac14 792 kN=m

                                                                    Figure Q610

                                                                    46 Lateral earth pressure

                                                                    P frac14 1

                                                                    2 s 3352 frac14 561s kN=m

                                                                    U frac14 1

                                                                    2 98 1852 cosec 543 frac14 206 kN=m

                                                                    Equations 630 and 631 then become

                                                                    561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                    792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                    ie

                                                                    561s 0616N 405 frac14 0

                                                                    792 0857N thorn 563 frac14 0

                                                                    N frac14 848

                                                                    0857frac14 989 kN=m

                                                                    Then

                                                                    561s 609 405 frac14 0

                                                                    s frac14 649

                                                                    561frac14 116 kN=m3

                                                                    The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                    F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                    20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                    s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                    Figure Q611

                                                                    Lateral earth pressure 47

                                                                    612

                                                                    For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                    45 thorn 0

                                                                    2frac14 63

                                                                    For the retained material between the surface and a depth of 36m

                                                                    Pa frac14 1

                                                                    2 030 18 362 frac14 350 kN=m

                                                                    Weight of reinforced fill between the surface and a depth of 36m is

                                                                    Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                    eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                    Eccentricity of Rv

                                                                    e frac14 263 250 frac14 013m

                                                                    The average vertical stress at a depth of 36m is

                                                                    z frac14 Rv

                                                                    L 2efrac14 324

                                                                    474frac14 68 kN=m2

                                                                    (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                    Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                    Tensile stress in the element frac14 138 103

                                                                    65 3frac14 71N=mm2

                                                                    Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                    Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                    Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                    The weight of ABC is

                                                                    W frac14 1

                                                                    2 18 52 265 frac14 124 kN=m

                                                                    From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                    48 Lateral earth pressure

                                                                    (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                    Tp frac14 032 68 120 065 frac14 170 kN

                                                                    Tr frac14 213 420

                                                                    418frac14 214 kN

                                                                    Again the tensile failure and slipping limit states are satisfied for this element

                                                                    Figure Q612

                                                                    Lateral earth pressure 49

                                                                    Chapter 7

                                                                    Consolidation theory

                                                                    71

                                                                    Total change in thickness

                                                                    H frac14 782 602 frac14 180mm

                                                                    Average thickness frac14 1530thorn 180

                                                                    2frac14 1620mm

                                                                    Length of drainage path d frac14 1620

                                                                    2frac14 810mm

                                                                    Root time plot (Figure Q71a)

                                                                    ffiffiffiffiffiffit90p frac14 33

                                                                    t90 frac14 109min

                                                                    cv frac14 0848d2

                                                                    t90frac14 0848 8102

                                                                    109 1440 365

                                                                    106frac14 27m2=year

                                                                    r0 frac14 782 764

                                                                    782 602frac14 018

                                                                    180frac14 0100

                                                                    rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                    10 119

                                                                    9 180frac14 0735

                                                                    rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                    Log time plot (Figure Q71b)

                                                                    t50 frac14 26min

                                                                    cv frac14 0196d2

                                                                    t50frac14 0196 8102

                                                                    26 1440 365

                                                                    106frac14 26m2=year

                                                                    r0 frac14 782 763

                                                                    782 602frac14 019

                                                                    180frac14 0106

                                                                    rp frac14 763 623

                                                                    782 602frac14 140

                                                                    180frac14 0778

                                                                    rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                    Figure Q71(a)

                                                                    Figure Q71(b)

                                                                    Final void ratio

                                                                    e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                    e

                                                                    Hfrac14 1thorn e0

                                                                    H0frac14 1thorn e1 thorne

                                                                    H0

                                                                    ie

                                                                    e

                                                                    180frac14 1631thorne

                                                                    1710

                                                                    e frac14 2936

                                                                    1530frac14 0192

                                                                    Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                    mv frac14 1

                                                                    1thorn e0 e0 e101 00

                                                                    frac14 1

                                                                    1823 0192

                                                                    0107frac14 098m2=MN

                                                                    k frac14 cvmvw frac14 265 098 98

                                                                    60 1440 365 103frac14 81 1010 m=s

                                                                    72

                                                                    Using Equation 77 (one-dimensional method)

                                                                    sc frac14 e0 e11thorn e0 H

                                                                    Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                    Figure Q72

                                                                    52 Consolidation theory

                                                                    Settlement

                                                                    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                    1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                    318

                                                                    Notes 5 92y 460thorn 84

                                                                    Heave

                                                                    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                    1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                    38

                                                                    73

                                                                    U frac14 f ethTvTHORN frac14 f cvt

                                                                    d2

                                                                    Hence if cv is constant

                                                                    t1

                                                                    t2frac14 d

                                                                    21

                                                                    d22

                                                                    where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                    d1 frac14 95mm and d2 frac14 2500mm

                                                                    for U frac14 050 t2 frac14 t1 d22

                                                                    d21

                                                                    frac14 20

                                                                    60 24 365 25002

                                                                    952frac14 263 years

                                                                    for U lt 060 Tv frac14

                                                                    4U2 (Equation 724(a))

                                                                    t030 frac14 t050 0302

                                                                    0502

                                                                    frac14 263 036 frac14 095 years

                                                                    Consolidation theory 53

                                                                    74

                                                                    The layer is open

                                                                    d frac14 8

                                                                    2frac14 4m

                                                                    Tv frac14 cvtd2frac14 24 3

                                                                    42frac14 0450

                                                                    ui frac14 frac14 84 kN=m2

                                                                    The excess pore water pressure is given by Equation 721

                                                                    ue frac14Xmfrac141mfrac140

                                                                    2ui

                                                                    Msin

                                                                    Mz

                                                                    d

                                                                    expethM2TvTHORN

                                                                    In this case z frac14 d

                                                                    sinMz

                                                                    d

                                                                    frac14 sinM

                                                                    where

                                                                    M frac14

                                                                    23

                                                                    25

                                                                    2

                                                                    M sin M M2Tv exp (M2Tv)

                                                                    2thorn1 1110 0329

                                                                    3

                                                                    21 9993 457 105

                                                                    ue frac14 2 84 2

                                                                    1 0329 ethother terms negligibleTHORN

                                                                    frac14 352 kN=m2

                                                                    75

                                                                    The layer is open

                                                                    d frac14 6

                                                                    2frac14 3m

                                                                    Tv frac14 cvtd2frac14 10 3

                                                                    32frac14 0333

                                                                    The layer thickness will be divided into six equal parts ie m frac14 6

                                                                    54 Consolidation theory

                                                                    For an open layer

                                                                    Tv frac14 4n

                                                                    m2

                                                                    n frac14 0333 62

                                                                    4frac14 300

                                                                    The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                    ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                    i j

                                                                    0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                    0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                    The initial and 3-year isochrones are plotted in Figure Q75

                                                                    Area under initial isochrone frac14 180 units

                                                                    Area under 3-year isochrone frac14 63 units

                                                                    The average degree of consolidation is given by Equation 725Thus

                                                                    U frac14 1 63

                                                                    180frac14 065

                                                                    Figure Q75

                                                                    Consolidation theory 55

                                                                    76

                                                                    At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                    0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                    The final consolidation settlement (one-dimensional method) is

                                                                    sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                    Corrected time t frac14 2 1

                                                                    2

                                                                    40

                                                                    52

                                                                    frac14 1615 years

                                                                    Tv frac14 cvtd2frac14 44 1615

                                                                    42frac14 0444

                                                                    From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                    77

                                                                    The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                    Figure Q77

                                                                    56 Consolidation theory

                                                                    Point m n Ir (kNm2) sc (mm)

                                                                    13020frac14 15 20

                                                                    20frac14 10 0194 (4) 113 124

                                                                    260

                                                                    20frac14 30

                                                                    20

                                                                    20frac14 10 0204 (2) 59 65

                                                                    360

                                                                    20frac14 30

                                                                    40

                                                                    20frac14 20 0238 (1) 35 38

                                                                    430

                                                                    20frac14 15

                                                                    40

                                                                    20frac14 20 0224 (2) 65 72

                                                                    Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                    78

                                                                    Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                    (a) Immediate settlement

                                                                    H

                                                                    Bfrac14 30

                                                                    35frac14 086

                                                                    D

                                                                    Bfrac14 2

                                                                    35frac14 006

                                                                    Figure Q78

                                                                    Consolidation theory 57

                                                                    From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                    si frac14 130131qB

                                                                    Eufrac14 10 032 105 35

                                                                    40frac14 30mm

                                                                    (b) Consolidation settlement

                                                                    Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                    1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                    3150

                                                                    Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                    Now

                                                                    H

                                                                    Bfrac14 30

                                                                    35frac14 086 and A frac14 065

                                                                    from Figure 712 13 frac14 079

                                                                    sc frac14 13sod frac14 079 315 frac14 250mm

                                                                    Total settlement

                                                                    s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                    79

                                                                    Without sand drains

                                                                    Uv frac14 025

                                                                    Tv frac14 0049 ethfrom Figure 718THORN

                                                                    t frac14 Tvd2

                                                                    cvfrac14 0049 82

                                                                    cvWith sand drains

                                                                    R frac14 0564S frac14 0564 3 frac14 169m

                                                                    n frac14 Rrfrac14 169

                                                                    015frac14 113

                                                                    Tr frac14 cht

                                                                    4R2frac14 ch

                                                                    4 1692 0049 82

                                                                    cvethand ch frac14 cvTHORN

                                                                    frac14 0275

                                                                    Ur frac14 073 (from Figure 730)

                                                                    58 Consolidation theory

                                                                    Using Equation 740

                                                                    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                    U frac14 080

                                                                    710

                                                                    Without sand drains

                                                                    Uv frac14 090

                                                                    Tv frac14 0848

                                                                    t frac14 Tvd2

                                                                    cvfrac14 0848 102

                                                                    96frac14 88 years

                                                                    With sand drains

                                                                    R frac14 0564S frac14 0564 4 frac14 226m

                                                                    n frac14 Rrfrac14 226

                                                                    015frac14 15

                                                                    Tr

                                                                    Tvfrac14 chcv

                                                                    d2

                                                                    4R2ethsame tTHORN

                                                                    Tr

                                                                    Tvfrac14 140

                                                                    96 102

                                                                    4 2262frac14 714 eth1THORN

                                                                    Using Equation 740

                                                                    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                    An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                    Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                    040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                    Thus

                                                                    Uv frac14 0295 and Ur frac14 086

                                                                    t frac14 88 00683

                                                                    0848frac14 07 years

                                                                    Consolidation theory 59

                                                                    Chapter 8

                                                                    Bearing capacity

                                                                    81

                                                                    (a) The ultimate bearing capacity is given by Equation 83

                                                                    qf frac14 cNc thorn DNq thorn 1

                                                                    2BN

                                                                    For u frac14 0

                                                                    Nc frac14 514 Nq frac14 1 N frac14 0

                                                                    qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                    The net ultimate bearing capacity is

                                                                    qnf frac14 qf D frac14 540 kN=m2

                                                                    The net foundation pressure is

                                                                    qn frac14 q D frac14 425

                                                                    2 eth21 1THORN frac14 192 kN=m2

                                                                    The factor of safety (Equation 86) is

                                                                    F frac14 qnfqnfrac14 540

                                                                    192frac14 28

                                                                    (b) For 0 frac14 28

                                                                    Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                    qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                    2 112 2 13

                                                                    frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                    qnf frac14 574 112 frac14 563 kN=m2

                                                                    F frac14 563

                                                                    192frac14 29

                                                                    (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                    82

                                                                    For 0 frac14 38

                                                                    Nq frac14 49 N frac14 67

                                                                    qnf frac14 DethNq 1THORN thorn 1

                                                                    2BN ethfrom Equation 83THORN

                                                                    frac14 eth18 075 48THORN thorn 1

                                                                    2 18 15 67

                                                                    frac14 648thorn 905 frac14 1553 kN=m2

                                                                    qn frac14 500

                                                                    15 eth18 075THORN frac14 320 kN=m2

                                                                    F frac14 qnfqnfrac14 1553

                                                                    320frac14 48

                                                                    0d frac14 tan1tan 38

                                                                    125

                                                                    frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                    Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                    2 18 15 25

                                                                    frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                    Design load (action) Vd frac14 500 kN=m

                                                                    The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                    83

                                                                    D

                                                                    Bfrac14 350

                                                                    225frac14 155

                                                                    From Figure 85 for a square foundation

                                                                    Nc frac14 81

                                                                    Bearing capacity 61

                                                                    For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                    Nc frac14 084thorn 016B

                                                                    L

                                                                    81 frac14 745

                                                                    Using Equation 810

                                                                    qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                    For F frac14 3

                                                                    qn frac14 1006

                                                                    3frac14 335 kN=m2

                                                                    q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                    Design load frac14 405 450 225 frac14 4100 kN

                                                                    Design undrained strength cud frac14 135

                                                                    14frac14 96 kN=m2

                                                                    Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                    frac14 7241 kN

                                                                    Design load Vd frac14 4100 kN

                                                                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                    84

                                                                    For 0 frac14 40

                                                                    Nq frac14 64 N frac14 95

                                                                    qnf frac14 DethNq 1THORN thorn 04BN

                                                                    (a) Water table 5m below ground level

                                                                    qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                    qn frac14 400 17 frac14 383 kN=m2

                                                                    F frac14 2686

                                                                    383frac14 70

                                                                    (b) Water table 1m below ground level (ie at foundation level)

                                                                    0 frac14 20 98 frac14 102 kN=m3

                                                                    62 Bearing capacity

                                                                    qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                    F frac14 2040

                                                                    383frac14 53

                                                                    (c) Water table at ground level with upward hydraulic gradient 02

                                                                    eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                    qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                    F frac14 1296

                                                                    392frac14 33

                                                                    85

                                                                    The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                    Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                    Design value of 0 frac14 tan1tan 39

                                                                    125

                                                                    frac14 33

                                                                    For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                    Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                    86

                                                                    (a) Undrained shear for u frac14 0

                                                                    Nc frac14 514 Nq frac14 1 N frac14 0

                                                                    qnf frac14 12cuNc

                                                                    frac14 12 100 514 frac14 617 kN=m2

                                                                    qn frac14 qnfFfrac14 617

                                                                    3frac14 206 kN=m2

                                                                    q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                    Bearing capacity 63

                                                                    Drained shear for 0 frac14 32

                                                                    Nq frac14 23 N frac14 25

                                                                    0 frac14 21 98 frac14 112 kN=m3

                                                                    qnf frac14 0DethNq 1THORN thorn 040BN

                                                                    frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                    frac14 694 kN=m2

                                                                    q frac14 694

                                                                    3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                    Design load frac14 42 227 frac14 3632 kN

                                                                    (b) Design undrained strength cud frac14 100

                                                                    14frac14 71 kNm2

                                                                    Design bearing resistance Rd frac14 12cudNe area

                                                                    frac14 12 71 514 42

                                                                    frac14 7007 kN

                                                                    For drained shear 0d frac14 tan1tan 32

                                                                    125

                                                                    frac14 26

                                                                    Nq frac14 12 N frac14 10

                                                                    Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                    (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                    Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                    1 2 100 0175 0700qn 0182qn

                                                                    2 6 033 0044 0176qn 0046qn

                                                                    3 10 020 0017 0068qn 0018qn

                                                                    0246qn

                                                                    Diameter of equivalent circle B frac14 45m

                                                                    H

                                                                    Bfrac14 12

                                                                    45frac14 27 and A frac14 042

                                                                    13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                    64 Bearing capacity

                                                                    For sc frac14 30mm

                                                                    qn frac14 30

                                                                    0147frac14 204 kN=m2

                                                                    q frac14 204thorn 21 frac14 225 kN=m2

                                                                    Design load frac14 42 225 frac14 3600 kN

                                                                    The design load is 3600 kN settlement being the limiting criterion

                                                                    87

                                                                    D

                                                                    Bfrac14 8

                                                                    4frac14 20

                                                                    From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                    F frac14 cuNc

                                                                    Dfrac14 40 71

                                                                    20 8frac14 18

                                                                    88

                                                                    Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                    Design value of 0 frac14 tan1tan 38

                                                                    125

                                                                    frac14 32

                                                                    Figure Q86

                                                                    Bearing capacity 65

                                                                    For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                    Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                    The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                    Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                    For B frac14 250m qn frac14 3750

                                                                    2502 17 frac14 583 kN=m2

                                                                    From Figure 510 m frac14 n frac14 126

                                                                    6frac14 021

                                                                    Ir frac14 0019

                                                                    Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                    Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                    The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                    89

                                                                    Depth (m) N 0v (kNm2) CN N1

                                                                    070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                    Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                    (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                    Cw frac14 05thorn 0530

                                                                    47

                                                                    frac14 082

                                                                    66 Bearing capacity

                                                                    Thus

                                                                    qa frac14 150 082 frac14 120 kN=m2

                                                                    (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                    Thus

                                                                    qa frac14 90 15 frac14 135 kN=m2

                                                                    (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                    Ic frac14 171

                                                                    1014frac14 0068

                                                                    From Equation 819(a) with s frac14 25mm

                                                                    q frac14 25

                                                                    3507 0068frac14 150 kN=m2

                                                                    810

                                                                    Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                    Izp frac14 05thorn 01130

                                                                    16 27

                                                                    05

                                                                    frac14 067

                                                                    Refer to Figure Q810

                                                                    E frac14 25qc

                                                                    Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                    Ez (mm3MN)

                                                                    1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                    0203

                                                                    C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                    130frac14 093

                                                                    C2 frac14 1 ethsayTHORN

                                                                    s frac14 C1C2qnX Iz

                                                                    Ez frac14 093 1 130 0203 frac14 25mm

                                                                    Bearing capacity 67

                                                                    811

                                                                    At pile base level

                                                                    cu frac14 220 kN=m2

                                                                    qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                    Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                    00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                    qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                    Then

                                                                    Qf frac14 Abqb thorn Asqs

                                                                    frac14

                                                                    4 32 1980

                                                                    thorn eth 105 139 86THORN

                                                                    frac14 13 996thorn 3941 frac14 17 937 kN

                                                                    0 01 02 03 04 05 06 07

                                                                    0 2 4 6 8 10 12 14

                                                                    1

                                                                    2

                                                                    3

                                                                    4

                                                                    5

                                                                    6

                                                                    7

                                                                    8

                                                                    (1)

                                                                    (2)

                                                                    (3)

                                                                    (4)

                                                                    (5)

                                                                    qc

                                                                    qc

                                                                    Iz

                                                                    Iz

                                                                    (MNm2)

                                                                    z (m)

                                                                    Figure Q810

                                                                    68 Bearing capacity

                                                                    Allowable load

                                                                    ethaTHORN Qf

                                                                    2frac14 17 937

                                                                    2frac14 8968 kN

                                                                    ethbTHORN Abqb

                                                                    3thorn Asqs frac14 13 996

                                                                    3thorn 3941 frac14 8606 kN

                                                                    ie allowable load frac14 8600 kN

                                                                    Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                    According to the limit state method

                                                                    Characteristic undrained strength at base level cuk frac14 220

                                                                    150kN=m2

                                                                    Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                    150frac14 1320 kN=m2

                                                                    Characteristic shaft resistance qsk frac14 00150

                                                                    frac14 86

                                                                    150frac14 57 kN=m2

                                                                    Characteristic base and shaft resistances

                                                                    Rbk frac14

                                                                    4 32 1320 frac14 9330 kN

                                                                    Rsk frac14 105 139 86

                                                                    150frac14 2629 kN

                                                                    For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                    Design bearing resistance Rcd frac14 9330

                                                                    160thorn 2629

                                                                    130

                                                                    frac14 5831thorn 2022

                                                                    frac14 7850 kN

                                                                    Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                    812

                                                                    ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                    qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                    For a single pile

                                                                    Qf frac14 Abqb thorn Asqs

                                                                    frac14

                                                                    4 062 1305

                                                                    thorn eth 06 15 42THORN

                                                                    frac14 369thorn 1187 frac14 1556 kN

                                                                    Bearing capacity 69

                                                                    Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                    eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                    (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                    (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                    qbkfrac14 9cuk frac14 9 220

                                                                    150frac14 1320 kN=m2

                                                                    qskfrac14cuk frac14 040 105

                                                                    150frac14 28 kN=m2

                                                                    Rbkfrac14

                                                                    4 0602 1320 frac14 373 kN

                                                                    Rskfrac14 060 15 28 frac14 791 kN

                                                                    Rcdfrac14 373

                                                                    160thorn 791

                                                                    130frac14 233thorn 608 frac14 841 kN

                                                                    Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                    Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                    (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                    q frac14 21 000

                                                                    1762frac14 68 kN=m2

                                                                    Immediate settlement

                                                                    H

                                                                    Bfrac14 15

                                                                    176frac14 085

                                                                    D

                                                                    Bfrac14 13

                                                                    176frac14 074

                                                                    L

                                                                    Bfrac14 1

                                                                    Hence from Figure 515

                                                                    130 frac14 078 and 131 frac14 041

                                                                    70 Bearing capacity

                                                                    Thus using Equation 528

                                                                    si frac14 078 041 68 176

                                                                    65frac14 6mm

                                                                    Consolidation settlement

                                                                    Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                    1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                    434 (sod)

                                                                    Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                    sc frac14 056 434 frac14 24mm

                                                                    The total settlement is (6thorn 24) frac14 30mm

                                                                    813

                                                                    At base level N frac14 26 Then using Equation 830

                                                                    qb frac14 40NDb

                                                                    Bfrac14 40 26 2

                                                                    025frac14 8320 kN=m2

                                                                    ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                    Figure Q812

                                                                    Bearing capacity 71

                                                                    Over the length embedded in sand

                                                                    N frac14 21 ie18thorn 24

                                                                    2

                                                                    Using Equation 831

                                                                    qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                    For a single pile

                                                                    Qf frac14 Abqb thorn Asqs

                                                                    frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                    For the pile group assuming a group efficiency of 12

                                                                    XQf frac14 12 9 604 frac14 6523 kN

                                                                    Then the load factor is

                                                                    F frac14 6523

                                                                    2000thorn 1000frac14 21

                                                                    (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                    Characteristic base resistance per unit area qbk frac14 8320

                                                                    150frac14 5547 kNm2

                                                                    Characteristic shaft resistance per unit area qsk frac14 42

                                                                    150frac14 28 kNm2

                                                                    Characteristic base and shaft resistances for a single pile

                                                                    Rbk frac14 0252 5547 frac14 347 kN

                                                                    Rsk frac14 4 025 2 28 frac14 56 kN

                                                                    For a driven pile the partial factors are b frac14 s frac14 130

                                                                    Design bearing resistance Rcd frac14 347

                                                                    130thorn 56

                                                                    130frac14 310 kN

                                                                    For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                    Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                    (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                    From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                    72 Bearing capacity

                                                                    N frac14 24thorn 26thorn 34

                                                                    3frac14 28

                                                                    Ic frac14 171

                                                                    2814frac14 0016 ethEquation 818THORN

                                                                    s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                    814

                                                                    Using Equation 841

                                                                    Tf frac14 DLcu thorn

                                                                    4ethD2 d2THORNcuNc

                                                                    frac14 eth 02 5 06 110THORN thorn

                                                                    4eth022 012THORN110 9

                                                                    frac14 207thorn 23 frac14 230 kN

                                                                    Figure Q813

                                                                    Bearing capacity 73

                                                                    Chapter 9

                                                                    Stability of slopes

                                                                    91

                                                                    Referring to Figure Q91

                                                                    W frac14 417 19 frac14 792 kN=m

                                                                    Q frac14 20 28 frac14 56 kN=m

                                                                    Arc lengthAB frac14

                                                                    180 73 90 frac14 115m

                                                                    Arc length BC frac14

                                                                    180 28 90 frac14 44m

                                                                    The factor of safety is given by

                                                                    F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                    90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                    Depth of tension crack z0 frac14 2cu

                                                                    frac14 2 20

                                                                    19frac14 21m

                                                                    Arc length BD frac14

                                                                    180 13

                                                                    1

                                                                    2 90 frac14 21m

                                                                    F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                    The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                    Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                    14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                    Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                    The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                    92

                                                                    u frac14 0

                                                                    Depth factor D frac14 11

                                                                    9frac14 122

                                                                    Using Equation 92 with F frac14 10

                                                                    Ns frac14 cu

                                                                    FHfrac14 30

                                                                    10 19 9frac14 0175

                                                                    Hence from Figure 93

                                                                    frac14 50

                                                                    For F frac14 12

                                                                    Ns frac14 30

                                                                    12 19 9frac14 0146

                                                                    frac14 27

                                                                    93

                                                                    Refer to Figure Q93

                                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                    1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                    74 m

                                                                    214 1deg

                                                                    213 1deg

                                                                    39 m

                                                                    WB

                                                                    D

                                                                    C

                                                                    28 m

                                                                    21 m

                                                                    A

                                                                    Q

                                                                    Soil (1)Soil (2)

                                                                    73deg

                                                                    Figure Q91

                                                                    Stability of slopes 75

                                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                    9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                    599 256 328 1372

                                                                    Figure Q93

                                                                    76 Stability of slopes

                                                                    XW cos frac14 b

                                                                    Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                    W sin frac14 bX

                                                                    h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                    Arc length La frac14

                                                                    180 57

                                                                    1

                                                                    2 326 frac14 327m

                                                                    The factor of safety is given by

                                                                    F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                    W sin

                                                                    frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                    frac14 091

                                                                    According to the limit state method

                                                                    0d frac14 tan1tan 32

                                                                    125

                                                                    frac14 265

                                                                    c0 frac14 8

                                                                    160frac14 5 kN=m2

                                                                    Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                    Design disturbing moment frac14 1075 kN=m

                                                                    The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                    94

                                                                    F frac14 1

                                                                    W sin

                                                                    Xfc0bthorn ethW ubTHORN tan0g sec

                                                                    1thorn ethtan tan0=FTHORN

                                                                    c0 frac14 8 kN=m2

                                                                    0 frac14 32

                                                                    c0b frac14 8 2 frac14 16 kN=m

                                                                    W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                    Try F frac14 100

                                                                    tan0

                                                                    Ffrac14 0625

                                                                    Stability of slopes 77

                                                                    Values of u are as obtained in Figure Q93

                                                                    SliceNo

                                                                    h(m)

                                                                    W frac14 bh(kNm)

                                                                    W sin(kNm)

                                                                    ub(kNm)

                                                                    c0bthorn (W ub) tan0(kNm)

                                                                    sec

                                                                    1thorn (tan tan0)FProduct(kNm)

                                                                    1 05 21 6 2 8 24 1078 262 13 55 31

                                                                    23 33 30 1042 31

                                                                    3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                    224 92 72 0931 67

                                                                    6 50 210 11 40 100 85 0907 777 55 231 14

                                                                    12 58 112 90 0889 80

                                                                    8 60 252 1812

                                                                    80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                    10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                    2154 88 116 0853 99

                                                                    14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                    1074 1091

                                                                    F frac14 1091

                                                                    1074frac14 102 (assumed value 100)

                                                                    Thus

                                                                    F frac14 101

                                                                    95

                                                                    F frac14 1

                                                                    W sin

                                                                    XfWeth1 ruTHORN tan0g sec

                                                                    1thorn ethtan tan0THORN=F

                                                                    0 frac14 33

                                                                    ru frac14 020

                                                                    W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                    eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                    Try F frac14 110

                                                                    tan 0

                                                                    Ffrac14 tan 33

                                                                    110frac14 0590

                                                                    78 Stability of slopes

                                                                    Referring to Figure Q95

                                                                    SliceNo

                                                                    h(m)

                                                                    W frac14 bh(kNm)

                                                                    W sin(kNm)

                                                                    W(1 ru) tan0(kNm)

                                                                    sec

                                                                    1thorn ( tan tan0)FProduct(kNm)

                                                                    1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                    2120 234 0892 209

                                                                    4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                    1185 1271

                                                                    Figure Q95

                                                                    Stability of slopes 79

                                                                    F frac14 1271

                                                                    1185frac14 107

                                                                    The trial value was 110 therefore take F to be 108

                                                                    96

                                                                    (a) Water table at surface the factor of safety is given by Equation 912

                                                                    F frac14 0

                                                                    sat

                                                                    tan0

                                                                    tan

                                                                    ptie 15 frac14 92

                                                                    19

                                                                    tan 36

                                                                    tan

                                                                    tan frac14 0234

                                                                    frac14 13

                                                                    Water table well below surface the factor of safety is given by Equation 911

                                                                    F frac14 tan0

                                                                    tan

                                                                    frac14 tan 36

                                                                    tan 13

                                                                    frac14 31

                                                                    (b) 0d frac14 tan1tan 36

                                                                    125

                                                                    frac14 30

                                                                    Depth of potential failure surface frac14 z

                                                                    Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                    frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                    frac14 504z kN

                                                                    Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                    frac14 19 z sin 13 cos 13

                                                                    frac14 416z kN

                                                                    Rd gtSd therefore the limit state for overall stability is satisfied

                                                                    80 Stability of slopes

                                                                    • Book Cover
                                                                    • Title
                                                                    • Contents
                                                                    • Basic characteristics of soils
                                                                    • Seepage
                                                                    • Effective stress
                                                                    • Shear strength
                                                                    • Stresses and displacements
                                                                    • Lateral earth pressure
                                                                    • Consolidation theory
                                                                    • Bearing capacity
                                                                    • Stability of slopes

                                                                      52

                                                                      Below the centre load (Figure Q52)

                                                                      r

                                                                      zfrac14 0 for the 7500-kN load

                                                                      Ip frac14 0478

                                                                      r

                                                                      zfrac14 5

                                                                      4frac14 125 for the 10 000- and 9000-kN loads

                                                                      Ip frac14 0045

                                                                      Then

                                                                      z frac14X Q

                                                                      z2Ip

                                                                      frac14 7500 0478

                                                                      42thorn 10 000 0045

                                                                      42thorn 9000 0045

                                                                      42

                                                                      frac14 224thorn 28thorn 25 frac14 277 kN=m2

                                                                      53

                                                                      The vertical stress under a corner of a rectangular area is given by

                                                                      z frac14 qIr

                                                                      where values of Ir are obtained from Figure 510 In this case

                                                                      z frac14 4 250 Ir ethkN=m2THORNm frac14 n frac14 1

                                                                      z

                                                                      Figure Q52

                                                                      Stresses and displacements 29

                                                                      z (m) m n Ir z (kNm2)

                                                                      0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                                                      10 010 0005 5

                                                                      z is plotted against z in Figure Q53

                                                                      54

                                                                      (a)

                                                                      m frac14 125

                                                                      12frac14 104

                                                                      n frac14 18

                                                                      12frac14 150

                                                                      From Figure 510 Irfrac14 0196

                                                                      z frac14 2 175 0196 frac14 68 kN=m2

                                                                      Figure Q53

                                                                      30 Stresses and displacements

                                                                      (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                                                      z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                                                      55

                                                                      Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                                                      Px frac14 2Q

                                                                      1

                                                                      m2 thorn 1frac14 2 150

                                                                      125frac14 76 kN=m

                                                                      Equation 517 is used to obtain the pressure distribution

                                                                      px frac14 4Q

                                                                      h

                                                                      m2n

                                                                      ethm2 thorn n2THORN2 frac14150

                                                                      m2n

                                                                      ethm2 thorn n2THORN2 ethkN=m2THORN

                                                                      Figure Q54

                                                                      Stresses and displacements 31

                                                                      n m2n

                                                                      (m2 thorn n2)2

                                                                      px(kNm2)

                                                                      0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                                      The pressure distribution is plotted in Figure Q55

                                                                      56

                                                                      H

                                                                      Bfrac14 10

                                                                      2frac14 5

                                                                      L

                                                                      Bfrac14 4

                                                                      2frac14 2

                                                                      D

                                                                      Bfrac14 1

                                                                      2frac14 05

                                                                      Hence from Figure 515

                                                                      131 frac14 082

                                                                      130 frac14 094

                                                                      Figure Q55

                                                                      32 Stresses and displacements

                                                                      The immediate settlement is given by Equation 528

                                                                      si frac14 130131qB

                                                                      Eu

                                                                      frac14 094 082 200 2

                                                                      45frac14 7mm

                                                                      Stresses and displacements 33

                                                                      Chapter 6

                                                                      Lateral earth pressure

                                                                      61

                                                                      For 0 frac14 37 the active pressure coefficient is given by

                                                                      Ka frac14 1 sin 37

                                                                      1thorn sin 37frac14 025

                                                                      The total active thrust (Equation 66a with c0 frac14 0) is

                                                                      Pa frac14 1

                                                                      2KaH

                                                                      2 frac14 1

                                                                      2 025 17 62 frac14 765 kN=m

                                                                      If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                                      K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                                      and the thrust on the wall is

                                                                      P0 frac14 1

                                                                      2K0H

                                                                      2 frac14 1

                                                                      2 040 17 62 frac14 122 kN=m

                                                                      62

                                                                      The active pressure coefficients for the three soil types are as follows

                                                                      Ka1 frac141 sin 35

                                                                      1thorn sin 35frac14 0271

                                                                      Ka2 frac141 sin 27

                                                                      1thorn sin 27frac14 0375

                                                                      ffiffiffiffiffiffiffiKa2

                                                                      p frac14 0613

                                                                      Ka3 frac141 sin 42

                                                                      1thorn sin 42frac14 0198

                                                                      Distribution of active pressure (plotted in Figure Q62)

                                                                      Depth (m) Soil Active pressure (kNm2)

                                                                      3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                                      12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                                      At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                                      Total thrust frac14 571 kNm

                                                                      Point of application is (4893571) m from the top of the wall ie 857m

                                                                      Force (kN) Arm (m) Moment (kN m)

                                                                      (1)1

                                                                      2 0271 16 32 frac14 195 20 390

                                                                      (2) 0271 16 3 2 frac14 260 40 1040

                                                                      (3)1

                                                                      2 0271 92 22 frac14 50 433 217

                                                                      (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                                      (5)1

                                                                      2 0375 102 32 frac14 172 70 1204

                                                                      (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                                      (7)1

                                                                      2 0198 112 42 frac14 177 1067 1889

                                                                      (8)1

                                                                      2 98 92 frac14 3969 90 35721

                                                                      5713 48934

                                                                      Figure Q62

                                                                      Lateral earth pressure 35

                                                                      63

                                                                      (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                                      Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                                      frac14 245

                                                                      At the lower end of the piling

                                                                      pa frac14 Kaqthorn Kasatz Kaccu

                                                                      frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                                      frac14 115 kN=m2

                                                                      pp frac14 Kpsatzthorn Kpccu

                                                                      frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                                      frac14 202 kN=m2

                                                                      (b) For 0 frac14 26 and frac14 1

                                                                      20

                                                                      Ka frac14 035

                                                                      Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                                      pfrac14 145 ethEquation 619THORN

                                                                      Kp frac14 37

                                                                      Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                                      pfrac14 47 ethEquation 624THORN

                                                                      At the lower end of the piling

                                                                      pa frac14 Kaqthorn Ka0z Kacc

                                                                      0

                                                                      frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                                      frac14 187 kN=m2

                                                                      pp frac14 Kp0zthorn Kpcc

                                                                      0

                                                                      frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                                      frac14 198 kN=m2

                                                                      36 Lateral earth pressure

                                                                      64

                                                                      (a) For 0 frac14 38 Ka frac14 024

                                                                      0 frac14 20 98 frac14 102 kN=m3

                                                                      The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                                      Force (kN) Arm (m) Moment (kN m)

                                                                      (1) 024 10 66 frac14 159 33 525

                                                                      (2)1

                                                                      2 024 17 392 frac14 310 400 1240

                                                                      (3) 024 17 39 27 frac14 430 135 580

                                                                      (4)1

                                                                      2 024 102 272 frac14 89 090 80

                                                                      (5)1

                                                                      2 98 272 frac14 357 090 321

                                                                      Hfrac14 1345 MH frac14 2746

                                                                      (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                                      (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                                      XM frac14MV MH frac14 7790 kNm

                                                                      Lever arm of base resultant

                                                                      M

                                                                      Vfrac14 779

                                                                      488frac14 160

                                                                      Eccentricity of base resultant

                                                                      e frac14 200 160 frac14 040m

                                                                      39 m

                                                                      27 m

                                                                      40 m

                                                                      04 m

                                                                      04 m

                                                                      26 m

                                                                      (7)

                                                                      (9)

                                                                      (1)(2)

                                                                      (3)

                                                                      (4)

                                                                      (5)

                                                                      (8)(6)

                                                                      (10)

                                                                      WT

                                                                      10 kNm2

                                                                      Hydrostatic

                                                                      Figure Q64

                                                                      Lateral earth pressure 37

                                                                      Base pressures (Equation 627)

                                                                      p frac14 VB

                                                                      1 6e

                                                                      B

                                                                      frac14 488

                                                                      4eth1 060THORN

                                                                      frac14 195 kN=m2 and 49 kN=m2

                                                                      Factor of safety against sliding (Equation 628)

                                                                      F frac14 V tan

                                                                      Hfrac14 488 tan 25

                                                                      1345frac14 17

                                                                      (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                                      Hfrac14 1633 kN

                                                                      V frac14 4879 kN

                                                                      MH frac14 3453 kNm

                                                                      MV frac14 10536 kNm

                                                                      The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                                      65

                                                                      For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                      Kp

                                                                      Ffrac14 385

                                                                      2

                                                                      0 frac14 20 98 frac14 102 kN=m3

                                                                      The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                                      Force (kN) Arm (m) Moment (kN m)

                                                                      (1)1

                                                                      2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                                      (2) 026 17 45 d frac14 199d d2 995d2

                                                                      (3)1

                                                                      2 026 102 d2 frac14 133d2 d3 044d3

                                                                      (4)1

                                                                      2 385

                                                                      2 17 152 frac14 368 dthorn 05 368d 184

                                                                      (5)385

                                                                      2 17 15 d frac14 491d d2 2455d2

                                                                      (6)1

                                                                      2 385

                                                                      2 102 d2 frac14 982d2 d3 327d3

                                                                      38 Lateral earth pressure

                                                                      XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                                      d3 thorn 516d2 283d 1724 frac14 0

                                                                      d frac14 179m

                                                                      Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                                      Over additional 20 embedded depth

                                                                      pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                                      Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                                      66

                                                                      The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                                      Ka frac14 sin 69=sin 105

                                                                      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                                      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                                      pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                                      26664

                                                                      37775

                                                                      2

                                                                      frac14 050

                                                                      The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                                      Pa frac14 1

                                                                      2 050 19 7502 frac14 267 kN=m

                                                                      Figure Q65

                                                                      Lateral earth pressure 39

                                                                      Horizontal component

                                                                      Ph frac14 267 cos 40 frac14 205 kN=m

                                                                      Vertical component

                                                                      Pv frac14 267 sin 40 frac14 172 kN=m

                                                                      Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                      Force (kN) Arm (m) Moment (kN m)

                                                                      (1)1

                                                                      2 175 650 235 frac14 1337 258 345

                                                                      (2) 050 650 235 frac14 764 175 134

                                                                      (3)1

                                                                      2 070 650 235 frac14 535 127 68

                                                                      (4) 100 400 235 frac14 940 200 188

                                                                      (5) 1

                                                                      2 080 050 235 frac14 47 027 1

                                                                      Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                      Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                      Lever arm of base resultant

                                                                      M

                                                                      Vfrac14 795

                                                                      525frac14 151m

                                                                      Eccentricity of base resultant

                                                                      e frac14 200 151 frac14 049m

                                                                      Figure Q66

                                                                      40 Lateral earth pressure

                                                                      Base pressures (Equation 627)

                                                                      p frac14 525

                                                                      41 6 049

                                                                      4

                                                                      frac14 228 kN=m2 and 35 kN=m2

                                                                      The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                      The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                      The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                      67

                                                                      For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                      Force (kN) Arm (m) Moment (kNm)

                                                                      (1)1

                                                                      2 027 17 52 frac14 574 183 1050

                                                                      (2) 027 17 5 3 frac14 689 500 3445

                                                                      (3)1

                                                                      2 027 102 32 frac14 124 550 682

                                                                      (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                      (5)1

                                                                      2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                      (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                      (7) 1

                                                                      2 267

                                                                      2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                      (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                      2 d frac14 163d d2thorn 650 82d2 1060d

                                                                      Tie rod force per m frac14 T 0 0

                                                                      XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                      d3 thorn 77d2 269d 1438 frac14 0

                                                                      d frac14 467m

                                                                      Depth of penetration frac14 12d frac14 560m

                                                                      Lateral earth pressure 41

                                                                      Algebraic sum of forces for d frac14 467m isX

                                                                      F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                      T frac14 905 kN=m

                                                                      Force in each tie rod frac14 25T frac14 226 kN

                                                                      68

                                                                      (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                      0 frac14 21 98 frac14 112 kN=m3

                                                                      The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                      uC frac14 150

                                                                      165 15 98 frac14 134 kN=m2

                                                                      The average seepage pressure is

                                                                      j frac14 15

                                                                      165 98 frac14 09 kN=m3

                                                                      Hence

                                                                      0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                      0 j frac14 112 09 frac14 103 kN=m3

                                                                      Figure Q67

                                                                      42 Lateral earth pressure

                                                                      Consider moments about the anchor point A (per m)

                                                                      Force (kN) Arm (m) Moment (kN m)

                                                                      (1) 10 026 150 frac14 390 60 2340

                                                                      (2)1

                                                                      2 026 18 452 frac14 474 15 711

                                                                      (3) 026 18 45 105 frac14 2211 825 18240

                                                                      (4)1

                                                                      2 026 121 1052 frac14 1734 100 17340

                                                                      (5)1

                                                                      2 134 15 frac14 101 40 404

                                                                      (6) 134 30 frac14 402 60 2412

                                                                      (7)1

                                                                      2 134 60 frac14 402 95 3819

                                                                      571 4527(8) Ppm

                                                                      115 115PPm

                                                                      XM frac14 0

                                                                      Ppm frac144527

                                                                      115frac14 394 kN=m

                                                                      Available passive resistance

                                                                      Pp frac14 1

                                                                      2 385 103 62 frac14 714 kN=m

                                                                      Factor of safety

                                                                      Fp frac14 Pp

                                                                      Ppm

                                                                      frac14 714

                                                                      394frac14 18

                                                                      Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                      Figure Q68

                                                                      Lateral earth pressure 43

                                                                      (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                      Consider moments (per m) about the tie point A

                                                                      Force (kN) Arm (m)

                                                                      (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                      (2)1

                                                                      2 033 18 452 frac14 601 15

                                                                      (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                      (4)1

                                                                      2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                      (5)1

                                                                      2 134 15 frac14 101 40

                                                                      (6) 134 30 frac14 402 60

                                                                      (7)1

                                                                      2 134 d frac14 67d d3thorn 75

                                                                      (8) 1

                                                                      2 30 103 d2 frac141545d2 2d3thorn 75

                                                                      Moment (kN m)

                                                                      (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                      XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                      d3 thorn 827d2 466d 1518 frac14 0

                                                                      By trial

                                                                      d frac14 544m

                                                                      The minimum depth of embedment required is 544m

                                                                      69

                                                                      For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                      0 frac14 20 98 frac14 102 kN=m3

                                                                      The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                      44 Lateral earth pressure

                                                                      uC frac14 147

                                                                      173 26 98 frac14 216 kN=m2

                                                                      and the average seepage pressure around the wall is

                                                                      j frac14 26

                                                                      173 98 frac14 15 kN=m3

                                                                      Consider moments about the prop (A) (per m)

                                                                      Force (kN) Arm (m) Moment (kN m)

                                                                      (1)1

                                                                      2 03 17 272 frac14 186 020 37

                                                                      (2) 03 17 27 53 frac14 730 335 2445

                                                                      (3)1

                                                                      2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                      (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                      (5)1

                                                                      2 216 26 frac14 281 243 684

                                                                      (6) 216 27 frac14 583 465 2712

                                                                      (7)1

                                                                      2 216 60 frac14 648 800 5184

                                                                      3055(8)

                                                                      1

                                                                      2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                      Factor of safety

                                                                      Fr frac14 6885

                                                                      3055frac14 225

                                                                      Figure Q69

                                                                      Lateral earth pressure 45

                                                                      610

                                                                      For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                      p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                      Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                      Using the recommendations of Twine and Roscoe

                                                                      p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                      Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                      611

                                                                      frac14 18 kN=m3 0 frac14 34

                                                                      H frac14 350m nH frac14 335m mH frac14 185m

                                                                      Consider a trial value of F frac14 20 Refer to Figure 635

                                                                      0m frac14 tan1tan 34

                                                                      20

                                                                      frac14 186

                                                                      Then

                                                                      frac14 45 thorn 0m2frac14 543

                                                                      W frac14 1

                                                                      2 18 3502 cot 543 frac14 792 kN=m

                                                                      Figure Q610

                                                                      46 Lateral earth pressure

                                                                      P frac14 1

                                                                      2 s 3352 frac14 561s kN=m

                                                                      U frac14 1

                                                                      2 98 1852 cosec 543 frac14 206 kN=m

                                                                      Equations 630 and 631 then become

                                                                      561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                      792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                      ie

                                                                      561s 0616N 405 frac14 0

                                                                      792 0857N thorn 563 frac14 0

                                                                      N frac14 848

                                                                      0857frac14 989 kN=m

                                                                      Then

                                                                      561s 609 405 frac14 0

                                                                      s frac14 649

                                                                      561frac14 116 kN=m3

                                                                      The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                      F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                      20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                      s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                      Figure Q611

                                                                      Lateral earth pressure 47

                                                                      612

                                                                      For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                      45 thorn 0

                                                                      2frac14 63

                                                                      For the retained material between the surface and a depth of 36m

                                                                      Pa frac14 1

                                                                      2 030 18 362 frac14 350 kN=m

                                                                      Weight of reinforced fill between the surface and a depth of 36m is

                                                                      Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                      eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                      Eccentricity of Rv

                                                                      e frac14 263 250 frac14 013m

                                                                      The average vertical stress at a depth of 36m is

                                                                      z frac14 Rv

                                                                      L 2efrac14 324

                                                                      474frac14 68 kN=m2

                                                                      (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                      Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                      Tensile stress in the element frac14 138 103

                                                                      65 3frac14 71N=mm2

                                                                      Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                      Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                      Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                      The weight of ABC is

                                                                      W frac14 1

                                                                      2 18 52 265 frac14 124 kN=m

                                                                      From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                      48 Lateral earth pressure

                                                                      (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                      Tp frac14 032 68 120 065 frac14 170 kN

                                                                      Tr frac14 213 420

                                                                      418frac14 214 kN

                                                                      Again the tensile failure and slipping limit states are satisfied for this element

                                                                      Figure Q612

                                                                      Lateral earth pressure 49

                                                                      Chapter 7

                                                                      Consolidation theory

                                                                      71

                                                                      Total change in thickness

                                                                      H frac14 782 602 frac14 180mm

                                                                      Average thickness frac14 1530thorn 180

                                                                      2frac14 1620mm

                                                                      Length of drainage path d frac14 1620

                                                                      2frac14 810mm

                                                                      Root time plot (Figure Q71a)

                                                                      ffiffiffiffiffiffit90p frac14 33

                                                                      t90 frac14 109min

                                                                      cv frac14 0848d2

                                                                      t90frac14 0848 8102

                                                                      109 1440 365

                                                                      106frac14 27m2=year

                                                                      r0 frac14 782 764

                                                                      782 602frac14 018

                                                                      180frac14 0100

                                                                      rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                      10 119

                                                                      9 180frac14 0735

                                                                      rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                      Log time plot (Figure Q71b)

                                                                      t50 frac14 26min

                                                                      cv frac14 0196d2

                                                                      t50frac14 0196 8102

                                                                      26 1440 365

                                                                      106frac14 26m2=year

                                                                      r0 frac14 782 763

                                                                      782 602frac14 019

                                                                      180frac14 0106

                                                                      rp frac14 763 623

                                                                      782 602frac14 140

                                                                      180frac14 0778

                                                                      rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                      Figure Q71(a)

                                                                      Figure Q71(b)

                                                                      Final void ratio

                                                                      e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                      e

                                                                      Hfrac14 1thorn e0

                                                                      H0frac14 1thorn e1 thorne

                                                                      H0

                                                                      ie

                                                                      e

                                                                      180frac14 1631thorne

                                                                      1710

                                                                      e frac14 2936

                                                                      1530frac14 0192

                                                                      Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                      mv frac14 1

                                                                      1thorn e0 e0 e101 00

                                                                      frac14 1

                                                                      1823 0192

                                                                      0107frac14 098m2=MN

                                                                      k frac14 cvmvw frac14 265 098 98

                                                                      60 1440 365 103frac14 81 1010 m=s

                                                                      72

                                                                      Using Equation 77 (one-dimensional method)

                                                                      sc frac14 e0 e11thorn e0 H

                                                                      Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                      Figure Q72

                                                                      52 Consolidation theory

                                                                      Settlement

                                                                      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                      1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                      318

                                                                      Notes 5 92y 460thorn 84

                                                                      Heave

                                                                      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                      1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                      38

                                                                      73

                                                                      U frac14 f ethTvTHORN frac14 f cvt

                                                                      d2

                                                                      Hence if cv is constant

                                                                      t1

                                                                      t2frac14 d

                                                                      21

                                                                      d22

                                                                      where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                      d1 frac14 95mm and d2 frac14 2500mm

                                                                      for U frac14 050 t2 frac14 t1 d22

                                                                      d21

                                                                      frac14 20

                                                                      60 24 365 25002

                                                                      952frac14 263 years

                                                                      for U lt 060 Tv frac14

                                                                      4U2 (Equation 724(a))

                                                                      t030 frac14 t050 0302

                                                                      0502

                                                                      frac14 263 036 frac14 095 years

                                                                      Consolidation theory 53

                                                                      74

                                                                      The layer is open

                                                                      d frac14 8

                                                                      2frac14 4m

                                                                      Tv frac14 cvtd2frac14 24 3

                                                                      42frac14 0450

                                                                      ui frac14 frac14 84 kN=m2

                                                                      The excess pore water pressure is given by Equation 721

                                                                      ue frac14Xmfrac141mfrac140

                                                                      2ui

                                                                      Msin

                                                                      Mz

                                                                      d

                                                                      expethM2TvTHORN

                                                                      In this case z frac14 d

                                                                      sinMz

                                                                      d

                                                                      frac14 sinM

                                                                      where

                                                                      M frac14

                                                                      23

                                                                      25

                                                                      2

                                                                      M sin M M2Tv exp (M2Tv)

                                                                      2thorn1 1110 0329

                                                                      3

                                                                      21 9993 457 105

                                                                      ue frac14 2 84 2

                                                                      1 0329 ethother terms negligibleTHORN

                                                                      frac14 352 kN=m2

                                                                      75

                                                                      The layer is open

                                                                      d frac14 6

                                                                      2frac14 3m

                                                                      Tv frac14 cvtd2frac14 10 3

                                                                      32frac14 0333

                                                                      The layer thickness will be divided into six equal parts ie m frac14 6

                                                                      54 Consolidation theory

                                                                      For an open layer

                                                                      Tv frac14 4n

                                                                      m2

                                                                      n frac14 0333 62

                                                                      4frac14 300

                                                                      The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                      ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                      i j

                                                                      0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                      0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                      The initial and 3-year isochrones are plotted in Figure Q75

                                                                      Area under initial isochrone frac14 180 units

                                                                      Area under 3-year isochrone frac14 63 units

                                                                      The average degree of consolidation is given by Equation 725Thus

                                                                      U frac14 1 63

                                                                      180frac14 065

                                                                      Figure Q75

                                                                      Consolidation theory 55

                                                                      76

                                                                      At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                      0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                      The final consolidation settlement (one-dimensional method) is

                                                                      sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                      Corrected time t frac14 2 1

                                                                      2

                                                                      40

                                                                      52

                                                                      frac14 1615 years

                                                                      Tv frac14 cvtd2frac14 44 1615

                                                                      42frac14 0444

                                                                      From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                      77

                                                                      The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                      Figure Q77

                                                                      56 Consolidation theory

                                                                      Point m n Ir (kNm2) sc (mm)

                                                                      13020frac14 15 20

                                                                      20frac14 10 0194 (4) 113 124

                                                                      260

                                                                      20frac14 30

                                                                      20

                                                                      20frac14 10 0204 (2) 59 65

                                                                      360

                                                                      20frac14 30

                                                                      40

                                                                      20frac14 20 0238 (1) 35 38

                                                                      430

                                                                      20frac14 15

                                                                      40

                                                                      20frac14 20 0224 (2) 65 72

                                                                      Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                      78

                                                                      Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                      (a) Immediate settlement

                                                                      H

                                                                      Bfrac14 30

                                                                      35frac14 086

                                                                      D

                                                                      Bfrac14 2

                                                                      35frac14 006

                                                                      Figure Q78

                                                                      Consolidation theory 57

                                                                      From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                      si frac14 130131qB

                                                                      Eufrac14 10 032 105 35

                                                                      40frac14 30mm

                                                                      (b) Consolidation settlement

                                                                      Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                      1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                      3150

                                                                      Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                      Now

                                                                      H

                                                                      Bfrac14 30

                                                                      35frac14 086 and A frac14 065

                                                                      from Figure 712 13 frac14 079

                                                                      sc frac14 13sod frac14 079 315 frac14 250mm

                                                                      Total settlement

                                                                      s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                      79

                                                                      Without sand drains

                                                                      Uv frac14 025

                                                                      Tv frac14 0049 ethfrom Figure 718THORN

                                                                      t frac14 Tvd2

                                                                      cvfrac14 0049 82

                                                                      cvWith sand drains

                                                                      R frac14 0564S frac14 0564 3 frac14 169m

                                                                      n frac14 Rrfrac14 169

                                                                      015frac14 113

                                                                      Tr frac14 cht

                                                                      4R2frac14 ch

                                                                      4 1692 0049 82

                                                                      cvethand ch frac14 cvTHORN

                                                                      frac14 0275

                                                                      Ur frac14 073 (from Figure 730)

                                                                      58 Consolidation theory

                                                                      Using Equation 740

                                                                      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                      U frac14 080

                                                                      710

                                                                      Without sand drains

                                                                      Uv frac14 090

                                                                      Tv frac14 0848

                                                                      t frac14 Tvd2

                                                                      cvfrac14 0848 102

                                                                      96frac14 88 years

                                                                      With sand drains

                                                                      R frac14 0564S frac14 0564 4 frac14 226m

                                                                      n frac14 Rrfrac14 226

                                                                      015frac14 15

                                                                      Tr

                                                                      Tvfrac14 chcv

                                                                      d2

                                                                      4R2ethsame tTHORN

                                                                      Tr

                                                                      Tvfrac14 140

                                                                      96 102

                                                                      4 2262frac14 714 eth1THORN

                                                                      Using Equation 740

                                                                      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                      An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                      Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                      040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                      Thus

                                                                      Uv frac14 0295 and Ur frac14 086

                                                                      t frac14 88 00683

                                                                      0848frac14 07 years

                                                                      Consolidation theory 59

                                                                      Chapter 8

                                                                      Bearing capacity

                                                                      81

                                                                      (a) The ultimate bearing capacity is given by Equation 83

                                                                      qf frac14 cNc thorn DNq thorn 1

                                                                      2BN

                                                                      For u frac14 0

                                                                      Nc frac14 514 Nq frac14 1 N frac14 0

                                                                      qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                      The net ultimate bearing capacity is

                                                                      qnf frac14 qf D frac14 540 kN=m2

                                                                      The net foundation pressure is

                                                                      qn frac14 q D frac14 425

                                                                      2 eth21 1THORN frac14 192 kN=m2

                                                                      The factor of safety (Equation 86) is

                                                                      F frac14 qnfqnfrac14 540

                                                                      192frac14 28

                                                                      (b) For 0 frac14 28

                                                                      Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                      qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                      2 112 2 13

                                                                      frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                      qnf frac14 574 112 frac14 563 kN=m2

                                                                      F frac14 563

                                                                      192frac14 29

                                                                      (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                      82

                                                                      For 0 frac14 38

                                                                      Nq frac14 49 N frac14 67

                                                                      qnf frac14 DethNq 1THORN thorn 1

                                                                      2BN ethfrom Equation 83THORN

                                                                      frac14 eth18 075 48THORN thorn 1

                                                                      2 18 15 67

                                                                      frac14 648thorn 905 frac14 1553 kN=m2

                                                                      qn frac14 500

                                                                      15 eth18 075THORN frac14 320 kN=m2

                                                                      F frac14 qnfqnfrac14 1553

                                                                      320frac14 48

                                                                      0d frac14 tan1tan 38

                                                                      125

                                                                      frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                      Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                      2 18 15 25

                                                                      frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                      Design load (action) Vd frac14 500 kN=m

                                                                      The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                      83

                                                                      D

                                                                      Bfrac14 350

                                                                      225frac14 155

                                                                      From Figure 85 for a square foundation

                                                                      Nc frac14 81

                                                                      Bearing capacity 61

                                                                      For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                      Nc frac14 084thorn 016B

                                                                      L

                                                                      81 frac14 745

                                                                      Using Equation 810

                                                                      qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                      For F frac14 3

                                                                      qn frac14 1006

                                                                      3frac14 335 kN=m2

                                                                      q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                      Design load frac14 405 450 225 frac14 4100 kN

                                                                      Design undrained strength cud frac14 135

                                                                      14frac14 96 kN=m2

                                                                      Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                      frac14 7241 kN

                                                                      Design load Vd frac14 4100 kN

                                                                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                      84

                                                                      For 0 frac14 40

                                                                      Nq frac14 64 N frac14 95

                                                                      qnf frac14 DethNq 1THORN thorn 04BN

                                                                      (a) Water table 5m below ground level

                                                                      qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                      qn frac14 400 17 frac14 383 kN=m2

                                                                      F frac14 2686

                                                                      383frac14 70

                                                                      (b) Water table 1m below ground level (ie at foundation level)

                                                                      0 frac14 20 98 frac14 102 kN=m3

                                                                      62 Bearing capacity

                                                                      qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                      F frac14 2040

                                                                      383frac14 53

                                                                      (c) Water table at ground level with upward hydraulic gradient 02

                                                                      eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                      qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                      F frac14 1296

                                                                      392frac14 33

                                                                      85

                                                                      The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                      Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                      Design value of 0 frac14 tan1tan 39

                                                                      125

                                                                      frac14 33

                                                                      For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                      Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                      86

                                                                      (a) Undrained shear for u frac14 0

                                                                      Nc frac14 514 Nq frac14 1 N frac14 0

                                                                      qnf frac14 12cuNc

                                                                      frac14 12 100 514 frac14 617 kN=m2

                                                                      qn frac14 qnfFfrac14 617

                                                                      3frac14 206 kN=m2

                                                                      q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                      Bearing capacity 63

                                                                      Drained shear for 0 frac14 32

                                                                      Nq frac14 23 N frac14 25

                                                                      0 frac14 21 98 frac14 112 kN=m3

                                                                      qnf frac14 0DethNq 1THORN thorn 040BN

                                                                      frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                      frac14 694 kN=m2

                                                                      q frac14 694

                                                                      3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                      Design load frac14 42 227 frac14 3632 kN

                                                                      (b) Design undrained strength cud frac14 100

                                                                      14frac14 71 kNm2

                                                                      Design bearing resistance Rd frac14 12cudNe area

                                                                      frac14 12 71 514 42

                                                                      frac14 7007 kN

                                                                      For drained shear 0d frac14 tan1tan 32

                                                                      125

                                                                      frac14 26

                                                                      Nq frac14 12 N frac14 10

                                                                      Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                      (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                      Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                      1 2 100 0175 0700qn 0182qn

                                                                      2 6 033 0044 0176qn 0046qn

                                                                      3 10 020 0017 0068qn 0018qn

                                                                      0246qn

                                                                      Diameter of equivalent circle B frac14 45m

                                                                      H

                                                                      Bfrac14 12

                                                                      45frac14 27 and A frac14 042

                                                                      13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                      64 Bearing capacity

                                                                      For sc frac14 30mm

                                                                      qn frac14 30

                                                                      0147frac14 204 kN=m2

                                                                      q frac14 204thorn 21 frac14 225 kN=m2

                                                                      Design load frac14 42 225 frac14 3600 kN

                                                                      The design load is 3600 kN settlement being the limiting criterion

                                                                      87

                                                                      D

                                                                      Bfrac14 8

                                                                      4frac14 20

                                                                      From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                      F frac14 cuNc

                                                                      Dfrac14 40 71

                                                                      20 8frac14 18

                                                                      88

                                                                      Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                      Design value of 0 frac14 tan1tan 38

                                                                      125

                                                                      frac14 32

                                                                      Figure Q86

                                                                      Bearing capacity 65

                                                                      For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                      Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                      The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                      Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                      For B frac14 250m qn frac14 3750

                                                                      2502 17 frac14 583 kN=m2

                                                                      From Figure 510 m frac14 n frac14 126

                                                                      6frac14 021

                                                                      Ir frac14 0019

                                                                      Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                      Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                      The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                      89

                                                                      Depth (m) N 0v (kNm2) CN N1

                                                                      070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                      Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                      (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                      Cw frac14 05thorn 0530

                                                                      47

                                                                      frac14 082

                                                                      66 Bearing capacity

                                                                      Thus

                                                                      qa frac14 150 082 frac14 120 kN=m2

                                                                      (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                      Thus

                                                                      qa frac14 90 15 frac14 135 kN=m2

                                                                      (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                      Ic frac14 171

                                                                      1014frac14 0068

                                                                      From Equation 819(a) with s frac14 25mm

                                                                      q frac14 25

                                                                      3507 0068frac14 150 kN=m2

                                                                      810

                                                                      Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                      Izp frac14 05thorn 01130

                                                                      16 27

                                                                      05

                                                                      frac14 067

                                                                      Refer to Figure Q810

                                                                      E frac14 25qc

                                                                      Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                      Ez (mm3MN)

                                                                      1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                      0203

                                                                      C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                      130frac14 093

                                                                      C2 frac14 1 ethsayTHORN

                                                                      s frac14 C1C2qnX Iz

                                                                      Ez frac14 093 1 130 0203 frac14 25mm

                                                                      Bearing capacity 67

                                                                      811

                                                                      At pile base level

                                                                      cu frac14 220 kN=m2

                                                                      qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                      Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                      00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                      qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                      Then

                                                                      Qf frac14 Abqb thorn Asqs

                                                                      frac14

                                                                      4 32 1980

                                                                      thorn eth 105 139 86THORN

                                                                      frac14 13 996thorn 3941 frac14 17 937 kN

                                                                      0 01 02 03 04 05 06 07

                                                                      0 2 4 6 8 10 12 14

                                                                      1

                                                                      2

                                                                      3

                                                                      4

                                                                      5

                                                                      6

                                                                      7

                                                                      8

                                                                      (1)

                                                                      (2)

                                                                      (3)

                                                                      (4)

                                                                      (5)

                                                                      qc

                                                                      qc

                                                                      Iz

                                                                      Iz

                                                                      (MNm2)

                                                                      z (m)

                                                                      Figure Q810

                                                                      68 Bearing capacity

                                                                      Allowable load

                                                                      ethaTHORN Qf

                                                                      2frac14 17 937

                                                                      2frac14 8968 kN

                                                                      ethbTHORN Abqb

                                                                      3thorn Asqs frac14 13 996

                                                                      3thorn 3941 frac14 8606 kN

                                                                      ie allowable load frac14 8600 kN

                                                                      Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                      According to the limit state method

                                                                      Characteristic undrained strength at base level cuk frac14 220

                                                                      150kN=m2

                                                                      Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                      150frac14 1320 kN=m2

                                                                      Characteristic shaft resistance qsk frac14 00150

                                                                      frac14 86

                                                                      150frac14 57 kN=m2

                                                                      Characteristic base and shaft resistances

                                                                      Rbk frac14

                                                                      4 32 1320 frac14 9330 kN

                                                                      Rsk frac14 105 139 86

                                                                      150frac14 2629 kN

                                                                      For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                      Design bearing resistance Rcd frac14 9330

                                                                      160thorn 2629

                                                                      130

                                                                      frac14 5831thorn 2022

                                                                      frac14 7850 kN

                                                                      Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                      812

                                                                      ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                      qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                      For a single pile

                                                                      Qf frac14 Abqb thorn Asqs

                                                                      frac14

                                                                      4 062 1305

                                                                      thorn eth 06 15 42THORN

                                                                      frac14 369thorn 1187 frac14 1556 kN

                                                                      Bearing capacity 69

                                                                      Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                      eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                      (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                      (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                      qbkfrac14 9cuk frac14 9 220

                                                                      150frac14 1320 kN=m2

                                                                      qskfrac14cuk frac14 040 105

                                                                      150frac14 28 kN=m2

                                                                      Rbkfrac14

                                                                      4 0602 1320 frac14 373 kN

                                                                      Rskfrac14 060 15 28 frac14 791 kN

                                                                      Rcdfrac14 373

                                                                      160thorn 791

                                                                      130frac14 233thorn 608 frac14 841 kN

                                                                      Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                      Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                      (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                      q frac14 21 000

                                                                      1762frac14 68 kN=m2

                                                                      Immediate settlement

                                                                      H

                                                                      Bfrac14 15

                                                                      176frac14 085

                                                                      D

                                                                      Bfrac14 13

                                                                      176frac14 074

                                                                      L

                                                                      Bfrac14 1

                                                                      Hence from Figure 515

                                                                      130 frac14 078 and 131 frac14 041

                                                                      70 Bearing capacity

                                                                      Thus using Equation 528

                                                                      si frac14 078 041 68 176

                                                                      65frac14 6mm

                                                                      Consolidation settlement

                                                                      Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                      1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                      434 (sod)

                                                                      Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                      sc frac14 056 434 frac14 24mm

                                                                      The total settlement is (6thorn 24) frac14 30mm

                                                                      813

                                                                      At base level N frac14 26 Then using Equation 830

                                                                      qb frac14 40NDb

                                                                      Bfrac14 40 26 2

                                                                      025frac14 8320 kN=m2

                                                                      ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                      Figure Q812

                                                                      Bearing capacity 71

                                                                      Over the length embedded in sand

                                                                      N frac14 21 ie18thorn 24

                                                                      2

                                                                      Using Equation 831

                                                                      qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                      For a single pile

                                                                      Qf frac14 Abqb thorn Asqs

                                                                      frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                      For the pile group assuming a group efficiency of 12

                                                                      XQf frac14 12 9 604 frac14 6523 kN

                                                                      Then the load factor is

                                                                      F frac14 6523

                                                                      2000thorn 1000frac14 21

                                                                      (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                      Characteristic base resistance per unit area qbk frac14 8320

                                                                      150frac14 5547 kNm2

                                                                      Characteristic shaft resistance per unit area qsk frac14 42

                                                                      150frac14 28 kNm2

                                                                      Characteristic base and shaft resistances for a single pile

                                                                      Rbk frac14 0252 5547 frac14 347 kN

                                                                      Rsk frac14 4 025 2 28 frac14 56 kN

                                                                      For a driven pile the partial factors are b frac14 s frac14 130

                                                                      Design bearing resistance Rcd frac14 347

                                                                      130thorn 56

                                                                      130frac14 310 kN

                                                                      For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                      Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                      (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                      From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                      72 Bearing capacity

                                                                      N frac14 24thorn 26thorn 34

                                                                      3frac14 28

                                                                      Ic frac14 171

                                                                      2814frac14 0016 ethEquation 818THORN

                                                                      s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                      814

                                                                      Using Equation 841

                                                                      Tf frac14 DLcu thorn

                                                                      4ethD2 d2THORNcuNc

                                                                      frac14 eth 02 5 06 110THORN thorn

                                                                      4eth022 012THORN110 9

                                                                      frac14 207thorn 23 frac14 230 kN

                                                                      Figure Q813

                                                                      Bearing capacity 73

                                                                      Chapter 9

                                                                      Stability of slopes

                                                                      91

                                                                      Referring to Figure Q91

                                                                      W frac14 417 19 frac14 792 kN=m

                                                                      Q frac14 20 28 frac14 56 kN=m

                                                                      Arc lengthAB frac14

                                                                      180 73 90 frac14 115m

                                                                      Arc length BC frac14

                                                                      180 28 90 frac14 44m

                                                                      The factor of safety is given by

                                                                      F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                      90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                      Depth of tension crack z0 frac14 2cu

                                                                      frac14 2 20

                                                                      19frac14 21m

                                                                      Arc length BD frac14

                                                                      180 13

                                                                      1

                                                                      2 90 frac14 21m

                                                                      F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                      The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                      Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                      14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                      Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                      The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                      92

                                                                      u frac14 0

                                                                      Depth factor D frac14 11

                                                                      9frac14 122

                                                                      Using Equation 92 with F frac14 10

                                                                      Ns frac14 cu

                                                                      FHfrac14 30

                                                                      10 19 9frac14 0175

                                                                      Hence from Figure 93

                                                                      frac14 50

                                                                      For F frac14 12

                                                                      Ns frac14 30

                                                                      12 19 9frac14 0146

                                                                      frac14 27

                                                                      93

                                                                      Refer to Figure Q93

                                                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                      1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                      74 m

                                                                      214 1deg

                                                                      213 1deg

                                                                      39 m

                                                                      WB

                                                                      D

                                                                      C

                                                                      28 m

                                                                      21 m

                                                                      A

                                                                      Q

                                                                      Soil (1)Soil (2)

                                                                      73deg

                                                                      Figure Q91

                                                                      Stability of slopes 75

                                                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                      9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                      599 256 328 1372

                                                                      Figure Q93

                                                                      76 Stability of slopes

                                                                      XW cos frac14 b

                                                                      Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                      W sin frac14 bX

                                                                      h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                      Arc length La frac14

                                                                      180 57

                                                                      1

                                                                      2 326 frac14 327m

                                                                      The factor of safety is given by

                                                                      F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                      W sin

                                                                      frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                      frac14 091

                                                                      According to the limit state method

                                                                      0d frac14 tan1tan 32

                                                                      125

                                                                      frac14 265

                                                                      c0 frac14 8

                                                                      160frac14 5 kN=m2

                                                                      Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                      Design disturbing moment frac14 1075 kN=m

                                                                      The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                      94

                                                                      F frac14 1

                                                                      W sin

                                                                      Xfc0bthorn ethW ubTHORN tan0g sec

                                                                      1thorn ethtan tan0=FTHORN

                                                                      c0 frac14 8 kN=m2

                                                                      0 frac14 32

                                                                      c0b frac14 8 2 frac14 16 kN=m

                                                                      W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                      Try F frac14 100

                                                                      tan0

                                                                      Ffrac14 0625

                                                                      Stability of slopes 77

                                                                      Values of u are as obtained in Figure Q93

                                                                      SliceNo

                                                                      h(m)

                                                                      W frac14 bh(kNm)

                                                                      W sin(kNm)

                                                                      ub(kNm)

                                                                      c0bthorn (W ub) tan0(kNm)

                                                                      sec

                                                                      1thorn (tan tan0)FProduct(kNm)

                                                                      1 05 21 6 2 8 24 1078 262 13 55 31

                                                                      23 33 30 1042 31

                                                                      3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                      224 92 72 0931 67

                                                                      6 50 210 11 40 100 85 0907 777 55 231 14

                                                                      12 58 112 90 0889 80

                                                                      8 60 252 1812

                                                                      80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                      10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                      2154 88 116 0853 99

                                                                      14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                      1074 1091

                                                                      F frac14 1091

                                                                      1074frac14 102 (assumed value 100)

                                                                      Thus

                                                                      F frac14 101

                                                                      95

                                                                      F frac14 1

                                                                      W sin

                                                                      XfWeth1 ruTHORN tan0g sec

                                                                      1thorn ethtan tan0THORN=F

                                                                      0 frac14 33

                                                                      ru frac14 020

                                                                      W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                      eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                      Try F frac14 110

                                                                      tan 0

                                                                      Ffrac14 tan 33

                                                                      110frac14 0590

                                                                      78 Stability of slopes

                                                                      Referring to Figure Q95

                                                                      SliceNo

                                                                      h(m)

                                                                      W frac14 bh(kNm)

                                                                      W sin(kNm)

                                                                      W(1 ru) tan0(kNm)

                                                                      sec

                                                                      1thorn ( tan tan0)FProduct(kNm)

                                                                      1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                      2120 234 0892 209

                                                                      4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                      1185 1271

                                                                      Figure Q95

                                                                      Stability of slopes 79

                                                                      F frac14 1271

                                                                      1185frac14 107

                                                                      The trial value was 110 therefore take F to be 108

                                                                      96

                                                                      (a) Water table at surface the factor of safety is given by Equation 912

                                                                      F frac14 0

                                                                      sat

                                                                      tan0

                                                                      tan

                                                                      ptie 15 frac14 92

                                                                      19

                                                                      tan 36

                                                                      tan

                                                                      tan frac14 0234

                                                                      frac14 13

                                                                      Water table well below surface the factor of safety is given by Equation 911

                                                                      F frac14 tan0

                                                                      tan

                                                                      frac14 tan 36

                                                                      tan 13

                                                                      frac14 31

                                                                      (b) 0d frac14 tan1tan 36

                                                                      125

                                                                      frac14 30

                                                                      Depth of potential failure surface frac14 z

                                                                      Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                      frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                      frac14 504z kN

                                                                      Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                      frac14 19 z sin 13 cos 13

                                                                      frac14 416z kN

                                                                      Rd gtSd therefore the limit state for overall stability is satisfied

                                                                      80 Stability of slopes

                                                                      • Book Cover
                                                                      • Title
                                                                      • Contents
                                                                      • Basic characteristics of soils
                                                                      • Seepage
                                                                      • Effective stress
                                                                      • Shear strength
                                                                      • Stresses and displacements
                                                                      • Lateral earth pressure
                                                                      • Consolidation theory
                                                                      • Bearing capacity
                                                                      • Stability of slopes

                                                                        z (m) m n Ir z (kNm2)

                                                                        0 ndash ndash (250)05 200 0233 2331 100 0176 17615 067 0122 1222 050 0085 853 033 0045 45 4 025 0027 277 014 0009 9

                                                                        10 010 0005 5

                                                                        z is plotted against z in Figure Q53

                                                                        54

                                                                        (a)

                                                                        m frac14 125

                                                                        12frac14 104

                                                                        n frac14 18

                                                                        12frac14 150

                                                                        From Figure 510 Irfrac14 0196

                                                                        z frac14 2 175 0196 frac14 68 kN=m2

                                                                        Figure Q53

                                                                        30 Stresses and displacements

                                                                        (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                                                        z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                                                        55

                                                                        Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                                                        Px frac14 2Q

                                                                        1

                                                                        m2 thorn 1frac14 2 150

                                                                        125frac14 76 kN=m

                                                                        Equation 517 is used to obtain the pressure distribution

                                                                        px frac14 4Q

                                                                        h

                                                                        m2n

                                                                        ethm2 thorn n2THORN2 frac14150

                                                                        m2n

                                                                        ethm2 thorn n2THORN2 ethkN=m2THORN

                                                                        Figure Q54

                                                                        Stresses and displacements 31

                                                                        n m2n

                                                                        (m2 thorn n2)2

                                                                        px(kNm2)

                                                                        0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                                        The pressure distribution is plotted in Figure Q55

                                                                        56

                                                                        H

                                                                        Bfrac14 10

                                                                        2frac14 5

                                                                        L

                                                                        Bfrac14 4

                                                                        2frac14 2

                                                                        D

                                                                        Bfrac14 1

                                                                        2frac14 05

                                                                        Hence from Figure 515

                                                                        131 frac14 082

                                                                        130 frac14 094

                                                                        Figure Q55

                                                                        32 Stresses and displacements

                                                                        The immediate settlement is given by Equation 528

                                                                        si frac14 130131qB

                                                                        Eu

                                                                        frac14 094 082 200 2

                                                                        45frac14 7mm

                                                                        Stresses and displacements 33

                                                                        Chapter 6

                                                                        Lateral earth pressure

                                                                        61

                                                                        For 0 frac14 37 the active pressure coefficient is given by

                                                                        Ka frac14 1 sin 37

                                                                        1thorn sin 37frac14 025

                                                                        The total active thrust (Equation 66a with c0 frac14 0) is

                                                                        Pa frac14 1

                                                                        2KaH

                                                                        2 frac14 1

                                                                        2 025 17 62 frac14 765 kN=m

                                                                        If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                                        K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                                        and the thrust on the wall is

                                                                        P0 frac14 1

                                                                        2K0H

                                                                        2 frac14 1

                                                                        2 040 17 62 frac14 122 kN=m

                                                                        62

                                                                        The active pressure coefficients for the three soil types are as follows

                                                                        Ka1 frac141 sin 35

                                                                        1thorn sin 35frac14 0271

                                                                        Ka2 frac141 sin 27

                                                                        1thorn sin 27frac14 0375

                                                                        ffiffiffiffiffiffiffiKa2

                                                                        p frac14 0613

                                                                        Ka3 frac141 sin 42

                                                                        1thorn sin 42frac14 0198

                                                                        Distribution of active pressure (plotted in Figure Q62)

                                                                        Depth (m) Soil Active pressure (kNm2)

                                                                        3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                                        12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                                        At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                                        Total thrust frac14 571 kNm

                                                                        Point of application is (4893571) m from the top of the wall ie 857m

                                                                        Force (kN) Arm (m) Moment (kN m)

                                                                        (1)1

                                                                        2 0271 16 32 frac14 195 20 390

                                                                        (2) 0271 16 3 2 frac14 260 40 1040

                                                                        (3)1

                                                                        2 0271 92 22 frac14 50 433 217

                                                                        (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                                        (5)1

                                                                        2 0375 102 32 frac14 172 70 1204

                                                                        (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                                        (7)1

                                                                        2 0198 112 42 frac14 177 1067 1889

                                                                        (8)1

                                                                        2 98 92 frac14 3969 90 35721

                                                                        5713 48934

                                                                        Figure Q62

                                                                        Lateral earth pressure 35

                                                                        63

                                                                        (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                                        Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                                        frac14 245

                                                                        At the lower end of the piling

                                                                        pa frac14 Kaqthorn Kasatz Kaccu

                                                                        frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                                        frac14 115 kN=m2

                                                                        pp frac14 Kpsatzthorn Kpccu

                                                                        frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                                        frac14 202 kN=m2

                                                                        (b) For 0 frac14 26 and frac14 1

                                                                        20

                                                                        Ka frac14 035

                                                                        Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                                        pfrac14 145 ethEquation 619THORN

                                                                        Kp frac14 37

                                                                        Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                                        pfrac14 47 ethEquation 624THORN

                                                                        At the lower end of the piling

                                                                        pa frac14 Kaqthorn Ka0z Kacc

                                                                        0

                                                                        frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                                        frac14 187 kN=m2

                                                                        pp frac14 Kp0zthorn Kpcc

                                                                        0

                                                                        frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                                        frac14 198 kN=m2

                                                                        36 Lateral earth pressure

                                                                        64

                                                                        (a) For 0 frac14 38 Ka frac14 024

                                                                        0 frac14 20 98 frac14 102 kN=m3

                                                                        The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                                        Force (kN) Arm (m) Moment (kN m)

                                                                        (1) 024 10 66 frac14 159 33 525

                                                                        (2)1

                                                                        2 024 17 392 frac14 310 400 1240

                                                                        (3) 024 17 39 27 frac14 430 135 580

                                                                        (4)1

                                                                        2 024 102 272 frac14 89 090 80

                                                                        (5)1

                                                                        2 98 272 frac14 357 090 321

                                                                        Hfrac14 1345 MH frac14 2746

                                                                        (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                                        (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                                        XM frac14MV MH frac14 7790 kNm

                                                                        Lever arm of base resultant

                                                                        M

                                                                        Vfrac14 779

                                                                        488frac14 160

                                                                        Eccentricity of base resultant

                                                                        e frac14 200 160 frac14 040m

                                                                        39 m

                                                                        27 m

                                                                        40 m

                                                                        04 m

                                                                        04 m

                                                                        26 m

                                                                        (7)

                                                                        (9)

                                                                        (1)(2)

                                                                        (3)

                                                                        (4)

                                                                        (5)

                                                                        (8)(6)

                                                                        (10)

                                                                        WT

                                                                        10 kNm2

                                                                        Hydrostatic

                                                                        Figure Q64

                                                                        Lateral earth pressure 37

                                                                        Base pressures (Equation 627)

                                                                        p frac14 VB

                                                                        1 6e

                                                                        B

                                                                        frac14 488

                                                                        4eth1 060THORN

                                                                        frac14 195 kN=m2 and 49 kN=m2

                                                                        Factor of safety against sliding (Equation 628)

                                                                        F frac14 V tan

                                                                        Hfrac14 488 tan 25

                                                                        1345frac14 17

                                                                        (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                                        Hfrac14 1633 kN

                                                                        V frac14 4879 kN

                                                                        MH frac14 3453 kNm

                                                                        MV frac14 10536 kNm

                                                                        The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                                        65

                                                                        For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                        Kp

                                                                        Ffrac14 385

                                                                        2

                                                                        0 frac14 20 98 frac14 102 kN=m3

                                                                        The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                                        Force (kN) Arm (m) Moment (kN m)

                                                                        (1)1

                                                                        2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                                        (2) 026 17 45 d frac14 199d d2 995d2

                                                                        (3)1

                                                                        2 026 102 d2 frac14 133d2 d3 044d3

                                                                        (4)1

                                                                        2 385

                                                                        2 17 152 frac14 368 dthorn 05 368d 184

                                                                        (5)385

                                                                        2 17 15 d frac14 491d d2 2455d2

                                                                        (6)1

                                                                        2 385

                                                                        2 102 d2 frac14 982d2 d3 327d3

                                                                        38 Lateral earth pressure

                                                                        XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                                        d3 thorn 516d2 283d 1724 frac14 0

                                                                        d frac14 179m

                                                                        Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                                        Over additional 20 embedded depth

                                                                        pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                                        Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                                        66

                                                                        The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                                        Ka frac14 sin 69=sin 105

                                                                        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                                        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                                        pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                                        26664

                                                                        37775

                                                                        2

                                                                        frac14 050

                                                                        The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                                        Pa frac14 1

                                                                        2 050 19 7502 frac14 267 kN=m

                                                                        Figure Q65

                                                                        Lateral earth pressure 39

                                                                        Horizontal component

                                                                        Ph frac14 267 cos 40 frac14 205 kN=m

                                                                        Vertical component

                                                                        Pv frac14 267 sin 40 frac14 172 kN=m

                                                                        Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                        Force (kN) Arm (m) Moment (kN m)

                                                                        (1)1

                                                                        2 175 650 235 frac14 1337 258 345

                                                                        (2) 050 650 235 frac14 764 175 134

                                                                        (3)1

                                                                        2 070 650 235 frac14 535 127 68

                                                                        (4) 100 400 235 frac14 940 200 188

                                                                        (5) 1

                                                                        2 080 050 235 frac14 47 027 1

                                                                        Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                        Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                        Lever arm of base resultant

                                                                        M

                                                                        Vfrac14 795

                                                                        525frac14 151m

                                                                        Eccentricity of base resultant

                                                                        e frac14 200 151 frac14 049m

                                                                        Figure Q66

                                                                        40 Lateral earth pressure

                                                                        Base pressures (Equation 627)

                                                                        p frac14 525

                                                                        41 6 049

                                                                        4

                                                                        frac14 228 kN=m2 and 35 kN=m2

                                                                        The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                        The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                        The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                        67

                                                                        For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                        Force (kN) Arm (m) Moment (kNm)

                                                                        (1)1

                                                                        2 027 17 52 frac14 574 183 1050

                                                                        (2) 027 17 5 3 frac14 689 500 3445

                                                                        (3)1

                                                                        2 027 102 32 frac14 124 550 682

                                                                        (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                        (5)1

                                                                        2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                        (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                        (7) 1

                                                                        2 267

                                                                        2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                        (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                        2 d frac14 163d d2thorn 650 82d2 1060d

                                                                        Tie rod force per m frac14 T 0 0

                                                                        XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                        d3 thorn 77d2 269d 1438 frac14 0

                                                                        d frac14 467m

                                                                        Depth of penetration frac14 12d frac14 560m

                                                                        Lateral earth pressure 41

                                                                        Algebraic sum of forces for d frac14 467m isX

                                                                        F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                        T frac14 905 kN=m

                                                                        Force in each tie rod frac14 25T frac14 226 kN

                                                                        68

                                                                        (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                        0 frac14 21 98 frac14 112 kN=m3

                                                                        The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                        uC frac14 150

                                                                        165 15 98 frac14 134 kN=m2

                                                                        The average seepage pressure is

                                                                        j frac14 15

                                                                        165 98 frac14 09 kN=m3

                                                                        Hence

                                                                        0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                        0 j frac14 112 09 frac14 103 kN=m3

                                                                        Figure Q67

                                                                        42 Lateral earth pressure

                                                                        Consider moments about the anchor point A (per m)

                                                                        Force (kN) Arm (m) Moment (kN m)

                                                                        (1) 10 026 150 frac14 390 60 2340

                                                                        (2)1

                                                                        2 026 18 452 frac14 474 15 711

                                                                        (3) 026 18 45 105 frac14 2211 825 18240

                                                                        (4)1

                                                                        2 026 121 1052 frac14 1734 100 17340

                                                                        (5)1

                                                                        2 134 15 frac14 101 40 404

                                                                        (6) 134 30 frac14 402 60 2412

                                                                        (7)1

                                                                        2 134 60 frac14 402 95 3819

                                                                        571 4527(8) Ppm

                                                                        115 115PPm

                                                                        XM frac14 0

                                                                        Ppm frac144527

                                                                        115frac14 394 kN=m

                                                                        Available passive resistance

                                                                        Pp frac14 1

                                                                        2 385 103 62 frac14 714 kN=m

                                                                        Factor of safety

                                                                        Fp frac14 Pp

                                                                        Ppm

                                                                        frac14 714

                                                                        394frac14 18

                                                                        Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                        Figure Q68

                                                                        Lateral earth pressure 43

                                                                        (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                        Consider moments (per m) about the tie point A

                                                                        Force (kN) Arm (m)

                                                                        (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                        (2)1

                                                                        2 033 18 452 frac14 601 15

                                                                        (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                        (4)1

                                                                        2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                        (5)1

                                                                        2 134 15 frac14 101 40

                                                                        (6) 134 30 frac14 402 60

                                                                        (7)1

                                                                        2 134 d frac14 67d d3thorn 75

                                                                        (8) 1

                                                                        2 30 103 d2 frac141545d2 2d3thorn 75

                                                                        Moment (kN m)

                                                                        (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                        XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                        d3 thorn 827d2 466d 1518 frac14 0

                                                                        By trial

                                                                        d frac14 544m

                                                                        The minimum depth of embedment required is 544m

                                                                        69

                                                                        For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                        0 frac14 20 98 frac14 102 kN=m3

                                                                        The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                        44 Lateral earth pressure

                                                                        uC frac14 147

                                                                        173 26 98 frac14 216 kN=m2

                                                                        and the average seepage pressure around the wall is

                                                                        j frac14 26

                                                                        173 98 frac14 15 kN=m3

                                                                        Consider moments about the prop (A) (per m)

                                                                        Force (kN) Arm (m) Moment (kN m)

                                                                        (1)1

                                                                        2 03 17 272 frac14 186 020 37

                                                                        (2) 03 17 27 53 frac14 730 335 2445

                                                                        (3)1

                                                                        2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                        (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                        (5)1

                                                                        2 216 26 frac14 281 243 684

                                                                        (6) 216 27 frac14 583 465 2712

                                                                        (7)1

                                                                        2 216 60 frac14 648 800 5184

                                                                        3055(8)

                                                                        1

                                                                        2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                        Factor of safety

                                                                        Fr frac14 6885

                                                                        3055frac14 225

                                                                        Figure Q69

                                                                        Lateral earth pressure 45

                                                                        610

                                                                        For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                        p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                        Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                        Using the recommendations of Twine and Roscoe

                                                                        p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                        Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                        611

                                                                        frac14 18 kN=m3 0 frac14 34

                                                                        H frac14 350m nH frac14 335m mH frac14 185m

                                                                        Consider a trial value of F frac14 20 Refer to Figure 635

                                                                        0m frac14 tan1tan 34

                                                                        20

                                                                        frac14 186

                                                                        Then

                                                                        frac14 45 thorn 0m2frac14 543

                                                                        W frac14 1

                                                                        2 18 3502 cot 543 frac14 792 kN=m

                                                                        Figure Q610

                                                                        46 Lateral earth pressure

                                                                        P frac14 1

                                                                        2 s 3352 frac14 561s kN=m

                                                                        U frac14 1

                                                                        2 98 1852 cosec 543 frac14 206 kN=m

                                                                        Equations 630 and 631 then become

                                                                        561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                        792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                        ie

                                                                        561s 0616N 405 frac14 0

                                                                        792 0857N thorn 563 frac14 0

                                                                        N frac14 848

                                                                        0857frac14 989 kN=m

                                                                        Then

                                                                        561s 609 405 frac14 0

                                                                        s frac14 649

                                                                        561frac14 116 kN=m3

                                                                        The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                        F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                        20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                        s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                        Figure Q611

                                                                        Lateral earth pressure 47

                                                                        612

                                                                        For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                        45 thorn 0

                                                                        2frac14 63

                                                                        For the retained material between the surface and a depth of 36m

                                                                        Pa frac14 1

                                                                        2 030 18 362 frac14 350 kN=m

                                                                        Weight of reinforced fill between the surface and a depth of 36m is

                                                                        Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                        eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                        Eccentricity of Rv

                                                                        e frac14 263 250 frac14 013m

                                                                        The average vertical stress at a depth of 36m is

                                                                        z frac14 Rv

                                                                        L 2efrac14 324

                                                                        474frac14 68 kN=m2

                                                                        (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                        Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                        Tensile stress in the element frac14 138 103

                                                                        65 3frac14 71N=mm2

                                                                        Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                        Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                        Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                        The weight of ABC is

                                                                        W frac14 1

                                                                        2 18 52 265 frac14 124 kN=m

                                                                        From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                        48 Lateral earth pressure

                                                                        (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                        Tp frac14 032 68 120 065 frac14 170 kN

                                                                        Tr frac14 213 420

                                                                        418frac14 214 kN

                                                                        Again the tensile failure and slipping limit states are satisfied for this element

                                                                        Figure Q612

                                                                        Lateral earth pressure 49

                                                                        Chapter 7

                                                                        Consolidation theory

                                                                        71

                                                                        Total change in thickness

                                                                        H frac14 782 602 frac14 180mm

                                                                        Average thickness frac14 1530thorn 180

                                                                        2frac14 1620mm

                                                                        Length of drainage path d frac14 1620

                                                                        2frac14 810mm

                                                                        Root time plot (Figure Q71a)

                                                                        ffiffiffiffiffiffit90p frac14 33

                                                                        t90 frac14 109min

                                                                        cv frac14 0848d2

                                                                        t90frac14 0848 8102

                                                                        109 1440 365

                                                                        106frac14 27m2=year

                                                                        r0 frac14 782 764

                                                                        782 602frac14 018

                                                                        180frac14 0100

                                                                        rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                        10 119

                                                                        9 180frac14 0735

                                                                        rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                        Log time plot (Figure Q71b)

                                                                        t50 frac14 26min

                                                                        cv frac14 0196d2

                                                                        t50frac14 0196 8102

                                                                        26 1440 365

                                                                        106frac14 26m2=year

                                                                        r0 frac14 782 763

                                                                        782 602frac14 019

                                                                        180frac14 0106

                                                                        rp frac14 763 623

                                                                        782 602frac14 140

                                                                        180frac14 0778

                                                                        rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                        Figure Q71(a)

                                                                        Figure Q71(b)

                                                                        Final void ratio

                                                                        e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                        e

                                                                        Hfrac14 1thorn e0

                                                                        H0frac14 1thorn e1 thorne

                                                                        H0

                                                                        ie

                                                                        e

                                                                        180frac14 1631thorne

                                                                        1710

                                                                        e frac14 2936

                                                                        1530frac14 0192

                                                                        Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                        mv frac14 1

                                                                        1thorn e0 e0 e101 00

                                                                        frac14 1

                                                                        1823 0192

                                                                        0107frac14 098m2=MN

                                                                        k frac14 cvmvw frac14 265 098 98

                                                                        60 1440 365 103frac14 81 1010 m=s

                                                                        72

                                                                        Using Equation 77 (one-dimensional method)

                                                                        sc frac14 e0 e11thorn e0 H

                                                                        Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                        Figure Q72

                                                                        52 Consolidation theory

                                                                        Settlement

                                                                        Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                        1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                        318

                                                                        Notes 5 92y 460thorn 84

                                                                        Heave

                                                                        Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                        1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                        38

                                                                        73

                                                                        U frac14 f ethTvTHORN frac14 f cvt

                                                                        d2

                                                                        Hence if cv is constant

                                                                        t1

                                                                        t2frac14 d

                                                                        21

                                                                        d22

                                                                        where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                        d1 frac14 95mm and d2 frac14 2500mm

                                                                        for U frac14 050 t2 frac14 t1 d22

                                                                        d21

                                                                        frac14 20

                                                                        60 24 365 25002

                                                                        952frac14 263 years

                                                                        for U lt 060 Tv frac14

                                                                        4U2 (Equation 724(a))

                                                                        t030 frac14 t050 0302

                                                                        0502

                                                                        frac14 263 036 frac14 095 years

                                                                        Consolidation theory 53

                                                                        74

                                                                        The layer is open

                                                                        d frac14 8

                                                                        2frac14 4m

                                                                        Tv frac14 cvtd2frac14 24 3

                                                                        42frac14 0450

                                                                        ui frac14 frac14 84 kN=m2

                                                                        The excess pore water pressure is given by Equation 721

                                                                        ue frac14Xmfrac141mfrac140

                                                                        2ui

                                                                        Msin

                                                                        Mz

                                                                        d

                                                                        expethM2TvTHORN

                                                                        In this case z frac14 d

                                                                        sinMz

                                                                        d

                                                                        frac14 sinM

                                                                        where

                                                                        M frac14

                                                                        23

                                                                        25

                                                                        2

                                                                        M sin M M2Tv exp (M2Tv)

                                                                        2thorn1 1110 0329

                                                                        3

                                                                        21 9993 457 105

                                                                        ue frac14 2 84 2

                                                                        1 0329 ethother terms negligibleTHORN

                                                                        frac14 352 kN=m2

                                                                        75

                                                                        The layer is open

                                                                        d frac14 6

                                                                        2frac14 3m

                                                                        Tv frac14 cvtd2frac14 10 3

                                                                        32frac14 0333

                                                                        The layer thickness will be divided into six equal parts ie m frac14 6

                                                                        54 Consolidation theory

                                                                        For an open layer

                                                                        Tv frac14 4n

                                                                        m2

                                                                        n frac14 0333 62

                                                                        4frac14 300

                                                                        The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                        ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                        i j

                                                                        0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                        0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                        The initial and 3-year isochrones are plotted in Figure Q75

                                                                        Area under initial isochrone frac14 180 units

                                                                        Area under 3-year isochrone frac14 63 units

                                                                        The average degree of consolidation is given by Equation 725Thus

                                                                        U frac14 1 63

                                                                        180frac14 065

                                                                        Figure Q75

                                                                        Consolidation theory 55

                                                                        76

                                                                        At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                        0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                        The final consolidation settlement (one-dimensional method) is

                                                                        sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                        Corrected time t frac14 2 1

                                                                        2

                                                                        40

                                                                        52

                                                                        frac14 1615 years

                                                                        Tv frac14 cvtd2frac14 44 1615

                                                                        42frac14 0444

                                                                        From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                        77

                                                                        The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                        Figure Q77

                                                                        56 Consolidation theory

                                                                        Point m n Ir (kNm2) sc (mm)

                                                                        13020frac14 15 20

                                                                        20frac14 10 0194 (4) 113 124

                                                                        260

                                                                        20frac14 30

                                                                        20

                                                                        20frac14 10 0204 (2) 59 65

                                                                        360

                                                                        20frac14 30

                                                                        40

                                                                        20frac14 20 0238 (1) 35 38

                                                                        430

                                                                        20frac14 15

                                                                        40

                                                                        20frac14 20 0224 (2) 65 72

                                                                        Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                        78

                                                                        Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                        (a) Immediate settlement

                                                                        H

                                                                        Bfrac14 30

                                                                        35frac14 086

                                                                        D

                                                                        Bfrac14 2

                                                                        35frac14 006

                                                                        Figure Q78

                                                                        Consolidation theory 57

                                                                        From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                        si frac14 130131qB

                                                                        Eufrac14 10 032 105 35

                                                                        40frac14 30mm

                                                                        (b) Consolidation settlement

                                                                        Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                        1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                        3150

                                                                        Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                        Now

                                                                        H

                                                                        Bfrac14 30

                                                                        35frac14 086 and A frac14 065

                                                                        from Figure 712 13 frac14 079

                                                                        sc frac14 13sod frac14 079 315 frac14 250mm

                                                                        Total settlement

                                                                        s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                        79

                                                                        Without sand drains

                                                                        Uv frac14 025

                                                                        Tv frac14 0049 ethfrom Figure 718THORN

                                                                        t frac14 Tvd2

                                                                        cvfrac14 0049 82

                                                                        cvWith sand drains

                                                                        R frac14 0564S frac14 0564 3 frac14 169m

                                                                        n frac14 Rrfrac14 169

                                                                        015frac14 113

                                                                        Tr frac14 cht

                                                                        4R2frac14 ch

                                                                        4 1692 0049 82

                                                                        cvethand ch frac14 cvTHORN

                                                                        frac14 0275

                                                                        Ur frac14 073 (from Figure 730)

                                                                        58 Consolidation theory

                                                                        Using Equation 740

                                                                        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                        U frac14 080

                                                                        710

                                                                        Without sand drains

                                                                        Uv frac14 090

                                                                        Tv frac14 0848

                                                                        t frac14 Tvd2

                                                                        cvfrac14 0848 102

                                                                        96frac14 88 years

                                                                        With sand drains

                                                                        R frac14 0564S frac14 0564 4 frac14 226m

                                                                        n frac14 Rrfrac14 226

                                                                        015frac14 15

                                                                        Tr

                                                                        Tvfrac14 chcv

                                                                        d2

                                                                        4R2ethsame tTHORN

                                                                        Tr

                                                                        Tvfrac14 140

                                                                        96 102

                                                                        4 2262frac14 714 eth1THORN

                                                                        Using Equation 740

                                                                        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                        An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                        Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                        040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                        Thus

                                                                        Uv frac14 0295 and Ur frac14 086

                                                                        t frac14 88 00683

                                                                        0848frac14 07 years

                                                                        Consolidation theory 59

                                                                        Chapter 8

                                                                        Bearing capacity

                                                                        81

                                                                        (a) The ultimate bearing capacity is given by Equation 83

                                                                        qf frac14 cNc thorn DNq thorn 1

                                                                        2BN

                                                                        For u frac14 0

                                                                        Nc frac14 514 Nq frac14 1 N frac14 0

                                                                        qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                        The net ultimate bearing capacity is

                                                                        qnf frac14 qf D frac14 540 kN=m2

                                                                        The net foundation pressure is

                                                                        qn frac14 q D frac14 425

                                                                        2 eth21 1THORN frac14 192 kN=m2

                                                                        The factor of safety (Equation 86) is

                                                                        F frac14 qnfqnfrac14 540

                                                                        192frac14 28

                                                                        (b) For 0 frac14 28

                                                                        Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                        qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                        2 112 2 13

                                                                        frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                        qnf frac14 574 112 frac14 563 kN=m2

                                                                        F frac14 563

                                                                        192frac14 29

                                                                        (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                        82

                                                                        For 0 frac14 38

                                                                        Nq frac14 49 N frac14 67

                                                                        qnf frac14 DethNq 1THORN thorn 1

                                                                        2BN ethfrom Equation 83THORN

                                                                        frac14 eth18 075 48THORN thorn 1

                                                                        2 18 15 67

                                                                        frac14 648thorn 905 frac14 1553 kN=m2

                                                                        qn frac14 500

                                                                        15 eth18 075THORN frac14 320 kN=m2

                                                                        F frac14 qnfqnfrac14 1553

                                                                        320frac14 48

                                                                        0d frac14 tan1tan 38

                                                                        125

                                                                        frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                        Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                        2 18 15 25

                                                                        frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                        Design load (action) Vd frac14 500 kN=m

                                                                        The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                        83

                                                                        D

                                                                        Bfrac14 350

                                                                        225frac14 155

                                                                        From Figure 85 for a square foundation

                                                                        Nc frac14 81

                                                                        Bearing capacity 61

                                                                        For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                        Nc frac14 084thorn 016B

                                                                        L

                                                                        81 frac14 745

                                                                        Using Equation 810

                                                                        qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                        For F frac14 3

                                                                        qn frac14 1006

                                                                        3frac14 335 kN=m2

                                                                        q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                        Design load frac14 405 450 225 frac14 4100 kN

                                                                        Design undrained strength cud frac14 135

                                                                        14frac14 96 kN=m2

                                                                        Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                        frac14 7241 kN

                                                                        Design load Vd frac14 4100 kN

                                                                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                        84

                                                                        For 0 frac14 40

                                                                        Nq frac14 64 N frac14 95

                                                                        qnf frac14 DethNq 1THORN thorn 04BN

                                                                        (a) Water table 5m below ground level

                                                                        qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                        qn frac14 400 17 frac14 383 kN=m2

                                                                        F frac14 2686

                                                                        383frac14 70

                                                                        (b) Water table 1m below ground level (ie at foundation level)

                                                                        0 frac14 20 98 frac14 102 kN=m3

                                                                        62 Bearing capacity

                                                                        qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                        F frac14 2040

                                                                        383frac14 53

                                                                        (c) Water table at ground level with upward hydraulic gradient 02

                                                                        eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                        qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                        F frac14 1296

                                                                        392frac14 33

                                                                        85

                                                                        The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                        Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                        Design value of 0 frac14 tan1tan 39

                                                                        125

                                                                        frac14 33

                                                                        For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                        Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                        86

                                                                        (a) Undrained shear for u frac14 0

                                                                        Nc frac14 514 Nq frac14 1 N frac14 0

                                                                        qnf frac14 12cuNc

                                                                        frac14 12 100 514 frac14 617 kN=m2

                                                                        qn frac14 qnfFfrac14 617

                                                                        3frac14 206 kN=m2

                                                                        q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                        Bearing capacity 63

                                                                        Drained shear for 0 frac14 32

                                                                        Nq frac14 23 N frac14 25

                                                                        0 frac14 21 98 frac14 112 kN=m3

                                                                        qnf frac14 0DethNq 1THORN thorn 040BN

                                                                        frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                        frac14 694 kN=m2

                                                                        q frac14 694

                                                                        3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                        Design load frac14 42 227 frac14 3632 kN

                                                                        (b) Design undrained strength cud frac14 100

                                                                        14frac14 71 kNm2

                                                                        Design bearing resistance Rd frac14 12cudNe area

                                                                        frac14 12 71 514 42

                                                                        frac14 7007 kN

                                                                        For drained shear 0d frac14 tan1tan 32

                                                                        125

                                                                        frac14 26

                                                                        Nq frac14 12 N frac14 10

                                                                        Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                        (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                        Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                        1 2 100 0175 0700qn 0182qn

                                                                        2 6 033 0044 0176qn 0046qn

                                                                        3 10 020 0017 0068qn 0018qn

                                                                        0246qn

                                                                        Diameter of equivalent circle B frac14 45m

                                                                        H

                                                                        Bfrac14 12

                                                                        45frac14 27 and A frac14 042

                                                                        13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                        64 Bearing capacity

                                                                        For sc frac14 30mm

                                                                        qn frac14 30

                                                                        0147frac14 204 kN=m2

                                                                        q frac14 204thorn 21 frac14 225 kN=m2

                                                                        Design load frac14 42 225 frac14 3600 kN

                                                                        The design load is 3600 kN settlement being the limiting criterion

                                                                        87

                                                                        D

                                                                        Bfrac14 8

                                                                        4frac14 20

                                                                        From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                        F frac14 cuNc

                                                                        Dfrac14 40 71

                                                                        20 8frac14 18

                                                                        88

                                                                        Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                        Design value of 0 frac14 tan1tan 38

                                                                        125

                                                                        frac14 32

                                                                        Figure Q86

                                                                        Bearing capacity 65

                                                                        For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                        Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                        The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                        Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                        For B frac14 250m qn frac14 3750

                                                                        2502 17 frac14 583 kN=m2

                                                                        From Figure 510 m frac14 n frac14 126

                                                                        6frac14 021

                                                                        Ir frac14 0019

                                                                        Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                        Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                        The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                        89

                                                                        Depth (m) N 0v (kNm2) CN N1

                                                                        070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                        Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                        (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                        Cw frac14 05thorn 0530

                                                                        47

                                                                        frac14 082

                                                                        66 Bearing capacity

                                                                        Thus

                                                                        qa frac14 150 082 frac14 120 kN=m2

                                                                        (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                        Thus

                                                                        qa frac14 90 15 frac14 135 kN=m2

                                                                        (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                        Ic frac14 171

                                                                        1014frac14 0068

                                                                        From Equation 819(a) with s frac14 25mm

                                                                        q frac14 25

                                                                        3507 0068frac14 150 kN=m2

                                                                        810

                                                                        Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                        Izp frac14 05thorn 01130

                                                                        16 27

                                                                        05

                                                                        frac14 067

                                                                        Refer to Figure Q810

                                                                        E frac14 25qc

                                                                        Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                        Ez (mm3MN)

                                                                        1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                        0203

                                                                        C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                        130frac14 093

                                                                        C2 frac14 1 ethsayTHORN

                                                                        s frac14 C1C2qnX Iz

                                                                        Ez frac14 093 1 130 0203 frac14 25mm

                                                                        Bearing capacity 67

                                                                        811

                                                                        At pile base level

                                                                        cu frac14 220 kN=m2

                                                                        qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                        Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                        00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                        qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                        Then

                                                                        Qf frac14 Abqb thorn Asqs

                                                                        frac14

                                                                        4 32 1980

                                                                        thorn eth 105 139 86THORN

                                                                        frac14 13 996thorn 3941 frac14 17 937 kN

                                                                        0 01 02 03 04 05 06 07

                                                                        0 2 4 6 8 10 12 14

                                                                        1

                                                                        2

                                                                        3

                                                                        4

                                                                        5

                                                                        6

                                                                        7

                                                                        8

                                                                        (1)

                                                                        (2)

                                                                        (3)

                                                                        (4)

                                                                        (5)

                                                                        qc

                                                                        qc

                                                                        Iz

                                                                        Iz

                                                                        (MNm2)

                                                                        z (m)

                                                                        Figure Q810

                                                                        68 Bearing capacity

                                                                        Allowable load

                                                                        ethaTHORN Qf

                                                                        2frac14 17 937

                                                                        2frac14 8968 kN

                                                                        ethbTHORN Abqb

                                                                        3thorn Asqs frac14 13 996

                                                                        3thorn 3941 frac14 8606 kN

                                                                        ie allowable load frac14 8600 kN

                                                                        Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                        According to the limit state method

                                                                        Characteristic undrained strength at base level cuk frac14 220

                                                                        150kN=m2

                                                                        Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                        150frac14 1320 kN=m2

                                                                        Characteristic shaft resistance qsk frac14 00150

                                                                        frac14 86

                                                                        150frac14 57 kN=m2

                                                                        Characteristic base and shaft resistances

                                                                        Rbk frac14

                                                                        4 32 1320 frac14 9330 kN

                                                                        Rsk frac14 105 139 86

                                                                        150frac14 2629 kN

                                                                        For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                        Design bearing resistance Rcd frac14 9330

                                                                        160thorn 2629

                                                                        130

                                                                        frac14 5831thorn 2022

                                                                        frac14 7850 kN

                                                                        Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                        812

                                                                        ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                        qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                        For a single pile

                                                                        Qf frac14 Abqb thorn Asqs

                                                                        frac14

                                                                        4 062 1305

                                                                        thorn eth 06 15 42THORN

                                                                        frac14 369thorn 1187 frac14 1556 kN

                                                                        Bearing capacity 69

                                                                        Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                        eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                        (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                        (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                        qbkfrac14 9cuk frac14 9 220

                                                                        150frac14 1320 kN=m2

                                                                        qskfrac14cuk frac14 040 105

                                                                        150frac14 28 kN=m2

                                                                        Rbkfrac14

                                                                        4 0602 1320 frac14 373 kN

                                                                        Rskfrac14 060 15 28 frac14 791 kN

                                                                        Rcdfrac14 373

                                                                        160thorn 791

                                                                        130frac14 233thorn 608 frac14 841 kN

                                                                        Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                        Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                        (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                        q frac14 21 000

                                                                        1762frac14 68 kN=m2

                                                                        Immediate settlement

                                                                        H

                                                                        Bfrac14 15

                                                                        176frac14 085

                                                                        D

                                                                        Bfrac14 13

                                                                        176frac14 074

                                                                        L

                                                                        Bfrac14 1

                                                                        Hence from Figure 515

                                                                        130 frac14 078 and 131 frac14 041

                                                                        70 Bearing capacity

                                                                        Thus using Equation 528

                                                                        si frac14 078 041 68 176

                                                                        65frac14 6mm

                                                                        Consolidation settlement

                                                                        Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                        1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                        434 (sod)

                                                                        Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                        sc frac14 056 434 frac14 24mm

                                                                        The total settlement is (6thorn 24) frac14 30mm

                                                                        813

                                                                        At base level N frac14 26 Then using Equation 830

                                                                        qb frac14 40NDb

                                                                        Bfrac14 40 26 2

                                                                        025frac14 8320 kN=m2

                                                                        ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                        Figure Q812

                                                                        Bearing capacity 71

                                                                        Over the length embedded in sand

                                                                        N frac14 21 ie18thorn 24

                                                                        2

                                                                        Using Equation 831

                                                                        qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                        For a single pile

                                                                        Qf frac14 Abqb thorn Asqs

                                                                        frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                        For the pile group assuming a group efficiency of 12

                                                                        XQf frac14 12 9 604 frac14 6523 kN

                                                                        Then the load factor is

                                                                        F frac14 6523

                                                                        2000thorn 1000frac14 21

                                                                        (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                        Characteristic base resistance per unit area qbk frac14 8320

                                                                        150frac14 5547 kNm2

                                                                        Characteristic shaft resistance per unit area qsk frac14 42

                                                                        150frac14 28 kNm2

                                                                        Characteristic base and shaft resistances for a single pile

                                                                        Rbk frac14 0252 5547 frac14 347 kN

                                                                        Rsk frac14 4 025 2 28 frac14 56 kN

                                                                        For a driven pile the partial factors are b frac14 s frac14 130

                                                                        Design bearing resistance Rcd frac14 347

                                                                        130thorn 56

                                                                        130frac14 310 kN

                                                                        For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                        Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                        (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                        From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                        72 Bearing capacity

                                                                        N frac14 24thorn 26thorn 34

                                                                        3frac14 28

                                                                        Ic frac14 171

                                                                        2814frac14 0016 ethEquation 818THORN

                                                                        s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                        814

                                                                        Using Equation 841

                                                                        Tf frac14 DLcu thorn

                                                                        4ethD2 d2THORNcuNc

                                                                        frac14 eth 02 5 06 110THORN thorn

                                                                        4eth022 012THORN110 9

                                                                        frac14 207thorn 23 frac14 230 kN

                                                                        Figure Q813

                                                                        Bearing capacity 73

                                                                        Chapter 9

                                                                        Stability of slopes

                                                                        91

                                                                        Referring to Figure Q91

                                                                        W frac14 417 19 frac14 792 kN=m

                                                                        Q frac14 20 28 frac14 56 kN=m

                                                                        Arc lengthAB frac14

                                                                        180 73 90 frac14 115m

                                                                        Arc length BC frac14

                                                                        180 28 90 frac14 44m

                                                                        The factor of safety is given by

                                                                        F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                        90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                        Depth of tension crack z0 frac14 2cu

                                                                        frac14 2 20

                                                                        19frac14 21m

                                                                        Arc length BD frac14

                                                                        180 13

                                                                        1

                                                                        2 90 frac14 21m

                                                                        F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                        The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                        Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                        14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                        Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                        The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                        92

                                                                        u frac14 0

                                                                        Depth factor D frac14 11

                                                                        9frac14 122

                                                                        Using Equation 92 with F frac14 10

                                                                        Ns frac14 cu

                                                                        FHfrac14 30

                                                                        10 19 9frac14 0175

                                                                        Hence from Figure 93

                                                                        frac14 50

                                                                        For F frac14 12

                                                                        Ns frac14 30

                                                                        12 19 9frac14 0146

                                                                        frac14 27

                                                                        93

                                                                        Refer to Figure Q93

                                                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                        1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                        74 m

                                                                        214 1deg

                                                                        213 1deg

                                                                        39 m

                                                                        WB

                                                                        D

                                                                        C

                                                                        28 m

                                                                        21 m

                                                                        A

                                                                        Q

                                                                        Soil (1)Soil (2)

                                                                        73deg

                                                                        Figure Q91

                                                                        Stability of slopes 75

                                                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                        9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                        599 256 328 1372

                                                                        Figure Q93

                                                                        76 Stability of slopes

                                                                        XW cos frac14 b

                                                                        Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                        W sin frac14 bX

                                                                        h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                        Arc length La frac14

                                                                        180 57

                                                                        1

                                                                        2 326 frac14 327m

                                                                        The factor of safety is given by

                                                                        F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                        W sin

                                                                        frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                        frac14 091

                                                                        According to the limit state method

                                                                        0d frac14 tan1tan 32

                                                                        125

                                                                        frac14 265

                                                                        c0 frac14 8

                                                                        160frac14 5 kN=m2

                                                                        Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                        Design disturbing moment frac14 1075 kN=m

                                                                        The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                        94

                                                                        F frac14 1

                                                                        W sin

                                                                        Xfc0bthorn ethW ubTHORN tan0g sec

                                                                        1thorn ethtan tan0=FTHORN

                                                                        c0 frac14 8 kN=m2

                                                                        0 frac14 32

                                                                        c0b frac14 8 2 frac14 16 kN=m

                                                                        W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                        Try F frac14 100

                                                                        tan0

                                                                        Ffrac14 0625

                                                                        Stability of slopes 77

                                                                        Values of u are as obtained in Figure Q93

                                                                        SliceNo

                                                                        h(m)

                                                                        W frac14 bh(kNm)

                                                                        W sin(kNm)

                                                                        ub(kNm)

                                                                        c0bthorn (W ub) tan0(kNm)

                                                                        sec

                                                                        1thorn (tan tan0)FProduct(kNm)

                                                                        1 05 21 6 2 8 24 1078 262 13 55 31

                                                                        23 33 30 1042 31

                                                                        3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                        224 92 72 0931 67

                                                                        6 50 210 11 40 100 85 0907 777 55 231 14

                                                                        12 58 112 90 0889 80

                                                                        8 60 252 1812

                                                                        80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                        10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                        2154 88 116 0853 99

                                                                        14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                        1074 1091

                                                                        F frac14 1091

                                                                        1074frac14 102 (assumed value 100)

                                                                        Thus

                                                                        F frac14 101

                                                                        95

                                                                        F frac14 1

                                                                        W sin

                                                                        XfWeth1 ruTHORN tan0g sec

                                                                        1thorn ethtan tan0THORN=F

                                                                        0 frac14 33

                                                                        ru frac14 020

                                                                        W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                        eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                        Try F frac14 110

                                                                        tan 0

                                                                        Ffrac14 tan 33

                                                                        110frac14 0590

                                                                        78 Stability of slopes

                                                                        Referring to Figure Q95

                                                                        SliceNo

                                                                        h(m)

                                                                        W frac14 bh(kNm)

                                                                        W sin(kNm)

                                                                        W(1 ru) tan0(kNm)

                                                                        sec

                                                                        1thorn ( tan tan0)FProduct(kNm)

                                                                        1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                        2120 234 0892 209

                                                                        4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                        1185 1271

                                                                        Figure Q95

                                                                        Stability of slopes 79

                                                                        F frac14 1271

                                                                        1185frac14 107

                                                                        The trial value was 110 therefore take F to be 108

                                                                        96

                                                                        (a) Water table at surface the factor of safety is given by Equation 912

                                                                        F frac14 0

                                                                        sat

                                                                        tan0

                                                                        tan

                                                                        ptie 15 frac14 92

                                                                        19

                                                                        tan 36

                                                                        tan

                                                                        tan frac14 0234

                                                                        frac14 13

                                                                        Water table well below surface the factor of safety is given by Equation 911

                                                                        F frac14 tan0

                                                                        tan

                                                                        frac14 tan 36

                                                                        tan 13

                                                                        frac14 31

                                                                        (b) 0d frac14 tan1tan 36

                                                                        125

                                                                        frac14 30

                                                                        Depth of potential failure surface frac14 z

                                                                        Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                        frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                        frac14 504z kN

                                                                        Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                        frac14 19 z sin 13 cos 13

                                                                        frac14 416z kN

                                                                        Rd gtSd therefore the limit state for overall stability is satisfied

                                                                        80 Stability of slopes

                                                                        • Book Cover
                                                                        • Title
                                                                        • Contents
                                                                        • Basic characteristics of soils
                                                                        • Seepage
                                                                        • Effective stress
                                                                        • Shear strength
                                                                        • Stresses and displacements
                                                                        • Lateral earth pressure
                                                                        • Consolidation theory
                                                                        • Bearing capacity
                                                                        • Stability of slopes

                                                                          (b) The foundation is drawn on Newmarkrsquos chart as shown in Figure Q54 the scaleline representing 12m (z) Number of influence areas (N )frac14 78

                                                                          z frac14 0005Nq frac14 0005 78 175 frac14 68 kN=m2

                                                                          55

                                                                          Qfrac14 150 kNm hfrac14 400m mfrac14 05 The total thrust is given by Equation 518

                                                                          Px frac14 2Q

                                                                          1

                                                                          m2 thorn 1frac14 2 150

                                                                          125frac14 76 kN=m

                                                                          Equation 517 is used to obtain the pressure distribution

                                                                          px frac14 4Q

                                                                          h

                                                                          m2n

                                                                          ethm2 thorn n2THORN2 frac14150

                                                                          m2n

                                                                          ethm2 thorn n2THORN2 ethkN=m2THORN

                                                                          Figure Q54

                                                                          Stresses and displacements 31

                                                                          n m2n

                                                                          (m2 thorn n2)2

                                                                          px(kNm2)

                                                                          0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                                          The pressure distribution is plotted in Figure Q55

                                                                          56

                                                                          H

                                                                          Bfrac14 10

                                                                          2frac14 5

                                                                          L

                                                                          Bfrac14 4

                                                                          2frac14 2

                                                                          D

                                                                          Bfrac14 1

                                                                          2frac14 05

                                                                          Hence from Figure 515

                                                                          131 frac14 082

                                                                          130 frac14 094

                                                                          Figure Q55

                                                                          32 Stresses and displacements

                                                                          The immediate settlement is given by Equation 528

                                                                          si frac14 130131qB

                                                                          Eu

                                                                          frac14 094 082 200 2

                                                                          45frac14 7mm

                                                                          Stresses and displacements 33

                                                                          Chapter 6

                                                                          Lateral earth pressure

                                                                          61

                                                                          For 0 frac14 37 the active pressure coefficient is given by

                                                                          Ka frac14 1 sin 37

                                                                          1thorn sin 37frac14 025

                                                                          The total active thrust (Equation 66a with c0 frac14 0) is

                                                                          Pa frac14 1

                                                                          2KaH

                                                                          2 frac14 1

                                                                          2 025 17 62 frac14 765 kN=m

                                                                          If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                                          K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                                          and the thrust on the wall is

                                                                          P0 frac14 1

                                                                          2K0H

                                                                          2 frac14 1

                                                                          2 040 17 62 frac14 122 kN=m

                                                                          62

                                                                          The active pressure coefficients for the three soil types are as follows

                                                                          Ka1 frac141 sin 35

                                                                          1thorn sin 35frac14 0271

                                                                          Ka2 frac141 sin 27

                                                                          1thorn sin 27frac14 0375

                                                                          ffiffiffiffiffiffiffiKa2

                                                                          p frac14 0613

                                                                          Ka3 frac141 sin 42

                                                                          1thorn sin 42frac14 0198

                                                                          Distribution of active pressure (plotted in Figure Q62)

                                                                          Depth (m) Soil Active pressure (kNm2)

                                                                          3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                                          12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                                          At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                                          Total thrust frac14 571 kNm

                                                                          Point of application is (4893571) m from the top of the wall ie 857m

                                                                          Force (kN) Arm (m) Moment (kN m)

                                                                          (1)1

                                                                          2 0271 16 32 frac14 195 20 390

                                                                          (2) 0271 16 3 2 frac14 260 40 1040

                                                                          (3)1

                                                                          2 0271 92 22 frac14 50 433 217

                                                                          (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                                          (5)1

                                                                          2 0375 102 32 frac14 172 70 1204

                                                                          (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                                          (7)1

                                                                          2 0198 112 42 frac14 177 1067 1889

                                                                          (8)1

                                                                          2 98 92 frac14 3969 90 35721

                                                                          5713 48934

                                                                          Figure Q62

                                                                          Lateral earth pressure 35

                                                                          63

                                                                          (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                                          Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                                          frac14 245

                                                                          At the lower end of the piling

                                                                          pa frac14 Kaqthorn Kasatz Kaccu

                                                                          frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                                          frac14 115 kN=m2

                                                                          pp frac14 Kpsatzthorn Kpccu

                                                                          frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                                          frac14 202 kN=m2

                                                                          (b) For 0 frac14 26 and frac14 1

                                                                          20

                                                                          Ka frac14 035

                                                                          Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                                          pfrac14 145 ethEquation 619THORN

                                                                          Kp frac14 37

                                                                          Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                                          pfrac14 47 ethEquation 624THORN

                                                                          At the lower end of the piling

                                                                          pa frac14 Kaqthorn Ka0z Kacc

                                                                          0

                                                                          frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                                          frac14 187 kN=m2

                                                                          pp frac14 Kp0zthorn Kpcc

                                                                          0

                                                                          frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                                          frac14 198 kN=m2

                                                                          36 Lateral earth pressure

                                                                          64

                                                                          (a) For 0 frac14 38 Ka frac14 024

                                                                          0 frac14 20 98 frac14 102 kN=m3

                                                                          The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                                          Force (kN) Arm (m) Moment (kN m)

                                                                          (1) 024 10 66 frac14 159 33 525

                                                                          (2)1

                                                                          2 024 17 392 frac14 310 400 1240

                                                                          (3) 024 17 39 27 frac14 430 135 580

                                                                          (4)1

                                                                          2 024 102 272 frac14 89 090 80

                                                                          (5)1

                                                                          2 98 272 frac14 357 090 321

                                                                          Hfrac14 1345 MH frac14 2746

                                                                          (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                                          (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                                          XM frac14MV MH frac14 7790 kNm

                                                                          Lever arm of base resultant

                                                                          M

                                                                          Vfrac14 779

                                                                          488frac14 160

                                                                          Eccentricity of base resultant

                                                                          e frac14 200 160 frac14 040m

                                                                          39 m

                                                                          27 m

                                                                          40 m

                                                                          04 m

                                                                          04 m

                                                                          26 m

                                                                          (7)

                                                                          (9)

                                                                          (1)(2)

                                                                          (3)

                                                                          (4)

                                                                          (5)

                                                                          (8)(6)

                                                                          (10)

                                                                          WT

                                                                          10 kNm2

                                                                          Hydrostatic

                                                                          Figure Q64

                                                                          Lateral earth pressure 37

                                                                          Base pressures (Equation 627)

                                                                          p frac14 VB

                                                                          1 6e

                                                                          B

                                                                          frac14 488

                                                                          4eth1 060THORN

                                                                          frac14 195 kN=m2 and 49 kN=m2

                                                                          Factor of safety against sliding (Equation 628)

                                                                          F frac14 V tan

                                                                          Hfrac14 488 tan 25

                                                                          1345frac14 17

                                                                          (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                                          Hfrac14 1633 kN

                                                                          V frac14 4879 kN

                                                                          MH frac14 3453 kNm

                                                                          MV frac14 10536 kNm

                                                                          The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                                          65

                                                                          For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                          Kp

                                                                          Ffrac14 385

                                                                          2

                                                                          0 frac14 20 98 frac14 102 kN=m3

                                                                          The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                                          Force (kN) Arm (m) Moment (kN m)

                                                                          (1)1

                                                                          2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                                          (2) 026 17 45 d frac14 199d d2 995d2

                                                                          (3)1

                                                                          2 026 102 d2 frac14 133d2 d3 044d3

                                                                          (4)1

                                                                          2 385

                                                                          2 17 152 frac14 368 dthorn 05 368d 184

                                                                          (5)385

                                                                          2 17 15 d frac14 491d d2 2455d2

                                                                          (6)1

                                                                          2 385

                                                                          2 102 d2 frac14 982d2 d3 327d3

                                                                          38 Lateral earth pressure

                                                                          XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                                          d3 thorn 516d2 283d 1724 frac14 0

                                                                          d frac14 179m

                                                                          Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                                          Over additional 20 embedded depth

                                                                          pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                                          Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                                          66

                                                                          The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                                          Ka frac14 sin 69=sin 105

                                                                          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                                          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                                          pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                                          26664

                                                                          37775

                                                                          2

                                                                          frac14 050

                                                                          The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                                          Pa frac14 1

                                                                          2 050 19 7502 frac14 267 kN=m

                                                                          Figure Q65

                                                                          Lateral earth pressure 39

                                                                          Horizontal component

                                                                          Ph frac14 267 cos 40 frac14 205 kN=m

                                                                          Vertical component

                                                                          Pv frac14 267 sin 40 frac14 172 kN=m

                                                                          Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                          Force (kN) Arm (m) Moment (kN m)

                                                                          (1)1

                                                                          2 175 650 235 frac14 1337 258 345

                                                                          (2) 050 650 235 frac14 764 175 134

                                                                          (3)1

                                                                          2 070 650 235 frac14 535 127 68

                                                                          (4) 100 400 235 frac14 940 200 188

                                                                          (5) 1

                                                                          2 080 050 235 frac14 47 027 1

                                                                          Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                          Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                          Lever arm of base resultant

                                                                          M

                                                                          Vfrac14 795

                                                                          525frac14 151m

                                                                          Eccentricity of base resultant

                                                                          e frac14 200 151 frac14 049m

                                                                          Figure Q66

                                                                          40 Lateral earth pressure

                                                                          Base pressures (Equation 627)

                                                                          p frac14 525

                                                                          41 6 049

                                                                          4

                                                                          frac14 228 kN=m2 and 35 kN=m2

                                                                          The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                          The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                          The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                          67

                                                                          For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                          Force (kN) Arm (m) Moment (kNm)

                                                                          (1)1

                                                                          2 027 17 52 frac14 574 183 1050

                                                                          (2) 027 17 5 3 frac14 689 500 3445

                                                                          (3)1

                                                                          2 027 102 32 frac14 124 550 682

                                                                          (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                          (5)1

                                                                          2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                          (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                          (7) 1

                                                                          2 267

                                                                          2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                          (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                          2 d frac14 163d d2thorn 650 82d2 1060d

                                                                          Tie rod force per m frac14 T 0 0

                                                                          XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                          d3 thorn 77d2 269d 1438 frac14 0

                                                                          d frac14 467m

                                                                          Depth of penetration frac14 12d frac14 560m

                                                                          Lateral earth pressure 41

                                                                          Algebraic sum of forces for d frac14 467m isX

                                                                          F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                          T frac14 905 kN=m

                                                                          Force in each tie rod frac14 25T frac14 226 kN

                                                                          68

                                                                          (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                          0 frac14 21 98 frac14 112 kN=m3

                                                                          The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                          uC frac14 150

                                                                          165 15 98 frac14 134 kN=m2

                                                                          The average seepage pressure is

                                                                          j frac14 15

                                                                          165 98 frac14 09 kN=m3

                                                                          Hence

                                                                          0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                          0 j frac14 112 09 frac14 103 kN=m3

                                                                          Figure Q67

                                                                          42 Lateral earth pressure

                                                                          Consider moments about the anchor point A (per m)

                                                                          Force (kN) Arm (m) Moment (kN m)

                                                                          (1) 10 026 150 frac14 390 60 2340

                                                                          (2)1

                                                                          2 026 18 452 frac14 474 15 711

                                                                          (3) 026 18 45 105 frac14 2211 825 18240

                                                                          (4)1

                                                                          2 026 121 1052 frac14 1734 100 17340

                                                                          (5)1

                                                                          2 134 15 frac14 101 40 404

                                                                          (6) 134 30 frac14 402 60 2412

                                                                          (7)1

                                                                          2 134 60 frac14 402 95 3819

                                                                          571 4527(8) Ppm

                                                                          115 115PPm

                                                                          XM frac14 0

                                                                          Ppm frac144527

                                                                          115frac14 394 kN=m

                                                                          Available passive resistance

                                                                          Pp frac14 1

                                                                          2 385 103 62 frac14 714 kN=m

                                                                          Factor of safety

                                                                          Fp frac14 Pp

                                                                          Ppm

                                                                          frac14 714

                                                                          394frac14 18

                                                                          Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                          Figure Q68

                                                                          Lateral earth pressure 43

                                                                          (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                          Consider moments (per m) about the tie point A

                                                                          Force (kN) Arm (m)

                                                                          (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                          (2)1

                                                                          2 033 18 452 frac14 601 15

                                                                          (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                          (4)1

                                                                          2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                          (5)1

                                                                          2 134 15 frac14 101 40

                                                                          (6) 134 30 frac14 402 60

                                                                          (7)1

                                                                          2 134 d frac14 67d d3thorn 75

                                                                          (8) 1

                                                                          2 30 103 d2 frac141545d2 2d3thorn 75

                                                                          Moment (kN m)

                                                                          (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                          XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                          d3 thorn 827d2 466d 1518 frac14 0

                                                                          By trial

                                                                          d frac14 544m

                                                                          The minimum depth of embedment required is 544m

                                                                          69

                                                                          For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                          0 frac14 20 98 frac14 102 kN=m3

                                                                          The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                          44 Lateral earth pressure

                                                                          uC frac14 147

                                                                          173 26 98 frac14 216 kN=m2

                                                                          and the average seepage pressure around the wall is

                                                                          j frac14 26

                                                                          173 98 frac14 15 kN=m3

                                                                          Consider moments about the prop (A) (per m)

                                                                          Force (kN) Arm (m) Moment (kN m)

                                                                          (1)1

                                                                          2 03 17 272 frac14 186 020 37

                                                                          (2) 03 17 27 53 frac14 730 335 2445

                                                                          (3)1

                                                                          2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                          (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                          (5)1

                                                                          2 216 26 frac14 281 243 684

                                                                          (6) 216 27 frac14 583 465 2712

                                                                          (7)1

                                                                          2 216 60 frac14 648 800 5184

                                                                          3055(8)

                                                                          1

                                                                          2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                          Factor of safety

                                                                          Fr frac14 6885

                                                                          3055frac14 225

                                                                          Figure Q69

                                                                          Lateral earth pressure 45

                                                                          610

                                                                          For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                          p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                          Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                          Using the recommendations of Twine and Roscoe

                                                                          p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                          Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                          611

                                                                          frac14 18 kN=m3 0 frac14 34

                                                                          H frac14 350m nH frac14 335m mH frac14 185m

                                                                          Consider a trial value of F frac14 20 Refer to Figure 635

                                                                          0m frac14 tan1tan 34

                                                                          20

                                                                          frac14 186

                                                                          Then

                                                                          frac14 45 thorn 0m2frac14 543

                                                                          W frac14 1

                                                                          2 18 3502 cot 543 frac14 792 kN=m

                                                                          Figure Q610

                                                                          46 Lateral earth pressure

                                                                          P frac14 1

                                                                          2 s 3352 frac14 561s kN=m

                                                                          U frac14 1

                                                                          2 98 1852 cosec 543 frac14 206 kN=m

                                                                          Equations 630 and 631 then become

                                                                          561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                          792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                          ie

                                                                          561s 0616N 405 frac14 0

                                                                          792 0857N thorn 563 frac14 0

                                                                          N frac14 848

                                                                          0857frac14 989 kN=m

                                                                          Then

                                                                          561s 609 405 frac14 0

                                                                          s frac14 649

                                                                          561frac14 116 kN=m3

                                                                          The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                          F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                          20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                          s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                          Figure Q611

                                                                          Lateral earth pressure 47

                                                                          612

                                                                          For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                          45 thorn 0

                                                                          2frac14 63

                                                                          For the retained material between the surface and a depth of 36m

                                                                          Pa frac14 1

                                                                          2 030 18 362 frac14 350 kN=m

                                                                          Weight of reinforced fill between the surface and a depth of 36m is

                                                                          Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                          eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                          Eccentricity of Rv

                                                                          e frac14 263 250 frac14 013m

                                                                          The average vertical stress at a depth of 36m is

                                                                          z frac14 Rv

                                                                          L 2efrac14 324

                                                                          474frac14 68 kN=m2

                                                                          (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                          Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                          Tensile stress in the element frac14 138 103

                                                                          65 3frac14 71N=mm2

                                                                          Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                          Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                          Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                          The weight of ABC is

                                                                          W frac14 1

                                                                          2 18 52 265 frac14 124 kN=m

                                                                          From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                          48 Lateral earth pressure

                                                                          (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                          Tp frac14 032 68 120 065 frac14 170 kN

                                                                          Tr frac14 213 420

                                                                          418frac14 214 kN

                                                                          Again the tensile failure and slipping limit states are satisfied for this element

                                                                          Figure Q612

                                                                          Lateral earth pressure 49

                                                                          Chapter 7

                                                                          Consolidation theory

                                                                          71

                                                                          Total change in thickness

                                                                          H frac14 782 602 frac14 180mm

                                                                          Average thickness frac14 1530thorn 180

                                                                          2frac14 1620mm

                                                                          Length of drainage path d frac14 1620

                                                                          2frac14 810mm

                                                                          Root time plot (Figure Q71a)

                                                                          ffiffiffiffiffiffit90p frac14 33

                                                                          t90 frac14 109min

                                                                          cv frac14 0848d2

                                                                          t90frac14 0848 8102

                                                                          109 1440 365

                                                                          106frac14 27m2=year

                                                                          r0 frac14 782 764

                                                                          782 602frac14 018

                                                                          180frac14 0100

                                                                          rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                          10 119

                                                                          9 180frac14 0735

                                                                          rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                          Log time plot (Figure Q71b)

                                                                          t50 frac14 26min

                                                                          cv frac14 0196d2

                                                                          t50frac14 0196 8102

                                                                          26 1440 365

                                                                          106frac14 26m2=year

                                                                          r0 frac14 782 763

                                                                          782 602frac14 019

                                                                          180frac14 0106

                                                                          rp frac14 763 623

                                                                          782 602frac14 140

                                                                          180frac14 0778

                                                                          rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                          Figure Q71(a)

                                                                          Figure Q71(b)

                                                                          Final void ratio

                                                                          e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                          e

                                                                          Hfrac14 1thorn e0

                                                                          H0frac14 1thorn e1 thorne

                                                                          H0

                                                                          ie

                                                                          e

                                                                          180frac14 1631thorne

                                                                          1710

                                                                          e frac14 2936

                                                                          1530frac14 0192

                                                                          Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                          mv frac14 1

                                                                          1thorn e0 e0 e101 00

                                                                          frac14 1

                                                                          1823 0192

                                                                          0107frac14 098m2=MN

                                                                          k frac14 cvmvw frac14 265 098 98

                                                                          60 1440 365 103frac14 81 1010 m=s

                                                                          72

                                                                          Using Equation 77 (one-dimensional method)

                                                                          sc frac14 e0 e11thorn e0 H

                                                                          Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                          Figure Q72

                                                                          52 Consolidation theory

                                                                          Settlement

                                                                          Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                          1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                          318

                                                                          Notes 5 92y 460thorn 84

                                                                          Heave

                                                                          Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                          1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                          38

                                                                          73

                                                                          U frac14 f ethTvTHORN frac14 f cvt

                                                                          d2

                                                                          Hence if cv is constant

                                                                          t1

                                                                          t2frac14 d

                                                                          21

                                                                          d22

                                                                          where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                          d1 frac14 95mm and d2 frac14 2500mm

                                                                          for U frac14 050 t2 frac14 t1 d22

                                                                          d21

                                                                          frac14 20

                                                                          60 24 365 25002

                                                                          952frac14 263 years

                                                                          for U lt 060 Tv frac14

                                                                          4U2 (Equation 724(a))

                                                                          t030 frac14 t050 0302

                                                                          0502

                                                                          frac14 263 036 frac14 095 years

                                                                          Consolidation theory 53

                                                                          74

                                                                          The layer is open

                                                                          d frac14 8

                                                                          2frac14 4m

                                                                          Tv frac14 cvtd2frac14 24 3

                                                                          42frac14 0450

                                                                          ui frac14 frac14 84 kN=m2

                                                                          The excess pore water pressure is given by Equation 721

                                                                          ue frac14Xmfrac141mfrac140

                                                                          2ui

                                                                          Msin

                                                                          Mz

                                                                          d

                                                                          expethM2TvTHORN

                                                                          In this case z frac14 d

                                                                          sinMz

                                                                          d

                                                                          frac14 sinM

                                                                          where

                                                                          M frac14

                                                                          23

                                                                          25

                                                                          2

                                                                          M sin M M2Tv exp (M2Tv)

                                                                          2thorn1 1110 0329

                                                                          3

                                                                          21 9993 457 105

                                                                          ue frac14 2 84 2

                                                                          1 0329 ethother terms negligibleTHORN

                                                                          frac14 352 kN=m2

                                                                          75

                                                                          The layer is open

                                                                          d frac14 6

                                                                          2frac14 3m

                                                                          Tv frac14 cvtd2frac14 10 3

                                                                          32frac14 0333

                                                                          The layer thickness will be divided into six equal parts ie m frac14 6

                                                                          54 Consolidation theory

                                                                          For an open layer

                                                                          Tv frac14 4n

                                                                          m2

                                                                          n frac14 0333 62

                                                                          4frac14 300

                                                                          The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                          ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                          i j

                                                                          0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                          0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                          The initial and 3-year isochrones are plotted in Figure Q75

                                                                          Area under initial isochrone frac14 180 units

                                                                          Area under 3-year isochrone frac14 63 units

                                                                          The average degree of consolidation is given by Equation 725Thus

                                                                          U frac14 1 63

                                                                          180frac14 065

                                                                          Figure Q75

                                                                          Consolidation theory 55

                                                                          76

                                                                          At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                          0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                          The final consolidation settlement (one-dimensional method) is

                                                                          sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                          Corrected time t frac14 2 1

                                                                          2

                                                                          40

                                                                          52

                                                                          frac14 1615 years

                                                                          Tv frac14 cvtd2frac14 44 1615

                                                                          42frac14 0444

                                                                          From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                          77

                                                                          The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                          Figure Q77

                                                                          56 Consolidation theory

                                                                          Point m n Ir (kNm2) sc (mm)

                                                                          13020frac14 15 20

                                                                          20frac14 10 0194 (4) 113 124

                                                                          260

                                                                          20frac14 30

                                                                          20

                                                                          20frac14 10 0204 (2) 59 65

                                                                          360

                                                                          20frac14 30

                                                                          40

                                                                          20frac14 20 0238 (1) 35 38

                                                                          430

                                                                          20frac14 15

                                                                          40

                                                                          20frac14 20 0224 (2) 65 72

                                                                          Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                          78

                                                                          Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                          (a) Immediate settlement

                                                                          H

                                                                          Bfrac14 30

                                                                          35frac14 086

                                                                          D

                                                                          Bfrac14 2

                                                                          35frac14 006

                                                                          Figure Q78

                                                                          Consolidation theory 57

                                                                          From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                          si frac14 130131qB

                                                                          Eufrac14 10 032 105 35

                                                                          40frac14 30mm

                                                                          (b) Consolidation settlement

                                                                          Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                          1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                          3150

                                                                          Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                          Now

                                                                          H

                                                                          Bfrac14 30

                                                                          35frac14 086 and A frac14 065

                                                                          from Figure 712 13 frac14 079

                                                                          sc frac14 13sod frac14 079 315 frac14 250mm

                                                                          Total settlement

                                                                          s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                          79

                                                                          Without sand drains

                                                                          Uv frac14 025

                                                                          Tv frac14 0049 ethfrom Figure 718THORN

                                                                          t frac14 Tvd2

                                                                          cvfrac14 0049 82

                                                                          cvWith sand drains

                                                                          R frac14 0564S frac14 0564 3 frac14 169m

                                                                          n frac14 Rrfrac14 169

                                                                          015frac14 113

                                                                          Tr frac14 cht

                                                                          4R2frac14 ch

                                                                          4 1692 0049 82

                                                                          cvethand ch frac14 cvTHORN

                                                                          frac14 0275

                                                                          Ur frac14 073 (from Figure 730)

                                                                          58 Consolidation theory

                                                                          Using Equation 740

                                                                          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                          U frac14 080

                                                                          710

                                                                          Without sand drains

                                                                          Uv frac14 090

                                                                          Tv frac14 0848

                                                                          t frac14 Tvd2

                                                                          cvfrac14 0848 102

                                                                          96frac14 88 years

                                                                          With sand drains

                                                                          R frac14 0564S frac14 0564 4 frac14 226m

                                                                          n frac14 Rrfrac14 226

                                                                          015frac14 15

                                                                          Tr

                                                                          Tvfrac14 chcv

                                                                          d2

                                                                          4R2ethsame tTHORN

                                                                          Tr

                                                                          Tvfrac14 140

                                                                          96 102

                                                                          4 2262frac14 714 eth1THORN

                                                                          Using Equation 740

                                                                          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                          An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                          Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                          040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                          Thus

                                                                          Uv frac14 0295 and Ur frac14 086

                                                                          t frac14 88 00683

                                                                          0848frac14 07 years

                                                                          Consolidation theory 59

                                                                          Chapter 8

                                                                          Bearing capacity

                                                                          81

                                                                          (a) The ultimate bearing capacity is given by Equation 83

                                                                          qf frac14 cNc thorn DNq thorn 1

                                                                          2BN

                                                                          For u frac14 0

                                                                          Nc frac14 514 Nq frac14 1 N frac14 0

                                                                          qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                          The net ultimate bearing capacity is

                                                                          qnf frac14 qf D frac14 540 kN=m2

                                                                          The net foundation pressure is

                                                                          qn frac14 q D frac14 425

                                                                          2 eth21 1THORN frac14 192 kN=m2

                                                                          The factor of safety (Equation 86) is

                                                                          F frac14 qnfqnfrac14 540

                                                                          192frac14 28

                                                                          (b) For 0 frac14 28

                                                                          Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                          qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                          2 112 2 13

                                                                          frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                          qnf frac14 574 112 frac14 563 kN=m2

                                                                          F frac14 563

                                                                          192frac14 29

                                                                          (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                          82

                                                                          For 0 frac14 38

                                                                          Nq frac14 49 N frac14 67

                                                                          qnf frac14 DethNq 1THORN thorn 1

                                                                          2BN ethfrom Equation 83THORN

                                                                          frac14 eth18 075 48THORN thorn 1

                                                                          2 18 15 67

                                                                          frac14 648thorn 905 frac14 1553 kN=m2

                                                                          qn frac14 500

                                                                          15 eth18 075THORN frac14 320 kN=m2

                                                                          F frac14 qnfqnfrac14 1553

                                                                          320frac14 48

                                                                          0d frac14 tan1tan 38

                                                                          125

                                                                          frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                          Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                          2 18 15 25

                                                                          frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                          Design load (action) Vd frac14 500 kN=m

                                                                          The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                          83

                                                                          D

                                                                          Bfrac14 350

                                                                          225frac14 155

                                                                          From Figure 85 for a square foundation

                                                                          Nc frac14 81

                                                                          Bearing capacity 61

                                                                          For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                          Nc frac14 084thorn 016B

                                                                          L

                                                                          81 frac14 745

                                                                          Using Equation 810

                                                                          qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                          For F frac14 3

                                                                          qn frac14 1006

                                                                          3frac14 335 kN=m2

                                                                          q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                          Design load frac14 405 450 225 frac14 4100 kN

                                                                          Design undrained strength cud frac14 135

                                                                          14frac14 96 kN=m2

                                                                          Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                          frac14 7241 kN

                                                                          Design load Vd frac14 4100 kN

                                                                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                          84

                                                                          For 0 frac14 40

                                                                          Nq frac14 64 N frac14 95

                                                                          qnf frac14 DethNq 1THORN thorn 04BN

                                                                          (a) Water table 5m below ground level

                                                                          qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                          qn frac14 400 17 frac14 383 kN=m2

                                                                          F frac14 2686

                                                                          383frac14 70

                                                                          (b) Water table 1m below ground level (ie at foundation level)

                                                                          0 frac14 20 98 frac14 102 kN=m3

                                                                          62 Bearing capacity

                                                                          qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                          F frac14 2040

                                                                          383frac14 53

                                                                          (c) Water table at ground level with upward hydraulic gradient 02

                                                                          eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                          qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                          F frac14 1296

                                                                          392frac14 33

                                                                          85

                                                                          The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                          Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                          Design value of 0 frac14 tan1tan 39

                                                                          125

                                                                          frac14 33

                                                                          For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                          Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                          86

                                                                          (a) Undrained shear for u frac14 0

                                                                          Nc frac14 514 Nq frac14 1 N frac14 0

                                                                          qnf frac14 12cuNc

                                                                          frac14 12 100 514 frac14 617 kN=m2

                                                                          qn frac14 qnfFfrac14 617

                                                                          3frac14 206 kN=m2

                                                                          q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                          Bearing capacity 63

                                                                          Drained shear for 0 frac14 32

                                                                          Nq frac14 23 N frac14 25

                                                                          0 frac14 21 98 frac14 112 kN=m3

                                                                          qnf frac14 0DethNq 1THORN thorn 040BN

                                                                          frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                          frac14 694 kN=m2

                                                                          q frac14 694

                                                                          3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                          Design load frac14 42 227 frac14 3632 kN

                                                                          (b) Design undrained strength cud frac14 100

                                                                          14frac14 71 kNm2

                                                                          Design bearing resistance Rd frac14 12cudNe area

                                                                          frac14 12 71 514 42

                                                                          frac14 7007 kN

                                                                          For drained shear 0d frac14 tan1tan 32

                                                                          125

                                                                          frac14 26

                                                                          Nq frac14 12 N frac14 10

                                                                          Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                          (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                          Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                          1 2 100 0175 0700qn 0182qn

                                                                          2 6 033 0044 0176qn 0046qn

                                                                          3 10 020 0017 0068qn 0018qn

                                                                          0246qn

                                                                          Diameter of equivalent circle B frac14 45m

                                                                          H

                                                                          Bfrac14 12

                                                                          45frac14 27 and A frac14 042

                                                                          13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                          64 Bearing capacity

                                                                          For sc frac14 30mm

                                                                          qn frac14 30

                                                                          0147frac14 204 kN=m2

                                                                          q frac14 204thorn 21 frac14 225 kN=m2

                                                                          Design load frac14 42 225 frac14 3600 kN

                                                                          The design load is 3600 kN settlement being the limiting criterion

                                                                          87

                                                                          D

                                                                          Bfrac14 8

                                                                          4frac14 20

                                                                          From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                          F frac14 cuNc

                                                                          Dfrac14 40 71

                                                                          20 8frac14 18

                                                                          88

                                                                          Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                          Design value of 0 frac14 tan1tan 38

                                                                          125

                                                                          frac14 32

                                                                          Figure Q86

                                                                          Bearing capacity 65

                                                                          For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                          Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                          The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                          Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                          For B frac14 250m qn frac14 3750

                                                                          2502 17 frac14 583 kN=m2

                                                                          From Figure 510 m frac14 n frac14 126

                                                                          6frac14 021

                                                                          Ir frac14 0019

                                                                          Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                          Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                          The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                          89

                                                                          Depth (m) N 0v (kNm2) CN N1

                                                                          070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                          Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                          (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                          Cw frac14 05thorn 0530

                                                                          47

                                                                          frac14 082

                                                                          66 Bearing capacity

                                                                          Thus

                                                                          qa frac14 150 082 frac14 120 kN=m2

                                                                          (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                          Thus

                                                                          qa frac14 90 15 frac14 135 kN=m2

                                                                          (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                          Ic frac14 171

                                                                          1014frac14 0068

                                                                          From Equation 819(a) with s frac14 25mm

                                                                          q frac14 25

                                                                          3507 0068frac14 150 kN=m2

                                                                          810

                                                                          Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                          Izp frac14 05thorn 01130

                                                                          16 27

                                                                          05

                                                                          frac14 067

                                                                          Refer to Figure Q810

                                                                          E frac14 25qc

                                                                          Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                          Ez (mm3MN)

                                                                          1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                          0203

                                                                          C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                          130frac14 093

                                                                          C2 frac14 1 ethsayTHORN

                                                                          s frac14 C1C2qnX Iz

                                                                          Ez frac14 093 1 130 0203 frac14 25mm

                                                                          Bearing capacity 67

                                                                          811

                                                                          At pile base level

                                                                          cu frac14 220 kN=m2

                                                                          qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                          Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                          00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                          qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                          Then

                                                                          Qf frac14 Abqb thorn Asqs

                                                                          frac14

                                                                          4 32 1980

                                                                          thorn eth 105 139 86THORN

                                                                          frac14 13 996thorn 3941 frac14 17 937 kN

                                                                          0 01 02 03 04 05 06 07

                                                                          0 2 4 6 8 10 12 14

                                                                          1

                                                                          2

                                                                          3

                                                                          4

                                                                          5

                                                                          6

                                                                          7

                                                                          8

                                                                          (1)

                                                                          (2)

                                                                          (3)

                                                                          (4)

                                                                          (5)

                                                                          qc

                                                                          qc

                                                                          Iz

                                                                          Iz

                                                                          (MNm2)

                                                                          z (m)

                                                                          Figure Q810

                                                                          68 Bearing capacity

                                                                          Allowable load

                                                                          ethaTHORN Qf

                                                                          2frac14 17 937

                                                                          2frac14 8968 kN

                                                                          ethbTHORN Abqb

                                                                          3thorn Asqs frac14 13 996

                                                                          3thorn 3941 frac14 8606 kN

                                                                          ie allowable load frac14 8600 kN

                                                                          Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                          According to the limit state method

                                                                          Characteristic undrained strength at base level cuk frac14 220

                                                                          150kN=m2

                                                                          Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                          150frac14 1320 kN=m2

                                                                          Characteristic shaft resistance qsk frac14 00150

                                                                          frac14 86

                                                                          150frac14 57 kN=m2

                                                                          Characteristic base and shaft resistances

                                                                          Rbk frac14

                                                                          4 32 1320 frac14 9330 kN

                                                                          Rsk frac14 105 139 86

                                                                          150frac14 2629 kN

                                                                          For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                          Design bearing resistance Rcd frac14 9330

                                                                          160thorn 2629

                                                                          130

                                                                          frac14 5831thorn 2022

                                                                          frac14 7850 kN

                                                                          Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                          812

                                                                          ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                          qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                          For a single pile

                                                                          Qf frac14 Abqb thorn Asqs

                                                                          frac14

                                                                          4 062 1305

                                                                          thorn eth 06 15 42THORN

                                                                          frac14 369thorn 1187 frac14 1556 kN

                                                                          Bearing capacity 69

                                                                          Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                          eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                          (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                          (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                          qbkfrac14 9cuk frac14 9 220

                                                                          150frac14 1320 kN=m2

                                                                          qskfrac14cuk frac14 040 105

                                                                          150frac14 28 kN=m2

                                                                          Rbkfrac14

                                                                          4 0602 1320 frac14 373 kN

                                                                          Rskfrac14 060 15 28 frac14 791 kN

                                                                          Rcdfrac14 373

                                                                          160thorn 791

                                                                          130frac14 233thorn 608 frac14 841 kN

                                                                          Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                          Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                          (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                          q frac14 21 000

                                                                          1762frac14 68 kN=m2

                                                                          Immediate settlement

                                                                          H

                                                                          Bfrac14 15

                                                                          176frac14 085

                                                                          D

                                                                          Bfrac14 13

                                                                          176frac14 074

                                                                          L

                                                                          Bfrac14 1

                                                                          Hence from Figure 515

                                                                          130 frac14 078 and 131 frac14 041

                                                                          70 Bearing capacity

                                                                          Thus using Equation 528

                                                                          si frac14 078 041 68 176

                                                                          65frac14 6mm

                                                                          Consolidation settlement

                                                                          Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                          1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                          434 (sod)

                                                                          Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                          sc frac14 056 434 frac14 24mm

                                                                          The total settlement is (6thorn 24) frac14 30mm

                                                                          813

                                                                          At base level N frac14 26 Then using Equation 830

                                                                          qb frac14 40NDb

                                                                          Bfrac14 40 26 2

                                                                          025frac14 8320 kN=m2

                                                                          ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                          Figure Q812

                                                                          Bearing capacity 71

                                                                          Over the length embedded in sand

                                                                          N frac14 21 ie18thorn 24

                                                                          2

                                                                          Using Equation 831

                                                                          qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                          For a single pile

                                                                          Qf frac14 Abqb thorn Asqs

                                                                          frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                          For the pile group assuming a group efficiency of 12

                                                                          XQf frac14 12 9 604 frac14 6523 kN

                                                                          Then the load factor is

                                                                          F frac14 6523

                                                                          2000thorn 1000frac14 21

                                                                          (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                          Characteristic base resistance per unit area qbk frac14 8320

                                                                          150frac14 5547 kNm2

                                                                          Characteristic shaft resistance per unit area qsk frac14 42

                                                                          150frac14 28 kNm2

                                                                          Characteristic base and shaft resistances for a single pile

                                                                          Rbk frac14 0252 5547 frac14 347 kN

                                                                          Rsk frac14 4 025 2 28 frac14 56 kN

                                                                          For a driven pile the partial factors are b frac14 s frac14 130

                                                                          Design bearing resistance Rcd frac14 347

                                                                          130thorn 56

                                                                          130frac14 310 kN

                                                                          For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                          Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                          (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                          From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                          72 Bearing capacity

                                                                          N frac14 24thorn 26thorn 34

                                                                          3frac14 28

                                                                          Ic frac14 171

                                                                          2814frac14 0016 ethEquation 818THORN

                                                                          s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                          814

                                                                          Using Equation 841

                                                                          Tf frac14 DLcu thorn

                                                                          4ethD2 d2THORNcuNc

                                                                          frac14 eth 02 5 06 110THORN thorn

                                                                          4eth022 012THORN110 9

                                                                          frac14 207thorn 23 frac14 230 kN

                                                                          Figure Q813

                                                                          Bearing capacity 73

                                                                          Chapter 9

                                                                          Stability of slopes

                                                                          91

                                                                          Referring to Figure Q91

                                                                          W frac14 417 19 frac14 792 kN=m

                                                                          Q frac14 20 28 frac14 56 kN=m

                                                                          Arc lengthAB frac14

                                                                          180 73 90 frac14 115m

                                                                          Arc length BC frac14

                                                                          180 28 90 frac14 44m

                                                                          The factor of safety is given by

                                                                          F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                          90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                          Depth of tension crack z0 frac14 2cu

                                                                          frac14 2 20

                                                                          19frac14 21m

                                                                          Arc length BD frac14

                                                                          180 13

                                                                          1

                                                                          2 90 frac14 21m

                                                                          F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                          The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                          Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                          14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                          Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                          The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                          92

                                                                          u frac14 0

                                                                          Depth factor D frac14 11

                                                                          9frac14 122

                                                                          Using Equation 92 with F frac14 10

                                                                          Ns frac14 cu

                                                                          FHfrac14 30

                                                                          10 19 9frac14 0175

                                                                          Hence from Figure 93

                                                                          frac14 50

                                                                          For F frac14 12

                                                                          Ns frac14 30

                                                                          12 19 9frac14 0146

                                                                          frac14 27

                                                                          93

                                                                          Refer to Figure Q93

                                                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                          1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                          74 m

                                                                          214 1deg

                                                                          213 1deg

                                                                          39 m

                                                                          WB

                                                                          D

                                                                          C

                                                                          28 m

                                                                          21 m

                                                                          A

                                                                          Q

                                                                          Soil (1)Soil (2)

                                                                          73deg

                                                                          Figure Q91

                                                                          Stability of slopes 75

                                                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                          9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                          599 256 328 1372

                                                                          Figure Q93

                                                                          76 Stability of slopes

                                                                          XW cos frac14 b

                                                                          Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                          W sin frac14 bX

                                                                          h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                          Arc length La frac14

                                                                          180 57

                                                                          1

                                                                          2 326 frac14 327m

                                                                          The factor of safety is given by

                                                                          F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                          W sin

                                                                          frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                          frac14 091

                                                                          According to the limit state method

                                                                          0d frac14 tan1tan 32

                                                                          125

                                                                          frac14 265

                                                                          c0 frac14 8

                                                                          160frac14 5 kN=m2

                                                                          Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                          Design disturbing moment frac14 1075 kN=m

                                                                          The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                          94

                                                                          F frac14 1

                                                                          W sin

                                                                          Xfc0bthorn ethW ubTHORN tan0g sec

                                                                          1thorn ethtan tan0=FTHORN

                                                                          c0 frac14 8 kN=m2

                                                                          0 frac14 32

                                                                          c0b frac14 8 2 frac14 16 kN=m

                                                                          W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                          Try F frac14 100

                                                                          tan0

                                                                          Ffrac14 0625

                                                                          Stability of slopes 77

                                                                          Values of u are as obtained in Figure Q93

                                                                          SliceNo

                                                                          h(m)

                                                                          W frac14 bh(kNm)

                                                                          W sin(kNm)

                                                                          ub(kNm)

                                                                          c0bthorn (W ub) tan0(kNm)

                                                                          sec

                                                                          1thorn (tan tan0)FProduct(kNm)

                                                                          1 05 21 6 2 8 24 1078 262 13 55 31

                                                                          23 33 30 1042 31

                                                                          3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                          224 92 72 0931 67

                                                                          6 50 210 11 40 100 85 0907 777 55 231 14

                                                                          12 58 112 90 0889 80

                                                                          8 60 252 1812

                                                                          80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                          10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                          2154 88 116 0853 99

                                                                          14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                          1074 1091

                                                                          F frac14 1091

                                                                          1074frac14 102 (assumed value 100)

                                                                          Thus

                                                                          F frac14 101

                                                                          95

                                                                          F frac14 1

                                                                          W sin

                                                                          XfWeth1 ruTHORN tan0g sec

                                                                          1thorn ethtan tan0THORN=F

                                                                          0 frac14 33

                                                                          ru frac14 020

                                                                          W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                          eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                          Try F frac14 110

                                                                          tan 0

                                                                          Ffrac14 tan 33

                                                                          110frac14 0590

                                                                          78 Stability of slopes

                                                                          Referring to Figure Q95

                                                                          SliceNo

                                                                          h(m)

                                                                          W frac14 bh(kNm)

                                                                          W sin(kNm)

                                                                          W(1 ru) tan0(kNm)

                                                                          sec

                                                                          1thorn ( tan tan0)FProduct(kNm)

                                                                          1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                          2120 234 0892 209

                                                                          4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                          1185 1271

                                                                          Figure Q95

                                                                          Stability of slopes 79

                                                                          F frac14 1271

                                                                          1185frac14 107

                                                                          The trial value was 110 therefore take F to be 108

                                                                          96

                                                                          (a) Water table at surface the factor of safety is given by Equation 912

                                                                          F frac14 0

                                                                          sat

                                                                          tan0

                                                                          tan

                                                                          ptie 15 frac14 92

                                                                          19

                                                                          tan 36

                                                                          tan

                                                                          tan frac14 0234

                                                                          frac14 13

                                                                          Water table well below surface the factor of safety is given by Equation 911

                                                                          F frac14 tan0

                                                                          tan

                                                                          frac14 tan 36

                                                                          tan 13

                                                                          frac14 31

                                                                          (b) 0d frac14 tan1tan 36

                                                                          125

                                                                          frac14 30

                                                                          Depth of potential failure surface frac14 z

                                                                          Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                          frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                          frac14 504z kN

                                                                          Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                          frac14 19 z sin 13 cos 13

                                                                          frac14 416z kN

                                                                          Rd gtSd therefore the limit state for overall stability is satisfied

                                                                          80 Stability of slopes

                                                                          • Book Cover
                                                                          • Title
                                                                          • Contents
                                                                          • Basic characteristics of soils
                                                                          • Seepage
                                                                          • Effective stress
                                                                          • Shear strength
                                                                          • Stresses and displacements
                                                                          • Lateral earth pressure
                                                                          • Consolidation theory
                                                                          • Bearing capacity
                                                                          • Stability of slopes

                                                                            n m2n

                                                                            (m2 thorn n2)2

                                                                            px(kNm2)

                                                                            0 0 001 0370 17702 0595 28403 0649 31004 0595 28406 0403 19208 0252 12010 0160 76

                                                                            The pressure distribution is plotted in Figure Q55

                                                                            56

                                                                            H

                                                                            Bfrac14 10

                                                                            2frac14 5

                                                                            L

                                                                            Bfrac14 4

                                                                            2frac14 2

                                                                            D

                                                                            Bfrac14 1

                                                                            2frac14 05

                                                                            Hence from Figure 515

                                                                            131 frac14 082

                                                                            130 frac14 094

                                                                            Figure Q55

                                                                            32 Stresses and displacements

                                                                            The immediate settlement is given by Equation 528

                                                                            si frac14 130131qB

                                                                            Eu

                                                                            frac14 094 082 200 2

                                                                            45frac14 7mm

                                                                            Stresses and displacements 33

                                                                            Chapter 6

                                                                            Lateral earth pressure

                                                                            61

                                                                            For 0 frac14 37 the active pressure coefficient is given by

                                                                            Ka frac14 1 sin 37

                                                                            1thorn sin 37frac14 025

                                                                            The total active thrust (Equation 66a with c0 frac14 0) is

                                                                            Pa frac14 1

                                                                            2KaH

                                                                            2 frac14 1

                                                                            2 025 17 62 frac14 765 kN=m

                                                                            If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                                            K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                                            and the thrust on the wall is

                                                                            P0 frac14 1

                                                                            2K0H

                                                                            2 frac14 1

                                                                            2 040 17 62 frac14 122 kN=m

                                                                            62

                                                                            The active pressure coefficients for the three soil types are as follows

                                                                            Ka1 frac141 sin 35

                                                                            1thorn sin 35frac14 0271

                                                                            Ka2 frac141 sin 27

                                                                            1thorn sin 27frac14 0375

                                                                            ffiffiffiffiffiffiffiKa2

                                                                            p frac14 0613

                                                                            Ka3 frac141 sin 42

                                                                            1thorn sin 42frac14 0198

                                                                            Distribution of active pressure (plotted in Figure Q62)

                                                                            Depth (m) Soil Active pressure (kNm2)

                                                                            3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                                            12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                                            At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                                            Total thrust frac14 571 kNm

                                                                            Point of application is (4893571) m from the top of the wall ie 857m

                                                                            Force (kN) Arm (m) Moment (kN m)

                                                                            (1)1

                                                                            2 0271 16 32 frac14 195 20 390

                                                                            (2) 0271 16 3 2 frac14 260 40 1040

                                                                            (3)1

                                                                            2 0271 92 22 frac14 50 433 217

                                                                            (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                                            (5)1

                                                                            2 0375 102 32 frac14 172 70 1204

                                                                            (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                                            (7)1

                                                                            2 0198 112 42 frac14 177 1067 1889

                                                                            (8)1

                                                                            2 98 92 frac14 3969 90 35721

                                                                            5713 48934

                                                                            Figure Q62

                                                                            Lateral earth pressure 35

                                                                            63

                                                                            (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                                            Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                                            frac14 245

                                                                            At the lower end of the piling

                                                                            pa frac14 Kaqthorn Kasatz Kaccu

                                                                            frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                                            frac14 115 kN=m2

                                                                            pp frac14 Kpsatzthorn Kpccu

                                                                            frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                                            frac14 202 kN=m2

                                                                            (b) For 0 frac14 26 and frac14 1

                                                                            20

                                                                            Ka frac14 035

                                                                            Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                                            pfrac14 145 ethEquation 619THORN

                                                                            Kp frac14 37

                                                                            Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                                            pfrac14 47 ethEquation 624THORN

                                                                            At the lower end of the piling

                                                                            pa frac14 Kaqthorn Ka0z Kacc

                                                                            0

                                                                            frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                                            frac14 187 kN=m2

                                                                            pp frac14 Kp0zthorn Kpcc

                                                                            0

                                                                            frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                                            frac14 198 kN=m2

                                                                            36 Lateral earth pressure

                                                                            64

                                                                            (a) For 0 frac14 38 Ka frac14 024

                                                                            0 frac14 20 98 frac14 102 kN=m3

                                                                            The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                                            Force (kN) Arm (m) Moment (kN m)

                                                                            (1) 024 10 66 frac14 159 33 525

                                                                            (2)1

                                                                            2 024 17 392 frac14 310 400 1240

                                                                            (3) 024 17 39 27 frac14 430 135 580

                                                                            (4)1

                                                                            2 024 102 272 frac14 89 090 80

                                                                            (5)1

                                                                            2 98 272 frac14 357 090 321

                                                                            Hfrac14 1345 MH frac14 2746

                                                                            (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                                            (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                                            XM frac14MV MH frac14 7790 kNm

                                                                            Lever arm of base resultant

                                                                            M

                                                                            Vfrac14 779

                                                                            488frac14 160

                                                                            Eccentricity of base resultant

                                                                            e frac14 200 160 frac14 040m

                                                                            39 m

                                                                            27 m

                                                                            40 m

                                                                            04 m

                                                                            04 m

                                                                            26 m

                                                                            (7)

                                                                            (9)

                                                                            (1)(2)

                                                                            (3)

                                                                            (4)

                                                                            (5)

                                                                            (8)(6)

                                                                            (10)

                                                                            WT

                                                                            10 kNm2

                                                                            Hydrostatic

                                                                            Figure Q64

                                                                            Lateral earth pressure 37

                                                                            Base pressures (Equation 627)

                                                                            p frac14 VB

                                                                            1 6e

                                                                            B

                                                                            frac14 488

                                                                            4eth1 060THORN

                                                                            frac14 195 kN=m2 and 49 kN=m2

                                                                            Factor of safety against sliding (Equation 628)

                                                                            F frac14 V tan

                                                                            Hfrac14 488 tan 25

                                                                            1345frac14 17

                                                                            (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                                            Hfrac14 1633 kN

                                                                            V frac14 4879 kN

                                                                            MH frac14 3453 kNm

                                                                            MV frac14 10536 kNm

                                                                            The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                                            65

                                                                            For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                            Kp

                                                                            Ffrac14 385

                                                                            2

                                                                            0 frac14 20 98 frac14 102 kN=m3

                                                                            The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                                            Force (kN) Arm (m) Moment (kN m)

                                                                            (1)1

                                                                            2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                                            (2) 026 17 45 d frac14 199d d2 995d2

                                                                            (3)1

                                                                            2 026 102 d2 frac14 133d2 d3 044d3

                                                                            (4)1

                                                                            2 385

                                                                            2 17 152 frac14 368 dthorn 05 368d 184

                                                                            (5)385

                                                                            2 17 15 d frac14 491d d2 2455d2

                                                                            (6)1

                                                                            2 385

                                                                            2 102 d2 frac14 982d2 d3 327d3

                                                                            38 Lateral earth pressure

                                                                            XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                                            d3 thorn 516d2 283d 1724 frac14 0

                                                                            d frac14 179m

                                                                            Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                                            Over additional 20 embedded depth

                                                                            pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                                            Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                                            66

                                                                            The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                                            Ka frac14 sin 69=sin 105

                                                                            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                                            ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                                            pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                                            26664

                                                                            37775

                                                                            2

                                                                            frac14 050

                                                                            The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                                            Pa frac14 1

                                                                            2 050 19 7502 frac14 267 kN=m

                                                                            Figure Q65

                                                                            Lateral earth pressure 39

                                                                            Horizontal component

                                                                            Ph frac14 267 cos 40 frac14 205 kN=m

                                                                            Vertical component

                                                                            Pv frac14 267 sin 40 frac14 172 kN=m

                                                                            Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                            Force (kN) Arm (m) Moment (kN m)

                                                                            (1)1

                                                                            2 175 650 235 frac14 1337 258 345

                                                                            (2) 050 650 235 frac14 764 175 134

                                                                            (3)1

                                                                            2 070 650 235 frac14 535 127 68

                                                                            (4) 100 400 235 frac14 940 200 188

                                                                            (5) 1

                                                                            2 080 050 235 frac14 47 027 1

                                                                            Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                            Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                            Lever arm of base resultant

                                                                            M

                                                                            Vfrac14 795

                                                                            525frac14 151m

                                                                            Eccentricity of base resultant

                                                                            e frac14 200 151 frac14 049m

                                                                            Figure Q66

                                                                            40 Lateral earth pressure

                                                                            Base pressures (Equation 627)

                                                                            p frac14 525

                                                                            41 6 049

                                                                            4

                                                                            frac14 228 kN=m2 and 35 kN=m2

                                                                            The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                            The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                            The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                            67

                                                                            For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                            Force (kN) Arm (m) Moment (kNm)

                                                                            (1)1

                                                                            2 027 17 52 frac14 574 183 1050

                                                                            (2) 027 17 5 3 frac14 689 500 3445

                                                                            (3)1

                                                                            2 027 102 32 frac14 124 550 682

                                                                            (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                            (5)1

                                                                            2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                            (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                            (7) 1

                                                                            2 267

                                                                            2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                            (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                            2 d frac14 163d d2thorn 650 82d2 1060d

                                                                            Tie rod force per m frac14 T 0 0

                                                                            XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                            d3 thorn 77d2 269d 1438 frac14 0

                                                                            d frac14 467m

                                                                            Depth of penetration frac14 12d frac14 560m

                                                                            Lateral earth pressure 41

                                                                            Algebraic sum of forces for d frac14 467m isX

                                                                            F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                            T frac14 905 kN=m

                                                                            Force in each tie rod frac14 25T frac14 226 kN

                                                                            68

                                                                            (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                            0 frac14 21 98 frac14 112 kN=m3

                                                                            The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                            uC frac14 150

                                                                            165 15 98 frac14 134 kN=m2

                                                                            The average seepage pressure is

                                                                            j frac14 15

                                                                            165 98 frac14 09 kN=m3

                                                                            Hence

                                                                            0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                            0 j frac14 112 09 frac14 103 kN=m3

                                                                            Figure Q67

                                                                            42 Lateral earth pressure

                                                                            Consider moments about the anchor point A (per m)

                                                                            Force (kN) Arm (m) Moment (kN m)

                                                                            (1) 10 026 150 frac14 390 60 2340

                                                                            (2)1

                                                                            2 026 18 452 frac14 474 15 711

                                                                            (3) 026 18 45 105 frac14 2211 825 18240

                                                                            (4)1

                                                                            2 026 121 1052 frac14 1734 100 17340

                                                                            (5)1

                                                                            2 134 15 frac14 101 40 404

                                                                            (6) 134 30 frac14 402 60 2412

                                                                            (7)1

                                                                            2 134 60 frac14 402 95 3819

                                                                            571 4527(8) Ppm

                                                                            115 115PPm

                                                                            XM frac14 0

                                                                            Ppm frac144527

                                                                            115frac14 394 kN=m

                                                                            Available passive resistance

                                                                            Pp frac14 1

                                                                            2 385 103 62 frac14 714 kN=m

                                                                            Factor of safety

                                                                            Fp frac14 Pp

                                                                            Ppm

                                                                            frac14 714

                                                                            394frac14 18

                                                                            Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                            Figure Q68

                                                                            Lateral earth pressure 43

                                                                            (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                            Consider moments (per m) about the tie point A

                                                                            Force (kN) Arm (m)

                                                                            (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                            (2)1

                                                                            2 033 18 452 frac14 601 15

                                                                            (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                            (4)1

                                                                            2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                            (5)1

                                                                            2 134 15 frac14 101 40

                                                                            (6) 134 30 frac14 402 60

                                                                            (7)1

                                                                            2 134 d frac14 67d d3thorn 75

                                                                            (8) 1

                                                                            2 30 103 d2 frac141545d2 2d3thorn 75

                                                                            Moment (kN m)

                                                                            (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                            XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                            d3 thorn 827d2 466d 1518 frac14 0

                                                                            By trial

                                                                            d frac14 544m

                                                                            The minimum depth of embedment required is 544m

                                                                            69

                                                                            For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                            0 frac14 20 98 frac14 102 kN=m3

                                                                            The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                            44 Lateral earth pressure

                                                                            uC frac14 147

                                                                            173 26 98 frac14 216 kN=m2

                                                                            and the average seepage pressure around the wall is

                                                                            j frac14 26

                                                                            173 98 frac14 15 kN=m3

                                                                            Consider moments about the prop (A) (per m)

                                                                            Force (kN) Arm (m) Moment (kN m)

                                                                            (1)1

                                                                            2 03 17 272 frac14 186 020 37

                                                                            (2) 03 17 27 53 frac14 730 335 2445

                                                                            (3)1

                                                                            2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                            (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                            (5)1

                                                                            2 216 26 frac14 281 243 684

                                                                            (6) 216 27 frac14 583 465 2712

                                                                            (7)1

                                                                            2 216 60 frac14 648 800 5184

                                                                            3055(8)

                                                                            1

                                                                            2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                            Factor of safety

                                                                            Fr frac14 6885

                                                                            3055frac14 225

                                                                            Figure Q69

                                                                            Lateral earth pressure 45

                                                                            610

                                                                            For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                            p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                            Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                            Using the recommendations of Twine and Roscoe

                                                                            p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                            Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                            611

                                                                            frac14 18 kN=m3 0 frac14 34

                                                                            H frac14 350m nH frac14 335m mH frac14 185m

                                                                            Consider a trial value of F frac14 20 Refer to Figure 635

                                                                            0m frac14 tan1tan 34

                                                                            20

                                                                            frac14 186

                                                                            Then

                                                                            frac14 45 thorn 0m2frac14 543

                                                                            W frac14 1

                                                                            2 18 3502 cot 543 frac14 792 kN=m

                                                                            Figure Q610

                                                                            46 Lateral earth pressure

                                                                            P frac14 1

                                                                            2 s 3352 frac14 561s kN=m

                                                                            U frac14 1

                                                                            2 98 1852 cosec 543 frac14 206 kN=m

                                                                            Equations 630 and 631 then become

                                                                            561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                            792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                            ie

                                                                            561s 0616N 405 frac14 0

                                                                            792 0857N thorn 563 frac14 0

                                                                            N frac14 848

                                                                            0857frac14 989 kN=m

                                                                            Then

                                                                            561s 609 405 frac14 0

                                                                            s frac14 649

                                                                            561frac14 116 kN=m3

                                                                            The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                            F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                            20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                            s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                            Figure Q611

                                                                            Lateral earth pressure 47

                                                                            612

                                                                            For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                            45 thorn 0

                                                                            2frac14 63

                                                                            For the retained material between the surface and a depth of 36m

                                                                            Pa frac14 1

                                                                            2 030 18 362 frac14 350 kN=m

                                                                            Weight of reinforced fill between the surface and a depth of 36m is

                                                                            Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                            eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                            Eccentricity of Rv

                                                                            e frac14 263 250 frac14 013m

                                                                            The average vertical stress at a depth of 36m is

                                                                            z frac14 Rv

                                                                            L 2efrac14 324

                                                                            474frac14 68 kN=m2

                                                                            (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                            Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                            Tensile stress in the element frac14 138 103

                                                                            65 3frac14 71N=mm2

                                                                            Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                            Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                            Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                            The weight of ABC is

                                                                            W frac14 1

                                                                            2 18 52 265 frac14 124 kN=m

                                                                            From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                            48 Lateral earth pressure

                                                                            (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                            Tp frac14 032 68 120 065 frac14 170 kN

                                                                            Tr frac14 213 420

                                                                            418frac14 214 kN

                                                                            Again the tensile failure and slipping limit states are satisfied for this element

                                                                            Figure Q612

                                                                            Lateral earth pressure 49

                                                                            Chapter 7

                                                                            Consolidation theory

                                                                            71

                                                                            Total change in thickness

                                                                            H frac14 782 602 frac14 180mm

                                                                            Average thickness frac14 1530thorn 180

                                                                            2frac14 1620mm

                                                                            Length of drainage path d frac14 1620

                                                                            2frac14 810mm

                                                                            Root time plot (Figure Q71a)

                                                                            ffiffiffiffiffiffit90p frac14 33

                                                                            t90 frac14 109min

                                                                            cv frac14 0848d2

                                                                            t90frac14 0848 8102

                                                                            109 1440 365

                                                                            106frac14 27m2=year

                                                                            r0 frac14 782 764

                                                                            782 602frac14 018

                                                                            180frac14 0100

                                                                            rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                            10 119

                                                                            9 180frac14 0735

                                                                            rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                            Log time plot (Figure Q71b)

                                                                            t50 frac14 26min

                                                                            cv frac14 0196d2

                                                                            t50frac14 0196 8102

                                                                            26 1440 365

                                                                            106frac14 26m2=year

                                                                            r0 frac14 782 763

                                                                            782 602frac14 019

                                                                            180frac14 0106

                                                                            rp frac14 763 623

                                                                            782 602frac14 140

                                                                            180frac14 0778

                                                                            rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                            Figure Q71(a)

                                                                            Figure Q71(b)

                                                                            Final void ratio

                                                                            e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                            e

                                                                            Hfrac14 1thorn e0

                                                                            H0frac14 1thorn e1 thorne

                                                                            H0

                                                                            ie

                                                                            e

                                                                            180frac14 1631thorne

                                                                            1710

                                                                            e frac14 2936

                                                                            1530frac14 0192

                                                                            Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                            mv frac14 1

                                                                            1thorn e0 e0 e101 00

                                                                            frac14 1

                                                                            1823 0192

                                                                            0107frac14 098m2=MN

                                                                            k frac14 cvmvw frac14 265 098 98

                                                                            60 1440 365 103frac14 81 1010 m=s

                                                                            72

                                                                            Using Equation 77 (one-dimensional method)

                                                                            sc frac14 e0 e11thorn e0 H

                                                                            Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                            Figure Q72

                                                                            52 Consolidation theory

                                                                            Settlement

                                                                            Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                            1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                            318

                                                                            Notes 5 92y 460thorn 84

                                                                            Heave

                                                                            Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                            1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                            38

                                                                            73

                                                                            U frac14 f ethTvTHORN frac14 f cvt

                                                                            d2

                                                                            Hence if cv is constant

                                                                            t1

                                                                            t2frac14 d

                                                                            21

                                                                            d22

                                                                            where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                            d1 frac14 95mm and d2 frac14 2500mm

                                                                            for U frac14 050 t2 frac14 t1 d22

                                                                            d21

                                                                            frac14 20

                                                                            60 24 365 25002

                                                                            952frac14 263 years

                                                                            for U lt 060 Tv frac14

                                                                            4U2 (Equation 724(a))

                                                                            t030 frac14 t050 0302

                                                                            0502

                                                                            frac14 263 036 frac14 095 years

                                                                            Consolidation theory 53

                                                                            74

                                                                            The layer is open

                                                                            d frac14 8

                                                                            2frac14 4m

                                                                            Tv frac14 cvtd2frac14 24 3

                                                                            42frac14 0450

                                                                            ui frac14 frac14 84 kN=m2

                                                                            The excess pore water pressure is given by Equation 721

                                                                            ue frac14Xmfrac141mfrac140

                                                                            2ui

                                                                            Msin

                                                                            Mz

                                                                            d

                                                                            expethM2TvTHORN

                                                                            In this case z frac14 d

                                                                            sinMz

                                                                            d

                                                                            frac14 sinM

                                                                            where

                                                                            M frac14

                                                                            23

                                                                            25

                                                                            2

                                                                            M sin M M2Tv exp (M2Tv)

                                                                            2thorn1 1110 0329

                                                                            3

                                                                            21 9993 457 105

                                                                            ue frac14 2 84 2

                                                                            1 0329 ethother terms negligibleTHORN

                                                                            frac14 352 kN=m2

                                                                            75

                                                                            The layer is open

                                                                            d frac14 6

                                                                            2frac14 3m

                                                                            Tv frac14 cvtd2frac14 10 3

                                                                            32frac14 0333

                                                                            The layer thickness will be divided into six equal parts ie m frac14 6

                                                                            54 Consolidation theory

                                                                            For an open layer

                                                                            Tv frac14 4n

                                                                            m2

                                                                            n frac14 0333 62

                                                                            4frac14 300

                                                                            The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                            ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                            i j

                                                                            0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                            0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                            The initial and 3-year isochrones are plotted in Figure Q75

                                                                            Area under initial isochrone frac14 180 units

                                                                            Area under 3-year isochrone frac14 63 units

                                                                            The average degree of consolidation is given by Equation 725Thus

                                                                            U frac14 1 63

                                                                            180frac14 065

                                                                            Figure Q75

                                                                            Consolidation theory 55

                                                                            76

                                                                            At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                            0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                            The final consolidation settlement (one-dimensional method) is

                                                                            sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                            Corrected time t frac14 2 1

                                                                            2

                                                                            40

                                                                            52

                                                                            frac14 1615 years

                                                                            Tv frac14 cvtd2frac14 44 1615

                                                                            42frac14 0444

                                                                            From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                            77

                                                                            The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                            Figure Q77

                                                                            56 Consolidation theory

                                                                            Point m n Ir (kNm2) sc (mm)

                                                                            13020frac14 15 20

                                                                            20frac14 10 0194 (4) 113 124

                                                                            260

                                                                            20frac14 30

                                                                            20

                                                                            20frac14 10 0204 (2) 59 65

                                                                            360

                                                                            20frac14 30

                                                                            40

                                                                            20frac14 20 0238 (1) 35 38

                                                                            430

                                                                            20frac14 15

                                                                            40

                                                                            20frac14 20 0224 (2) 65 72

                                                                            Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                            78

                                                                            Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                            (a) Immediate settlement

                                                                            H

                                                                            Bfrac14 30

                                                                            35frac14 086

                                                                            D

                                                                            Bfrac14 2

                                                                            35frac14 006

                                                                            Figure Q78

                                                                            Consolidation theory 57

                                                                            From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                            si frac14 130131qB

                                                                            Eufrac14 10 032 105 35

                                                                            40frac14 30mm

                                                                            (b) Consolidation settlement

                                                                            Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                            1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                            3150

                                                                            Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                            Now

                                                                            H

                                                                            Bfrac14 30

                                                                            35frac14 086 and A frac14 065

                                                                            from Figure 712 13 frac14 079

                                                                            sc frac14 13sod frac14 079 315 frac14 250mm

                                                                            Total settlement

                                                                            s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                            79

                                                                            Without sand drains

                                                                            Uv frac14 025

                                                                            Tv frac14 0049 ethfrom Figure 718THORN

                                                                            t frac14 Tvd2

                                                                            cvfrac14 0049 82

                                                                            cvWith sand drains

                                                                            R frac14 0564S frac14 0564 3 frac14 169m

                                                                            n frac14 Rrfrac14 169

                                                                            015frac14 113

                                                                            Tr frac14 cht

                                                                            4R2frac14 ch

                                                                            4 1692 0049 82

                                                                            cvethand ch frac14 cvTHORN

                                                                            frac14 0275

                                                                            Ur frac14 073 (from Figure 730)

                                                                            58 Consolidation theory

                                                                            Using Equation 740

                                                                            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                            U frac14 080

                                                                            710

                                                                            Without sand drains

                                                                            Uv frac14 090

                                                                            Tv frac14 0848

                                                                            t frac14 Tvd2

                                                                            cvfrac14 0848 102

                                                                            96frac14 88 years

                                                                            With sand drains

                                                                            R frac14 0564S frac14 0564 4 frac14 226m

                                                                            n frac14 Rrfrac14 226

                                                                            015frac14 15

                                                                            Tr

                                                                            Tvfrac14 chcv

                                                                            d2

                                                                            4R2ethsame tTHORN

                                                                            Tr

                                                                            Tvfrac14 140

                                                                            96 102

                                                                            4 2262frac14 714 eth1THORN

                                                                            Using Equation 740

                                                                            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                            An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                            Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                            040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                            Thus

                                                                            Uv frac14 0295 and Ur frac14 086

                                                                            t frac14 88 00683

                                                                            0848frac14 07 years

                                                                            Consolidation theory 59

                                                                            Chapter 8

                                                                            Bearing capacity

                                                                            81

                                                                            (a) The ultimate bearing capacity is given by Equation 83

                                                                            qf frac14 cNc thorn DNq thorn 1

                                                                            2BN

                                                                            For u frac14 0

                                                                            Nc frac14 514 Nq frac14 1 N frac14 0

                                                                            qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                            The net ultimate bearing capacity is

                                                                            qnf frac14 qf D frac14 540 kN=m2

                                                                            The net foundation pressure is

                                                                            qn frac14 q D frac14 425

                                                                            2 eth21 1THORN frac14 192 kN=m2

                                                                            The factor of safety (Equation 86) is

                                                                            F frac14 qnfqnfrac14 540

                                                                            192frac14 28

                                                                            (b) For 0 frac14 28

                                                                            Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                            qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                            2 112 2 13

                                                                            frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                            qnf frac14 574 112 frac14 563 kN=m2

                                                                            F frac14 563

                                                                            192frac14 29

                                                                            (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                            82

                                                                            For 0 frac14 38

                                                                            Nq frac14 49 N frac14 67

                                                                            qnf frac14 DethNq 1THORN thorn 1

                                                                            2BN ethfrom Equation 83THORN

                                                                            frac14 eth18 075 48THORN thorn 1

                                                                            2 18 15 67

                                                                            frac14 648thorn 905 frac14 1553 kN=m2

                                                                            qn frac14 500

                                                                            15 eth18 075THORN frac14 320 kN=m2

                                                                            F frac14 qnfqnfrac14 1553

                                                                            320frac14 48

                                                                            0d frac14 tan1tan 38

                                                                            125

                                                                            frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                            Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                            2 18 15 25

                                                                            frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                            Design load (action) Vd frac14 500 kN=m

                                                                            The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                            83

                                                                            D

                                                                            Bfrac14 350

                                                                            225frac14 155

                                                                            From Figure 85 for a square foundation

                                                                            Nc frac14 81

                                                                            Bearing capacity 61

                                                                            For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                            Nc frac14 084thorn 016B

                                                                            L

                                                                            81 frac14 745

                                                                            Using Equation 810

                                                                            qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                            For F frac14 3

                                                                            qn frac14 1006

                                                                            3frac14 335 kN=m2

                                                                            q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                            Design load frac14 405 450 225 frac14 4100 kN

                                                                            Design undrained strength cud frac14 135

                                                                            14frac14 96 kN=m2

                                                                            Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                            frac14 7241 kN

                                                                            Design load Vd frac14 4100 kN

                                                                            Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                            84

                                                                            For 0 frac14 40

                                                                            Nq frac14 64 N frac14 95

                                                                            qnf frac14 DethNq 1THORN thorn 04BN

                                                                            (a) Water table 5m below ground level

                                                                            qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                            qn frac14 400 17 frac14 383 kN=m2

                                                                            F frac14 2686

                                                                            383frac14 70

                                                                            (b) Water table 1m below ground level (ie at foundation level)

                                                                            0 frac14 20 98 frac14 102 kN=m3

                                                                            62 Bearing capacity

                                                                            qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                            F frac14 2040

                                                                            383frac14 53

                                                                            (c) Water table at ground level with upward hydraulic gradient 02

                                                                            eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                            qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                            F frac14 1296

                                                                            392frac14 33

                                                                            85

                                                                            The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                            Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                            Design value of 0 frac14 tan1tan 39

                                                                            125

                                                                            frac14 33

                                                                            For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                            Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                            Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                            86

                                                                            (a) Undrained shear for u frac14 0

                                                                            Nc frac14 514 Nq frac14 1 N frac14 0

                                                                            qnf frac14 12cuNc

                                                                            frac14 12 100 514 frac14 617 kN=m2

                                                                            qn frac14 qnfFfrac14 617

                                                                            3frac14 206 kN=m2

                                                                            q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                            Bearing capacity 63

                                                                            Drained shear for 0 frac14 32

                                                                            Nq frac14 23 N frac14 25

                                                                            0 frac14 21 98 frac14 112 kN=m3

                                                                            qnf frac14 0DethNq 1THORN thorn 040BN

                                                                            frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                            frac14 694 kN=m2

                                                                            q frac14 694

                                                                            3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                            Design load frac14 42 227 frac14 3632 kN

                                                                            (b) Design undrained strength cud frac14 100

                                                                            14frac14 71 kNm2

                                                                            Design bearing resistance Rd frac14 12cudNe area

                                                                            frac14 12 71 514 42

                                                                            frac14 7007 kN

                                                                            For drained shear 0d frac14 tan1tan 32

                                                                            125

                                                                            frac14 26

                                                                            Nq frac14 12 N frac14 10

                                                                            Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                            (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                            Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                            1 2 100 0175 0700qn 0182qn

                                                                            2 6 033 0044 0176qn 0046qn

                                                                            3 10 020 0017 0068qn 0018qn

                                                                            0246qn

                                                                            Diameter of equivalent circle B frac14 45m

                                                                            H

                                                                            Bfrac14 12

                                                                            45frac14 27 and A frac14 042

                                                                            13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                            64 Bearing capacity

                                                                            For sc frac14 30mm

                                                                            qn frac14 30

                                                                            0147frac14 204 kN=m2

                                                                            q frac14 204thorn 21 frac14 225 kN=m2

                                                                            Design load frac14 42 225 frac14 3600 kN

                                                                            The design load is 3600 kN settlement being the limiting criterion

                                                                            87

                                                                            D

                                                                            Bfrac14 8

                                                                            4frac14 20

                                                                            From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                            F frac14 cuNc

                                                                            Dfrac14 40 71

                                                                            20 8frac14 18

                                                                            88

                                                                            Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                            Design value of 0 frac14 tan1tan 38

                                                                            125

                                                                            frac14 32

                                                                            Figure Q86

                                                                            Bearing capacity 65

                                                                            For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                            Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                            The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                            Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                            For B frac14 250m qn frac14 3750

                                                                            2502 17 frac14 583 kN=m2

                                                                            From Figure 510 m frac14 n frac14 126

                                                                            6frac14 021

                                                                            Ir frac14 0019

                                                                            Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                            Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                            The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                            89

                                                                            Depth (m) N 0v (kNm2) CN N1

                                                                            070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                            Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                            (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                            Cw frac14 05thorn 0530

                                                                            47

                                                                            frac14 082

                                                                            66 Bearing capacity

                                                                            Thus

                                                                            qa frac14 150 082 frac14 120 kN=m2

                                                                            (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                            Thus

                                                                            qa frac14 90 15 frac14 135 kN=m2

                                                                            (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                            Ic frac14 171

                                                                            1014frac14 0068

                                                                            From Equation 819(a) with s frac14 25mm

                                                                            q frac14 25

                                                                            3507 0068frac14 150 kN=m2

                                                                            810

                                                                            Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                            Izp frac14 05thorn 01130

                                                                            16 27

                                                                            05

                                                                            frac14 067

                                                                            Refer to Figure Q810

                                                                            E frac14 25qc

                                                                            Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                            Ez (mm3MN)

                                                                            1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                            0203

                                                                            C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                            130frac14 093

                                                                            C2 frac14 1 ethsayTHORN

                                                                            s frac14 C1C2qnX Iz

                                                                            Ez frac14 093 1 130 0203 frac14 25mm

                                                                            Bearing capacity 67

                                                                            811

                                                                            At pile base level

                                                                            cu frac14 220 kN=m2

                                                                            qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                            Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                            00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                            qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                            Then

                                                                            Qf frac14 Abqb thorn Asqs

                                                                            frac14

                                                                            4 32 1980

                                                                            thorn eth 105 139 86THORN

                                                                            frac14 13 996thorn 3941 frac14 17 937 kN

                                                                            0 01 02 03 04 05 06 07

                                                                            0 2 4 6 8 10 12 14

                                                                            1

                                                                            2

                                                                            3

                                                                            4

                                                                            5

                                                                            6

                                                                            7

                                                                            8

                                                                            (1)

                                                                            (2)

                                                                            (3)

                                                                            (4)

                                                                            (5)

                                                                            qc

                                                                            qc

                                                                            Iz

                                                                            Iz

                                                                            (MNm2)

                                                                            z (m)

                                                                            Figure Q810

                                                                            68 Bearing capacity

                                                                            Allowable load

                                                                            ethaTHORN Qf

                                                                            2frac14 17 937

                                                                            2frac14 8968 kN

                                                                            ethbTHORN Abqb

                                                                            3thorn Asqs frac14 13 996

                                                                            3thorn 3941 frac14 8606 kN

                                                                            ie allowable load frac14 8600 kN

                                                                            Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                            According to the limit state method

                                                                            Characteristic undrained strength at base level cuk frac14 220

                                                                            150kN=m2

                                                                            Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                            150frac14 1320 kN=m2

                                                                            Characteristic shaft resistance qsk frac14 00150

                                                                            frac14 86

                                                                            150frac14 57 kN=m2

                                                                            Characteristic base and shaft resistances

                                                                            Rbk frac14

                                                                            4 32 1320 frac14 9330 kN

                                                                            Rsk frac14 105 139 86

                                                                            150frac14 2629 kN

                                                                            For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                            Design bearing resistance Rcd frac14 9330

                                                                            160thorn 2629

                                                                            130

                                                                            frac14 5831thorn 2022

                                                                            frac14 7850 kN

                                                                            Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                            812

                                                                            ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                            qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                            For a single pile

                                                                            Qf frac14 Abqb thorn Asqs

                                                                            frac14

                                                                            4 062 1305

                                                                            thorn eth 06 15 42THORN

                                                                            frac14 369thorn 1187 frac14 1556 kN

                                                                            Bearing capacity 69

                                                                            Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                            eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                            (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                            (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                            qbkfrac14 9cuk frac14 9 220

                                                                            150frac14 1320 kN=m2

                                                                            qskfrac14cuk frac14 040 105

                                                                            150frac14 28 kN=m2

                                                                            Rbkfrac14

                                                                            4 0602 1320 frac14 373 kN

                                                                            Rskfrac14 060 15 28 frac14 791 kN

                                                                            Rcdfrac14 373

                                                                            160thorn 791

                                                                            130frac14 233thorn 608 frac14 841 kN

                                                                            Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                            Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                            (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                            q frac14 21 000

                                                                            1762frac14 68 kN=m2

                                                                            Immediate settlement

                                                                            H

                                                                            Bfrac14 15

                                                                            176frac14 085

                                                                            D

                                                                            Bfrac14 13

                                                                            176frac14 074

                                                                            L

                                                                            Bfrac14 1

                                                                            Hence from Figure 515

                                                                            130 frac14 078 and 131 frac14 041

                                                                            70 Bearing capacity

                                                                            Thus using Equation 528

                                                                            si frac14 078 041 68 176

                                                                            65frac14 6mm

                                                                            Consolidation settlement

                                                                            Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                            1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                            434 (sod)

                                                                            Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                            sc frac14 056 434 frac14 24mm

                                                                            The total settlement is (6thorn 24) frac14 30mm

                                                                            813

                                                                            At base level N frac14 26 Then using Equation 830

                                                                            qb frac14 40NDb

                                                                            Bfrac14 40 26 2

                                                                            025frac14 8320 kN=m2

                                                                            ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                            Figure Q812

                                                                            Bearing capacity 71

                                                                            Over the length embedded in sand

                                                                            N frac14 21 ie18thorn 24

                                                                            2

                                                                            Using Equation 831

                                                                            qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                            For a single pile

                                                                            Qf frac14 Abqb thorn Asqs

                                                                            frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                            For the pile group assuming a group efficiency of 12

                                                                            XQf frac14 12 9 604 frac14 6523 kN

                                                                            Then the load factor is

                                                                            F frac14 6523

                                                                            2000thorn 1000frac14 21

                                                                            (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                            Characteristic base resistance per unit area qbk frac14 8320

                                                                            150frac14 5547 kNm2

                                                                            Characteristic shaft resistance per unit area qsk frac14 42

                                                                            150frac14 28 kNm2

                                                                            Characteristic base and shaft resistances for a single pile

                                                                            Rbk frac14 0252 5547 frac14 347 kN

                                                                            Rsk frac14 4 025 2 28 frac14 56 kN

                                                                            For a driven pile the partial factors are b frac14 s frac14 130

                                                                            Design bearing resistance Rcd frac14 347

                                                                            130thorn 56

                                                                            130frac14 310 kN

                                                                            For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                            Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                            (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                            From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                            72 Bearing capacity

                                                                            N frac14 24thorn 26thorn 34

                                                                            3frac14 28

                                                                            Ic frac14 171

                                                                            2814frac14 0016 ethEquation 818THORN

                                                                            s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                            814

                                                                            Using Equation 841

                                                                            Tf frac14 DLcu thorn

                                                                            4ethD2 d2THORNcuNc

                                                                            frac14 eth 02 5 06 110THORN thorn

                                                                            4eth022 012THORN110 9

                                                                            frac14 207thorn 23 frac14 230 kN

                                                                            Figure Q813

                                                                            Bearing capacity 73

                                                                            Chapter 9

                                                                            Stability of slopes

                                                                            91

                                                                            Referring to Figure Q91

                                                                            W frac14 417 19 frac14 792 kN=m

                                                                            Q frac14 20 28 frac14 56 kN=m

                                                                            Arc lengthAB frac14

                                                                            180 73 90 frac14 115m

                                                                            Arc length BC frac14

                                                                            180 28 90 frac14 44m

                                                                            The factor of safety is given by

                                                                            F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                            90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                            Depth of tension crack z0 frac14 2cu

                                                                            frac14 2 20

                                                                            19frac14 21m

                                                                            Arc length BD frac14

                                                                            180 13

                                                                            1

                                                                            2 90 frac14 21m

                                                                            F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                            The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                            Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                            14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                            Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                            The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                            92

                                                                            u frac14 0

                                                                            Depth factor D frac14 11

                                                                            9frac14 122

                                                                            Using Equation 92 with F frac14 10

                                                                            Ns frac14 cu

                                                                            FHfrac14 30

                                                                            10 19 9frac14 0175

                                                                            Hence from Figure 93

                                                                            frac14 50

                                                                            For F frac14 12

                                                                            Ns frac14 30

                                                                            12 19 9frac14 0146

                                                                            frac14 27

                                                                            93

                                                                            Refer to Figure Q93

                                                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                            1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                            74 m

                                                                            214 1deg

                                                                            213 1deg

                                                                            39 m

                                                                            WB

                                                                            D

                                                                            C

                                                                            28 m

                                                                            21 m

                                                                            A

                                                                            Q

                                                                            Soil (1)Soil (2)

                                                                            73deg

                                                                            Figure Q91

                                                                            Stability of slopes 75

                                                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                            9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                            599 256 328 1372

                                                                            Figure Q93

                                                                            76 Stability of slopes

                                                                            XW cos frac14 b

                                                                            Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                            W sin frac14 bX

                                                                            h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                            Arc length La frac14

                                                                            180 57

                                                                            1

                                                                            2 326 frac14 327m

                                                                            The factor of safety is given by

                                                                            F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                            W sin

                                                                            frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                            frac14 091

                                                                            According to the limit state method

                                                                            0d frac14 tan1tan 32

                                                                            125

                                                                            frac14 265

                                                                            c0 frac14 8

                                                                            160frac14 5 kN=m2

                                                                            Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                            Design disturbing moment frac14 1075 kN=m

                                                                            The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                            94

                                                                            F frac14 1

                                                                            W sin

                                                                            Xfc0bthorn ethW ubTHORN tan0g sec

                                                                            1thorn ethtan tan0=FTHORN

                                                                            c0 frac14 8 kN=m2

                                                                            0 frac14 32

                                                                            c0b frac14 8 2 frac14 16 kN=m

                                                                            W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                            Try F frac14 100

                                                                            tan0

                                                                            Ffrac14 0625

                                                                            Stability of slopes 77

                                                                            Values of u are as obtained in Figure Q93

                                                                            SliceNo

                                                                            h(m)

                                                                            W frac14 bh(kNm)

                                                                            W sin(kNm)

                                                                            ub(kNm)

                                                                            c0bthorn (W ub) tan0(kNm)

                                                                            sec

                                                                            1thorn (tan tan0)FProduct(kNm)

                                                                            1 05 21 6 2 8 24 1078 262 13 55 31

                                                                            23 33 30 1042 31

                                                                            3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                            224 92 72 0931 67

                                                                            6 50 210 11 40 100 85 0907 777 55 231 14

                                                                            12 58 112 90 0889 80

                                                                            8 60 252 1812

                                                                            80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                            10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                            2154 88 116 0853 99

                                                                            14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                            1074 1091

                                                                            F frac14 1091

                                                                            1074frac14 102 (assumed value 100)

                                                                            Thus

                                                                            F frac14 101

                                                                            95

                                                                            F frac14 1

                                                                            W sin

                                                                            XfWeth1 ruTHORN tan0g sec

                                                                            1thorn ethtan tan0THORN=F

                                                                            0 frac14 33

                                                                            ru frac14 020

                                                                            W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                            eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                            Try F frac14 110

                                                                            tan 0

                                                                            Ffrac14 tan 33

                                                                            110frac14 0590

                                                                            78 Stability of slopes

                                                                            Referring to Figure Q95

                                                                            SliceNo

                                                                            h(m)

                                                                            W frac14 bh(kNm)

                                                                            W sin(kNm)

                                                                            W(1 ru) tan0(kNm)

                                                                            sec

                                                                            1thorn ( tan tan0)FProduct(kNm)

                                                                            1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                            2120 234 0892 209

                                                                            4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                            1185 1271

                                                                            Figure Q95

                                                                            Stability of slopes 79

                                                                            F frac14 1271

                                                                            1185frac14 107

                                                                            The trial value was 110 therefore take F to be 108

                                                                            96

                                                                            (a) Water table at surface the factor of safety is given by Equation 912

                                                                            F frac14 0

                                                                            sat

                                                                            tan0

                                                                            tan

                                                                            ptie 15 frac14 92

                                                                            19

                                                                            tan 36

                                                                            tan

                                                                            tan frac14 0234

                                                                            frac14 13

                                                                            Water table well below surface the factor of safety is given by Equation 911

                                                                            F frac14 tan0

                                                                            tan

                                                                            frac14 tan 36

                                                                            tan 13

                                                                            frac14 31

                                                                            (b) 0d frac14 tan1tan 36

                                                                            125

                                                                            frac14 30

                                                                            Depth of potential failure surface frac14 z

                                                                            Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                            frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                            frac14 504z kN

                                                                            Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                            frac14 19 z sin 13 cos 13

                                                                            frac14 416z kN

                                                                            Rd gtSd therefore the limit state for overall stability is satisfied

                                                                            80 Stability of slopes

                                                                            • Book Cover
                                                                            • Title
                                                                            • Contents
                                                                            • Basic characteristics of soils
                                                                            • Seepage
                                                                            • Effective stress
                                                                            • Shear strength
                                                                            • Stresses and displacements
                                                                            • Lateral earth pressure
                                                                            • Consolidation theory
                                                                            • Bearing capacity
                                                                            • Stability of slopes

                                                                              The immediate settlement is given by Equation 528

                                                                              si frac14 130131qB

                                                                              Eu

                                                                              frac14 094 082 200 2

                                                                              45frac14 7mm

                                                                              Stresses and displacements 33

                                                                              Chapter 6

                                                                              Lateral earth pressure

                                                                              61

                                                                              For 0 frac14 37 the active pressure coefficient is given by

                                                                              Ka frac14 1 sin 37

                                                                              1thorn sin 37frac14 025

                                                                              The total active thrust (Equation 66a with c0 frac14 0) is

                                                                              Pa frac14 1

                                                                              2KaH

                                                                              2 frac14 1

                                                                              2 025 17 62 frac14 765 kN=m

                                                                              If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                                              K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                                              and the thrust on the wall is

                                                                              P0 frac14 1

                                                                              2K0H

                                                                              2 frac14 1

                                                                              2 040 17 62 frac14 122 kN=m

                                                                              62

                                                                              The active pressure coefficients for the three soil types are as follows

                                                                              Ka1 frac141 sin 35

                                                                              1thorn sin 35frac14 0271

                                                                              Ka2 frac141 sin 27

                                                                              1thorn sin 27frac14 0375

                                                                              ffiffiffiffiffiffiffiKa2

                                                                              p frac14 0613

                                                                              Ka3 frac141 sin 42

                                                                              1thorn sin 42frac14 0198

                                                                              Distribution of active pressure (plotted in Figure Q62)

                                                                              Depth (m) Soil Active pressure (kNm2)

                                                                              3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                                              12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                                              At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                                              Total thrust frac14 571 kNm

                                                                              Point of application is (4893571) m from the top of the wall ie 857m

                                                                              Force (kN) Arm (m) Moment (kN m)

                                                                              (1)1

                                                                              2 0271 16 32 frac14 195 20 390

                                                                              (2) 0271 16 3 2 frac14 260 40 1040

                                                                              (3)1

                                                                              2 0271 92 22 frac14 50 433 217

                                                                              (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                                              (5)1

                                                                              2 0375 102 32 frac14 172 70 1204

                                                                              (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                                              (7)1

                                                                              2 0198 112 42 frac14 177 1067 1889

                                                                              (8)1

                                                                              2 98 92 frac14 3969 90 35721

                                                                              5713 48934

                                                                              Figure Q62

                                                                              Lateral earth pressure 35

                                                                              63

                                                                              (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                                              Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                                              frac14 245

                                                                              At the lower end of the piling

                                                                              pa frac14 Kaqthorn Kasatz Kaccu

                                                                              frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                                              frac14 115 kN=m2

                                                                              pp frac14 Kpsatzthorn Kpccu

                                                                              frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                                              frac14 202 kN=m2

                                                                              (b) For 0 frac14 26 and frac14 1

                                                                              20

                                                                              Ka frac14 035

                                                                              Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                                              pfrac14 145 ethEquation 619THORN

                                                                              Kp frac14 37

                                                                              Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                                              pfrac14 47 ethEquation 624THORN

                                                                              At the lower end of the piling

                                                                              pa frac14 Kaqthorn Ka0z Kacc

                                                                              0

                                                                              frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                                              frac14 187 kN=m2

                                                                              pp frac14 Kp0zthorn Kpcc

                                                                              0

                                                                              frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                                              frac14 198 kN=m2

                                                                              36 Lateral earth pressure

                                                                              64

                                                                              (a) For 0 frac14 38 Ka frac14 024

                                                                              0 frac14 20 98 frac14 102 kN=m3

                                                                              The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                                              Force (kN) Arm (m) Moment (kN m)

                                                                              (1) 024 10 66 frac14 159 33 525

                                                                              (2)1

                                                                              2 024 17 392 frac14 310 400 1240

                                                                              (3) 024 17 39 27 frac14 430 135 580

                                                                              (4)1

                                                                              2 024 102 272 frac14 89 090 80

                                                                              (5)1

                                                                              2 98 272 frac14 357 090 321

                                                                              Hfrac14 1345 MH frac14 2746

                                                                              (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                                              (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                                              XM frac14MV MH frac14 7790 kNm

                                                                              Lever arm of base resultant

                                                                              M

                                                                              Vfrac14 779

                                                                              488frac14 160

                                                                              Eccentricity of base resultant

                                                                              e frac14 200 160 frac14 040m

                                                                              39 m

                                                                              27 m

                                                                              40 m

                                                                              04 m

                                                                              04 m

                                                                              26 m

                                                                              (7)

                                                                              (9)

                                                                              (1)(2)

                                                                              (3)

                                                                              (4)

                                                                              (5)

                                                                              (8)(6)

                                                                              (10)

                                                                              WT

                                                                              10 kNm2

                                                                              Hydrostatic

                                                                              Figure Q64

                                                                              Lateral earth pressure 37

                                                                              Base pressures (Equation 627)

                                                                              p frac14 VB

                                                                              1 6e

                                                                              B

                                                                              frac14 488

                                                                              4eth1 060THORN

                                                                              frac14 195 kN=m2 and 49 kN=m2

                                                                              Factor of safety against sliding (Equation 628)

                                                                              F frac14 V tan

                                                                              Hfrac14 488 tan 25

                                                                              1345frac14 17

                                                                              (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                                              Hfrac14 1633 kN

                                                                              V frac14 4879 kN

                                                                              MH frac14 3453 kNm

                                                                              MV frac14 10536 kNm

                                                                              The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                                              65

                                                                              For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                              Kp

                                                                              Ffrac14 385

                                                                              2

                                                                              0 frac14 20 98 frac14 102 kN=m3

                                                                              The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                                              Force (kN) Arm (m) Moment (kN m)

                                                                              (1)1

                                                                              2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                                              (2) 026 17 45 d frac14 199d d2 995d2

                                                                              (3)1

                                                                              2 026 102 d2 frac14 133d2 d3 044d3

                                                                              (4)1

                                                                              2 385

                                                                              2 17 152 frac14 368 dthorn 05 368d 184

                                                                              (5)385

                                                                              2 17 15 d frac14 491d d2 2455d2

                                                                              (6)1

                                                                              2 385

                                                                              2 102 d2 frac14 982d2 d3 327d3

                                                                              38 Lateral earth pressure

                                                                              XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                                              d3 thorn 516d2 283d 1724 frac14 0

                                                                              d frac14 179m

                                                                              Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                                              Over additional 20 embedded depth

                                                                              pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                                              Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                                              66

                                                                              The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                                              Ka frac14 sin 69=sin 105

                                                                              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                                              ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                                              pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                                              26664

                                                                              37775

                                                                              2

                                                                              frac14 050

                                                                              The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                                              Pa frac14 1

                                                                              2 050 19 7502 frac14 267 kN=m

                                                                              Figure Q65

                                                                              Lateral earth pressure 39

                                                                              Horizontal component

                                                                              Ph frac14 267 cos 40 frac14 205 kN=m

                                                                              Vertical component

                                                                              Pv frac14 267 sin 40 frac14 172 kN=m

                                                                              Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                              Force (kN) Arm (m) Moment (kN m)

                                                                              (1)1

                                                                              2 175 650 235 frac14 1337 258 345

                                                                              (2) 050 650 235 frac14 764 175 134

                                                                              (3)1

                                                                              2 070 650 235 frac14 535 127 68

                                                                              (4) 100 400 235 frac14 940 200 188

                                                                              (5) 1

                                                                              2 080 050 235 frac14 47 027 1

                                                                              Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                              Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                              Lever arm of base resultant

                                                                              M

                                                                              Vfrac14 795

                                                                              525frac14 151m

                                                                              Eccentricity of base resultant

                                                                              e frac14 200 151 frac14 049m

                                                                              Figure Q66

                                                                              40 Lateral earth pressure

                                                                              Base pressures (Equation 627)

                                                                              p frac14 525

                                                                              41 6 049

                                                                              4

                                                                              frac14 228 kN=m2 and 35 kN=m2

                                                                              The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                              The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                              The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                              67

                                                                              For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                              Force (kN) Arm (m) Moment (kNm)

                                                                              (1)1

                                                                              2 027 17 52 frac14 574 183 1050

                                                                              (2) 027 17 5 3 frac14 689 500 3445

                                                                              (3)1

                                                                              2 027 102 32 frac14 124 550 682

                                                                              (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                              (5)1

                                                                              2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                              (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                              (7) 1

                                                                              2 267

                                                                              2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                              (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                              2 d frac14 163d d2thorn 650 82d2 1060d

                                                                              Tie rod force per m frac14 T 0 0

                                                                              XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                              d3 thorn 77d2 269d 1438 frac14 0

                                                                              d frac14 467m

                                                                              Depth of penetration frac14 12d frac14 560m

                                                                              Lateral earth pressure 41

                                                                              Algebraic sum of forces for d frac14 467m isX

                                                                              F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                              T frac14 905 kN=m

                                                                              Force in each tie rod frac14 25T frac14 226 kN

                                                                              68

                                                                              (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                              0 frac14 21 98 frac14 112 kN=m3

                                                                              The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                              uC frac14 150

                                                                              165 15 98 frac14 134 kN=m2

                                                                              The average seepage pressure is

                                                                              j frac14 15

                                                                              165 98 frac14 09 kN=m3

                                                                              Hence

                                                                              0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                              0 j frac14 112 09 frac14 103 kN=m3

                                                                              Figure Q67

                                                                              42 Lateral earth pressure

                                                                              Consider moments about the anchor point A (per m)

                                                                              Force (kN) Arm (m) Moment (kN m)

                                                                              (1) 10 026 150 frac14 390 60 2340

                                                                              (2)1

                                                                              2 026 18 452 frac14 474 15 711

                                                                              (3) 026 18 45 105 frac14 2211 825 18240

                                                                              (4)1

                                                                              2 026 121 1052 frac14 1734 100 17340

                                                                              (5)1

                                                                              2 134 15 frac14 101 40 404

                                                                              (6) 134 30 frac14 402 60 2412

                                                                              (7)1

                                                                              2 134 60 frac14 402 95 3819

                                                                              571 4527(8) Ppm

                                                                              115 115PPm

                                                                              XM frac14 0

                                                                              Ppm frac144527

                                                                              115frac14 394 kN=m

                                                                              Available passive resistance

                                                                              Pp frac14 1

                                                                              2 385 103 62 frac14 714 kN=m

                                                                              Factor of safety

                                                                              Fp frac14 Pp

                                                                              Ppm

                                                                              frac14 714

                                                                              394frac14 18

                                                                              Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                              Figure Q68

                                                                              Lateral earth pressure 43

                                                                              (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                              Consider moments (per m) about the tie point A

                                                                              Force (kN) Arm (m)

                                                                              (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                              (2)1

                                                                              2 033 18 452 frac14 601 15

                                                                              (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                              (4)1

                                                                              2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                              (5)1

                                                                              2 134 15 frac14 101 40

                                                                              (6) 134 30 frac14 402 60

                                                                              (7)1

                                                                              2 134 d frac14 67d d3thorn 75

                                                                              (8) 1

                                                                              2 30 103 d2 frac141545d2 2d3thorn 75

                                                                              Moment (kN m)

                                                                              (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                              XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                              d3 thorn 827d2 466d 1518 frac14 0

                                                                              By trial

                                                                              d frac14 544m

                                                                              The minimum depth of embedment required is 544m

                                                                              69

                                                                              For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                              0 frac14 20 98 frac14 102 kN=m3

                                                                              The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                              44 Lateral earth pressure

                                                                              uC frac14 147

                                                                              173 26 98 frac14 216 kN=m2

                                                                              and the average seepage pressure around the wall is

                                                                              j frac14 26

                                                                              173 98 frac14 15 kN=m3

                                                                              Consider moments about the prop (A) (per m)

                                                                              Force (kN) Arm (m) Moment (kN m)

                                                                              (1)1

                                                                              2 03 17 272 frac14 186 020 37

                                                                              (2) 03 17 27 53 frac14 730 335 2445

                                                                              (3)1

                                                                              2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                              (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                              (5)1

                                                                              2 216 26 frac14 281 243 684

                                                                              (6) 216 27 frac14 583 465 2712

                                                                              (7)1

                                                                              2 216 60 frac14 648 800 5184

                                                                              3055(8)

                                                                              1

                                                                              2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                              Factor of safety

                                                                              Fr frac14 6885

                                                                              3055frac14 225

                                                                              Figure Q69

                                                                              Lateral earth pressure 45

                                                                              610

                                                                              For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                              p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                              Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                              Using the recommendations of Twine and Roscoe

                                                                              p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                              Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                              611

                                                                              frac14 18 kN=m3 0 frac14 34

                                                                              H frac14 350m nH frac14 335m mH frac14 185m

                                                                              Consider a trial value of F frac14 20 Refer to Figure 635

                                                                              0m frac14 tan1tan 34

                                                                              20

                                                                              frac14 186

                                                                              Then

                                                                              frac14 45 thorn 0m2frac14 543

                                                                              W frac14 1

                                                                              2 18 3502 cot 543 frac14 792 kN=m

                                                                              Figure Q610

                                                                              46 Lateral earth pressure

                                                                              P frac14 1

                                                                              2 s 3352 frac14 561s kN=m

                                                                              U frac14 1

                                                                              2 98 1852 cosec 543 frac14 206 kN=m

                                                                              Equations 630 and 631 then become

                                                                              561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                              792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                              ie

                                                                              561s 0616N 405 frac14 0

                                                                              792 0857N thorn 563 frac14 0

                                                                              N frac14 848

                                                                              0857frac14 989 kN=m

                                                                              Then

                                                                              561s 609 405 frac14 0

                                                                              s frac14 649

                                                                              561frac14 116 kN=m3

                                                                              The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                              F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                              20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                              s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                              Figure Q611

                                                                              Lateral earth pressure 47

                                                                              612

                                                                              For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                              45 thorn 0

                                                                              2frac14 63

                                                                              For the retained material between the surface and a depth of 36m

                                                                              Pa frac14 1

                                                                              2 030 18 362 frac14 350 kN=m

                                                                              Weight of reinforced fill between the surface and a depth of 36m is

                                                                              Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                              eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                              Eccentricity of Rv

                                                                              e frac14 263 250 frac14 013m

                                                                              The average vertical stress at a depth of 36m is

                                                                              z frac14 Rv

                                                                              L 2efrac14 324

                                                                              474frac14 68 kN=m2

                                                                              (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                              Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                              Tensile stress in the element frac14 138 103

                                                                              65 3frac14 71N=mm2

                                                                              Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                              Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                              Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                              The weight of ABC is

                                                                              W frac14 1

                                                                              2 18 52 265 frac14 124 kN=m

                                                                              From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                              48 Lateral earth pressure

                                                                              (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                              Tp frac14 032 68 120 065 frac14 170 kN

                                                                              Tr frac14 213 420

                                                                              418frac14 214 kN

                                                                              Again the tensile failure and slipping limit states are satisfied for this element

                                                                              Figure Q612

                                                                              Lateral earth pressure 49

                                                                              Chapter 7

                                                                              Consolidation theory

                                                                              71

                                                                              Total change in thickness

                                                                              H frac14 782 602 frac14 180mm

                                                                              Average thickness frac14 1530thorn 180

                                                                              2frac14 1620mm

                                                                              Length of drainage path d frac14 1620

                                                                              2frac14 810mm

                                                                              Root time plot (Figure Q71a)

                                                                              ffiffiffiffiffiffit90p frac14 33

                                                                              t90 frac14 109min

                                                                              cv frac14 0848d2

                                                                              t90frac14 0848 8102

                                                                              109 1440 365

                                                                              106frac14 27m2=year

                                                                              r0 frac14 782 764

                                                                              782 602frac14 018

                                                                              180frac14 0100

                                                                              rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                              10 119

                                                                              9 180frac14 0735

                                                                              rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                              Log time plot (Figure Q71b)

                                                                              t50 frac14 26min

                                                                              cv frac14 0196d2

                                                                              t50frac14 0196 8102

                                                                              26 1440 365

                                                                              106frac14 26m2=year

                                                                              r0 frac14 782 763

                                                                              782 602frac14 019

                                                                              180frac14 0106

                                                                              rp frac14 763 623

                                                                              782 602frac14 140

                                                                              180frac14 0778

                                                                              rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                              Figure Q71(a)

                                                                              Figure Q71(b)

                                                                              Final void ratio

                                                                              e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                              e

                                                                              Hfrac14 1thorn e0

                                                                              H0frac14 1thorn e1 thorne

                                                                              H0

                                                                              ie

                                                                              e

                                                                              180frac14 1631thorne

                                                                              1710

                                                                              e frac14 2936

                                                                              1530frac14 0192

                                                                              Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                              mv frac14 1

                                                                              1thorn e0 e0 e101 00

                                                                              frac14 1

                                                                              1823 0192

                                                                              0107frac14 098m2=MN

                                                                              k frac14 cvmvw frac14 265 098 98

                                                                              60 1440 365 103frac14 81 1010 m=s

                                                                              72

                                                                              Using Equation 77 (one-dimensional method)

                                                                              sc frac14 e0 e11thorn e0 H

                                                                              Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                              Figure Q72

                                                                              52 Consolidation theory

                                                                              Settlement

                                                                              Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                              1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                              318

                                                                              Notes 5 92y 460thorn 84

                                                                              Heave

                                                                              Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                              1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                              38

                                                                              73

                                                                              U frac14 f ethTvTHORN frac14 f cvt

                                                                              d2

                                                                              Hence if cv is constant

                                                                              t1

                                                                              t2frac14 d

                                                                              21

                                                                              d22

                                                                              where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                              d1 frac14 95mm and d2 frac14 2500mm

                                                                              for U frac14 050 t2 frac14 t1 d22

                                                                              d21

                                                                              frac14 20

                                                                              60 24 365 25002

                                                                              952frac14 263 years

                                                                              for U lt 060 Tv frac14

                                                                              4U2 (Equation 724(a))

                                                                              t030 frac14 t050 0302

                                                                              0502

                                                                              frac14 263 036 frac14 095 years

                                                                              Consolidation theory 53

                                                                              74

                                                                              The layer is open

                                                                              d frac14 8

                                                                              2frac14 4m

                                                                              Tv frac14 cvtd2frac14 24 3

                                                                              42frac14 0450

                                                                              ui frac14 frac14 84 kN=m2

                                                                              The excess pore water pressure is given by Equation 721

                                                                              ue frac14Xmfrac141mfrac140

                                                                              2ui

                                                                              Msin

                                                                              Mz

                                                                              d

                                                                              expethM2TvTHORN

                                                                              In this case z frac14 d

                                                                              sinMz

                                                                              d

                                                                              frac14 sinM

                                                                              where

                                                                              M frac14

                                                                              23

                                                                              25

                                                                              2

                                                                              M sin M M2Tv exp (M2Tv)

                                                                              2thorn1 1110 0329

                                                                              3

                                                                              21 9993 457 105

                                                                              ue frac14 2 84 2

                                                                              1 0329 ethother terms negligibleTHORN

                                                                              frac14 352 kN=m2

                                                                              75

                                                                              The layer is open

                                                                              d frac14 6

                                                                              2frac14 3m

                                                                              Tv frac14 cvtd2frac14 10 3

                                                                              32frac14 0333

                                                                              The layer thickness will be divided into six equal parts ie m frac14 6

                                                                              54 Consolidation theory

                                                                              For an open layer

                                                                              Tv frac14 4n

                                                                              m2

                                                                              n frac14 0333 62

                                                                              4frac14 300

                                                                              The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                              ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                              i j

                                                                              0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                              0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                              The initial and 3-year isochrones are plotted in Figure Q75

                                                                              Area under initial isochrone frac14 180 units

                                                                              Area under 3-year isochrone frac14 63 units

                                                                              The average degree of consolidation is given by Equation 725Thus

                                                                              U frac14 1 63

                                                                              180frac14 065

                                                                              Figure Q75

                                                                              Consolidation theory 55

                                                                              76

                                                                              At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                              0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                              The final consolidation settlement (one-dimensional method) is

                                                                              sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                              Corrected time t frac14 2 1

                                                                              2

                                                                              40

                                                                              52

                                                                              frac14 1615 years

                                                                              Tv frac14 cvtd2frac14 44 1615

                                                                              42frac14 0444

                                                                              From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                              77

                                                                              The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                              Figure Q77

                                                                              56 Consolidation theory

                                                                              Point m n Ir (kNm2) sc (mm)

                                                                              13020frac14 15 20

                                                                              20frac14 10 0194 (4) 113 124

                                                                              260

                                                                              20frac14 30

                                                                              20

                                                                              20frac14 10 0204 (2) 59 65

                                                                              360

                                                                              20frac14 30

                                                                              40

                                                                              20frac14 20 0238 (1) 35 38

                                                                              430

                                                                              20frac14 15

                                                                              40

                                                                              20frac14 20 0224 (2) 65 72

                                                                              Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                              78

                                                                              Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                              (a) Immediate settlement

                                                                              H

                                                                              Bfrac14 30

                                                                              35frac14 086

                                                                              D

                                                                              Bfrac14 2

                                                                              35frac14 006

                                                                              Figure Q78

                                                                              Consolidation theory 57

                                                                              From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                              si frac14 130131qB

                                                                              Eufrac14 10 032 105 35

                                                                              40frac14 30mm

                                                                              (b) Consolidation settlement

                                                                              Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                              1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                              3150

                                                                              Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                              Now

                                                                              H

                                                                              Bfrac14 30

                                                                              35frac14 086 and A frac14 065

                                                                              from Figure 712 13 frac14 079

                                                                              sc frac14 13sod frac14 079 315 frac14 250mm

                                                                              Total settlement

                                                                              s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                              79

                                                                              Without sand drains

                                                                              Uv frac14 025

                                                                              Tv frac14 0049 ethfrom Figure 718THORN

                                                                              t frac14 Tvd2

                                                                              cvfrac14 0049 82

                                                                              cvWith sand drains

                                                                              R frac14 0564S frac14 0564 3 frac14 169m

                                                                              n frac14 Rrfrac14 169

                                                                              015frac14 113

                                                                              Tr frac14 cht

                                                                              4R2frac14 ch

                                                                              4 1692 0049 82

                                                                              cvethand ch frac14 cvTHORN

                                                                              frac14 0275

                                                                              Ur frac14 073 (from Figure 730)

                                                                              58 Consolidation theory

                                                                              Using Equation 740

                                                                              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                              U frac14 080

                                                                              710

                                                                              Without sand drains

                                                                              Uv frac14 090

                                                                              Tv frac14 0848

                                                                              t frac14 Tvd2

                                                                              cvfrac14 0848 102

                                                                              96frac14 88 years

                                                                              With sand drains

                                                                              R frac14 0564S frac14 0564 4 frac14 226m

                                                                              n frac14 Rrfrac14 226

                                                                              015frac14 15

                                                                              Tr

                                                                              Tvfrac14 chcv

                                                                              d2

                                                                              4R2ethsame tTHORN

                                                                              Tr

                                                                              Tvfrac14 140

                                                                              96 102

                                                                              4 2262frac14 714 eth1THORN

                                                                              Using Equation 740

                                                                              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                              An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                              Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                              040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                              Thus

                                                                              Uv frac14 0295 and Ur frac14 086

                                                                              t frac14 88 00683

                                                                              0848frac14 07 years

                                                                              Consolidation theory 59

                                                                              Chapter 8

                                                                              Bearing capacity

                                                                              81

                                                                              (a) The ultimate bearing capacity is given by Equation 83

                                                                              qf frac14 cNc thorn DNq thorn 1

                                                                              2BN

                                                                              For u frac14 0

                                                                              Nc frac14 514 Nq frac14 1 N frac14 0

                                                                              qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                              The net ultimate bearing capacity is

                                                                              qnf frac14 qf D frac14 540 kN=m2

                                                                              The net foundation pressure is

                                                                              qn frac14 q D frac14 425

                                                                              2 eth21 1THORN frac14 192 kN=m2

                                                                              The factor of safety (Equation 86) is

                                                                              F frac14 qnfqnfrac14 540

                                                                              192frac14 28

                                                                              (b) For 0 frac14 28

                                                                              Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                              qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                              2 112 2 13

                                                                              frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                              qnf frac14 574 112 frac14 563 kN=m2

                                                                              F frac14 563

                                                                              192frac14 29

                                                                              (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                              82

                                                                              For 0 frac14 38

                                                                              Nq frac14 49 N frac14 67

                                                                              qnf frac14 DethNq 1THORN thorn 1

                                                                              2BN ethfrom Equation 83THORN

                                                                              frac14 eth18 075 48THORN thorn 1

                                                                              2 18 15 67

                                                                              frac14 648thorn 905 frac14 1553 kN=m2

                                                                              qn frac14 500

                                                                              15 eth18 075THORN frac14 320 kN=m2

                                                                              F frac14 qnfqnfrac14 1553

                                                                              320frac14 48

                                                                              0d frac14 tan1tan 38

                                                                              125

                                                                              frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                              Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                              2 18 15 25

                                                                              frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                              Design load (action) Vd frac14 500 kN=m

                                                                              The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                              83

                                                                              D

                                                                              Bfrac14 350

                                                                              225frac14 155

                                                                              From Figure 85 for a square foundation

                                                                              Nc frac14 81

                                                                              Bearing capacity 61

                                                                              For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                              Nc frac14 084thorn 016B

                                                                              L

                                                                              81 frac14 745

                                                                              Using Equation 810

                                                                              qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                              For F frac14 3

                                                                              qn frac14 1006

                                                                              3frac14 335 kN=m2

                                                                              q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                              Design load frac14 405 450 225 frac14 4100 kN

                                                                              Design undrained strength cud frac14 135

                                                                              14frac14 96 kN=m2

                                                                              Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                              frac14 7241 kN

                                                                              Design load Vd frac14 4100 kN

                                                                              Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                              84

                                                                              For 0 frac14 40

                                                                              Nq frac14 64 N frac14 95

                                                                              qnf frac14 DethNq 1THORN thorn 04BN

                                                                              (a) Water table 5m below ground level

                                                                              qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                              qn frac14 400 17 frac14 383 kN=m2

                                                                              F frac14 2686

                                                                              383frac14 70

                                                                              (b) Water table 1m below ground level (ie at foundation level)

                                                                              0 frac14 20 98 frac14 102 kN=m3

                                                                              62 Bearing capacity

                                                                              qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                              F frac14 2040

                                                                              383frac14 53

                                                                              (c) Water table at ground level with upward hydraulic gradient 02

                                                                              eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                              qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                              F frac14 1296

                                                                              392frac14 33

                                                                              85

                                                                              The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                              Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                              Design value of 0 frac14 tan1tan 39

                                                                              125

                                                                              frac14 33

                                                                              For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                              Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                              Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                              86

                                                                              (a) Undrained shear for u frac14 0

                                                                              Nc frac14 514 Nq frac14 1 N frac14 0

                                                                              qnf frac14 12cuNc

                                                                              frac14 12 100 514 frac14 617 kN=m2

                                                                              qn frac14 qnfFfrac14 617

                                                                              3frac14 206 kN=m2

                                                                              q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                              Bearing capacity 63

                                                                              Drained shear for 0 frac14 32

                                                                              Nq frac14 23 N frac14 25

                                                                              0 frac14 21 98 frac14 112 kN=m3

                                                                              qnf frac14 0DethNq 1THORN thorn 040BN

                                                                              frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                              frac14 694 kN=m2

                                                                              q frac14 694

                                                                              3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                              Design load frac14 42 227 frac14 3632 kN

                                                                              (b) Design undrained strength cud frac14 100

                                                                              14frac14 71 kNm2

                                                                              Design bearing resistance Rd frac14 12cudNe area

                                                                              frac14 12 71 514 42

                                                                              frac14 7007 kN

                                                                              For drained shear 0d frac14 tan1tan 32

                                                                              125

                                                                              frac14 26

                                                                              Nq frac14 12 N frac14 10

                                                                              Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                              (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                              Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                              1 2 100 0175 0700qn 0182qn

                                                                              2 6 033 0044 0176qn 0046qn

                                                                              3 10 020 0017 0068qn 0018qn

                                                                              0246qn

                                                                              Diameter of equivalent circle B frac14 45m

                                                                              H

                                                                              Bfrac14 12

                                                                              45frac14 27 and A frac14 042

                                                                              13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                              64 Bearing capacity

                                                                              For sc frac14 30mm

                                                                              qn frac14 30

                                                                              0147frac14 204 kN=m2

                                                                              q frac14 204thorn 21 frac14 225 kN=m2

                                                                              Design load frac14 42 225 frac14 3600 kN

                                                                              The design load is 3600 kN settlement being the limiting criterion

                                                                              87

                                                                              D

                                                                              Bfrac14 8

                                                                              4frac14 20

                                                                              From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                              F frac14 cuNc

                                                                              Dfrac14 40 71

                                                                              20 8frac14 18

                                                                              88

                                                                              Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                              Design value of 0 frac14 tan1tan 38

                                                                              125

                                                                              frac14 32

                                                                              Figure Q86

                                                                              Bearing capacity 65

                                                                              For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                              Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                              The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                              Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                              For B frac14 250m qn frac14 3750

                                                                              2502 17 frac14 583 kN=m2

                                                                              From Figure 510 m frac14 n frac14 126

                                                                              6frac14 021

                                                                              Ir frac14 0019

                                                                              Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                              Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                              The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                              89

                                                                              Depth (m) N 0v (kNm2) CN N1

                                                                              070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                              Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                              (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                              Cw frac14 05thorn 0530

                                                                              47

                                                                              frac14 082

                                                                              66 Bearing capacity

                                                                              Thus

                                                                              qa frac14 150 082 frac14 120 kN=m2

                                                                              (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                              Thus

                                                                              qa frac14 90 15 frac14 135 kN=m2

                                                                              (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                              Ic frac14 171

                                                                              1014frac14 0068

                                                                              From Equation 819(a) with s frac14 25mm

                                                                              q frac14 25

                                                                              3507 0068frac14 150 kN=m2

                                                                              810

                                                                              Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                              Izp frac14 05thorn 01130

                                                                              16 27

                                                                              05

                                                                              frac14 067

                                                                              Refer to Figure Q810

                                                                              E frac14 25qc

                                                                              Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                              Ez (mm3MN)

                                                                              1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                              0203

                                                                              C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                              130frac14 093

                                                                              C2 frac14 1 ethsayTHORN

                                                                              s frac14 C1C2qnX Iz

                                                                              Ez frac14 093 1 130 0203 frac14 25mm

                                                                              Bearing capacity 67

                                                                              811

                                                                              At pile base level

                                                                              cu frac14 220 kN=m2

                                                                              qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                              Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                              00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                              qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                              Then

                                                                              Qf frac14 Abqb thorn Asqs

                                                                              frac14

                                                                              4 32 1980

                                                                              thorn eth 105 139 86THORN

                                                                              frac14 13 996thorn 3941 frac14 17 937 kN

                                                                              0 01 02 03 04 05 06 07

                                                                              0 2 4 6 8 10 12 14

                                                                              1

                                                                              2

                                                                              3

                                                                              4

                                                                              5

                                                                              6

                                                                              7

                                                                              8

                                                                              (1)

                                                                              (2)

                                                                              (3)

                                                                              (4)

                                                                              (5)

                                                                              qc

                                                                              qc

                                                                              Iz

                                                                              Iz

                                                                              (MNm2)

                                                                              z (m)

                                                                              Figure Q810

                                                                              68 Bearing capacity

                                                                              Allowable load

                                                                              ethaTHORN Qf

                                                                              2frac14 17 937

                                                                              2frac14 8968 kN

                                                                              ethbTHORN Abqb

                                                                              3thorn Asqs frac14 13 996

                                                                              3thorn 3941 frac14 8606 kN

                                                                              ie allowable load frac14 8600 kN

                                                                              Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                              According to the limit state method

                                                                              Characteristic undrained strength at base level cuk frac14 220

                                                                              150kN=m2

                                                                              Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                              150frac14 1320 kN=m2

                                                                              Characteristic shaft resistance qsk frac14 00150

                                                                              frac14 86

                                                                              150frac14 57 kN=m2

                                                                              Characteristic base and shaft resistances

                                                                              Rbk frac14

                                                                              4 32 1320 frac14 9330 kN

                                                                              Rsk frac14 105 139 86

                                                                              150frac14 2629 kN

                                                                              For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                              Design bearing resistance Rcd frac14 9330

                                                                              160thorn 2629

                                                                              130

                                                                              frac14 5831thorn 2022

                                                                              frac14 7850 kN

                                                                              Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                              812

                                                                              ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                              qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                              For a single pile

                                                                              Qf frac14 Abqb thorn Asqs

                                                                              frac14

                                                                              4 062 1305

                                                                              thorn eth 06 15 42THORN

                                                                              frac14 369thorn 1187 frac14 1556 kN

                                                                              Bearing capacity 69

                                                                              Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                              eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                              (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                              (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                              qbkfrac14 9cuk frac14 9 220

                                                                              150frac14 1320 kN=m2

                                                                              qskfrac14cuk frac14 040 105

                                                                              150frac14 28 kN=m2

                                                                              Rbkfrac14

                                                                              4 0602 1320 frac14 373 kN

                                                                              Rskfrac14 060 15 28 frac14 791 kN

                                                                              Rcdfrac14 373

                                                                              160thorn 791

                                                                              130frac14 233thorn 608 frac14 841 kN

                                                                              Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                              Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                              (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                              q frac14 21 000

                                                                              1762frac14 68 kN=m2

                                                                              Immediate settlement

                                                                              H

                                                                              Bfrac14 15

                                                                              176frac14 085

                                                                              D

                                                                              Bfrac14 13

                                                                              176frac14 074

                                                                              L

                                                                              Bfrac14 1

                                                                              Hence from Figure 515

                                                                              130 frac14 078 and 131 frac14 041

                                                                              70 Bearing capacity

                                                                              Thus using Equation 528

                                                                              si frac14 078 041 68 176

                                                                              65frac14 6mm

                                                                              Consolidation settlement

                                                                              Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                              1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                              434 (sod)

                                                                              Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                              sc frac14 056 434 frac14 24mm

                                                                              The total settlement is (6thorn 24) frac14 30mm

                                                                              813

                                                                              At base level N frac14 26 Then using Equation 830

                                                                              qb frac14 40NDb

                                                                              Bfrac14 40 26 2

                                                                              025frac14 8320 kN=m2

                                                                              ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                              Figure Q812

                                                                              Bearing capacity 71

                                                                              Over the length embedded in sand

                                                                              N frac14 21 ie18thorn 24

                                                                              2

                                                                              Using Equation 831

                                                                              qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                              For a single pile

                                                                              Qf frac14 Abqb thorn Asqs

                                                                              frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                              For the pile group assuming a group efficiency of 12

                                                                              XQf frac14 12 9 604 frac14 6523 kN

                                                                              Then the load factor is

                                                                              F frac14 6523

                                                                              2000thorn 1000frac14 21

                                                                              (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                              Characteristic base resistance per unit area qbk frac14 8320

                                                                              150frac14 5547 kNm2

                                                                              Characteristic shaft resistance per unit area qsk frac14 42

                                                                              150frac14 28 kNm2

                                                                              Characteristic base and shaft resistances for a single pile

                                                                              Rbk frac14 0252 5547 frac14 347 kN

                                                                              Rsk frac14 4 025 2 28 frac14 56 kN

                                                                              For a driven pile the partial factors are b frac14 s frac14 130

                                                                              Design bearing resistance Rcd frac14 347

                                                                              130thorn 56

                                                                              130frac14 310 kN

                                                                              For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                              Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                              (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                              From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                              72 Bearing capacity

                                                                              N frac14 24thorn 26thorn 34

                                                                              3frac14 28

                                                                              Ic frac14 171

                                                                              2814frac14 0016 ethEquation 818THORN

                                                                              s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                              814

                                                                              Using Equation 841

                                                                              Tf frac14 DLcu thorn

                                                                              4ethD2 d2THORNcuNc

                                                                              frac14 eth 02 5 06 110THORN thorn

                                                                              4eth022 012THORN110 9

                                                                              frac14 207thorn 23 frac14 230 kN

                                                                              Figure Q813

                                                                              Bearing capacity 73

                                                                              Chapter 9

                                                                              Stability of slopes

                                                                              91

                                                                              Referring to Figure Q91

                                                                              W frac14 417 19 frac14 792 kN=m

                                                                              Q frac14 20 28 frac14 56 kN=m

                                                                              Arc lengthAB frac14

                                                                              180 73 90 frac14 115m

                                                                              Arc length BC frac14

                                                                              180 28 90 frac14 44m

                                                                              The factor of safety is given by

                                                                              F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                              90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                              Depth of tension crack z0 frac14 2cu

                                                                              frac14 2 20

                                                                              19frac14 21m

                                                                              Arc length BD frac14

                                                                              180 13

                                                                              1

                                                                              2 90 frac14 21m

                                                                              F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                              The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                              Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                              14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                              Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                              The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                              92

                                                                              u frac14 0

                                                                              Depth factor D frac14 11

                                                                              9frac14 122

                                                                              Using Equation 92 with F frac14 10

                                                                              Ns frac14 cu

                                                                              FHfrac14 30

                                                                              10 19 9frac14 0175

                                                                              Hence from Figure 93

                                                                              frac14 50

                                                                              For F frac14 12

                                                                              Ns frac14 30

                                                                              12 19 9frac14 0146

                                                                              frac14 27

                                                                              93

                                                                              Refer to Figure Q93

                                                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                              1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                              74 m

                                                                              214 1deg

                                                                              213 1deg

                                                                              39 m

                                                                              WB

                                                                              D

                                                                              C

                                                                              28 m

                                                                              21 m

                                                                              A

                                                                              Q

                                                                              Soil (1)Soil (2)

                                                                              73deg

                                                                              Figure Q91

                                                                              Stability of slopes 75

                                                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                              9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                              599 256 328 1372

                                                                              Figure Q93

                                                                              76 Stability of slopes

                                                                              XW cos frac14 b

                                                                              Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                              W sin frac14 bX

                                                                              h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                              Arc length La frac14

                                                                              180 57

                                                                              1

                                                                              2 326 frac14 327m

                                                                              The factor of safety is given by

                                                                              F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                              W sin

                                                                              frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                              frac14 091

                                                                              According to the limit state method

                                                                              0d frac14 tan1tan 32

                                                                              125

                                                                              frac14 265

                                                                              c0 frac14 8

                                                                              160frac14 5 kN=m2

                                                                              Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                              Design disturbing moment frac14 1075 kN=m

                                                                              The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                              94

                                                                              F frac14 1

                                                                              W sin

                                                                              Xfc0bthorn ethW ubTHORN tan0g sec

                                                                              1thorn ethtan tan0=FTHORN

                                                                              c0 frac14 8 kN=m2

                                                                              0 frac14 32

                                                                              c0b frac14 8 2 frac14 16 kN=m

                                                                              W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                              Try F frac14 100

                                                                              tan0

                                                                              Ffrac14 0625

                                                                              Stability of slopes 77

                                                                              Values of u are as obtained in Figure Q93

                                                                              SliceNo

                                                                              h(m)

                                                                              W frac14 bh(kNm)

                                                                              W sin(kNm)

                                                                              ub(kNm)

                                                                              c0bthorn (W ub) tan0(kNm)

                                                                              sec

                                                                              1thorn (tan tan0)FProduct(kNm)

                                                                              1 05 21 6 2 8 24 1078 262 13 55 31

                                                                              23 33 30 1042 31

                                                                              3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                              224 92 72 0931 67

                                                                              6 50 210 11 40 100 85 0907 777 55 231 14

                                                                              12 58 112 90 0889 80

                                                                              8 60 252 1812

                                                                              80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                              10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                              2154 88 116 0853 99

                                                                              14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                              1074 1091

                                                                              F frac14 1091

                                                                              1074frac14 102 (assumed value 100)

                                                                              Thus

                                                                              F frac14 101

                                                                              95

                                                                              F frac14 1

                                                                              W sin

                                                                              XfWeth1 ruTHORN tan0g sec

                                                                              1thorn ethtan tan0THORN=F

                                                                              0 frac14 33

                                                                              ru frac14 020

                                                                              W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                              eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                              Try F frac14 110

                                                                              tan 0

                                                                              Ffrac14 tan 33

                                                                              110frac14 0590

                                                                              78 Stability of slopes

                                                                              Referring to Figure Q95

                                                                              SliceNo

                                                                              h(m)

                                                                              W frac14 bh(kNm)

                                                                              W sin(kNm)

                                                                              W(1 ru) tan0(kNm)

                                                                              sec

                                                                              1thorn ( tan tan0)FProduct(kNm)

                                                                              1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                              2120 234 0892 209

                                                                              4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                              1185 1271

                                                                              Figure Q95

                                                                              Stability of slopes 79

                                                                              F frac14 1271

                                                                              1185frac14 107

                                                                              The trial value was 110 therefore take F to be 108

                                                                              96

                                                                              (a) Water table at surface the factor of safety is given by Equation 912

                                                                              F frac14 0

                                                                              sat

                                                                              tan0

                                                                              tan

                                                                              ptie 15 frac14 92

                                                                              19

                                                                              tan 36

                                                                              tan

                                                                              tan frac14 0234

                                                                              frac14 13

                                                                              Water table well below surface the factor of safety is given by Equation 911

                                                                              F frac14 tan0

                                                                              tan

                                                                              frac14 tan 36

                                                                              tan 13

                                                                              frac14 31

                                                                              (b) 0d frac14 tan1tan 36

                                                                              125

                                                                              frac14 30

                                                                              Depth of potential failure surface frac14 z

                                                                              Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                              frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                              frac14 504z kN

                                                                              Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                              frac14 19 z sin 13 cos 13

                                                                              frac14 416z kN

                                                                              Rd gtSd therefore the limit state for overall stability is satisfied

                                                                              80 Stability of slopes

                                                                              • Book Cover
                                                                              • Title
                                                                              • Contents
                                                                              • Basic characteristics of soils
                                                                              • Seepage
                                                                              • Effective stress
                                                                              • Shear strength
                                                                              • Stresses and displacements
                                                                              • Lateral earth pressure
                                                                              • Consolidation theory
                                                                              • Bearing capacity
                                                                              • Stability of slopes

                                                                                Chapter 6

                                                                                Lateral earth pressure

                                                                                61

                                                                                For 0 frac14 37 the active pressure coefficient is given by

                                                                                Ka frac14 1 sin 37

                                                                                1thorn sin 37frac14 025

                                                                                The total active thrust (Equation 66a with c0 frac14 0) is

                                                                                Pa frac14 1

                                                                                2KaH

                                                                                2 frac14 1

                                                                                2 025 17 62 frac14 765 kN=m

                                                                                If the wall is prevented from yielding the at-rest condition applies The approximatevalue of the coefficient of earth pressure at-rest is given by Equation 615a

                                                                                K0 frac14 1 sin0 frac14 1 sin 37 frac14 040

                                                                                and the thrust on the wall is

                                                                                P0 frac14 1

                                                                                2K0H

                                                                                2 frac14 1

                                                                                2 040 17 62 frac14 122 kN=m

                                                                                62

                                                                                The active pressure coefficients for the three soil types are as follows

                                                                                Ka1 frac141 sin 35

                                                                                1thorn sin 35frac14 0271

                                                                                Ka2 frac141 sin 27

                                                                                1thorn sin 27frac14 0375

                                                                                ffiffiffiffiffiffiffiKa2

                                                                                p frac14 0613

                                                                                Ka3 frac141 sin 42

                                                                                1thorn sin 42frac14 0198

                                                                                Distribution of active pressure (plotted in Figure Q62)

                                                                                Depth (m) Soil Active pressure (kNm2)

                                                                                3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                                                12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                                                At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                                                Total thrust frac14 571 kNm

                                                                                Point of application is (4893571) m from the top of the wall ie 857m

                                                                                Force (kN) Arm (m) Moment (kN m)

                                                                                (1)1

                                                                                2 0271 16 32 frac14 195 20 390

                                                                                (2) 0271 16 3 2 frac14 260 40 1040

                                                                                (3)1

                                                                                2 0271 92 22 frac14 50 433 217

                                                                                (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                                                (5)1

                                                                                2 0375 102 32 frac14 172 70 1204

                                                                                (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                                                (7)1

                                                                                2 0198 112 42 frac14 177 1067 1889

                                                                                (8)1

                                                                                2 98 92 frac14 3969 90 35721

                                                                                5713 48934

                                                                                Figure Q62

                                                                                Lateral earth pressure 35

                                                                                63

                                                                                (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                                                Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                                                frac14 245

                                                                                At the lower end of the piling

                                                                                pa frac14 Kaqthorn Kasatz Kaccu

                                                                                frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                                                frac14 115 kN=m2

                                                                                pp frac14 Kpsatzthorn Kpccu

                                                                                frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                                                frac14 202 kN=m2

                                                                                (b) For 0 frac14 26 and frac14 1

                                                                                20

                                                                                Ka frac14 035

                                                                                Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                                                pfrac14 145 ethEquation 619THORN

                                                                                Kp frac14 37

                                                                                Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                                                pfrac14 47 ethEquation 624THORN

                                                                                At the lower end of the piling

                                                                                pa frac14 Kaqthorn Ka0z Kacc

                                                                                0

                                                                                frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                                                frac14 187 kN=m2

                                                                                pp frac14 Kp0zthorn Kpcc

                                                                                0

                                                                                frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                                                frac14 198 kN=m2

                                                                                36 Lateral earth pressure

                                                                                64

                                                                                (a) For 0 frac14 38 Ka frac14 024

                                                                                0 frac14 20 98 frac14 102 kN=m3

                                                                                The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                                                Force (kN) Arm (m) Moment (kN m)

                                                                                (1) 024 10 66 frac14 159 33 525

                                                                                (2)1

                                                                                2 024 17 392 frac14 310 400 1240

                                                                                (3) 024 17 39 27 frac14 430 135 580

                                                                                (4)1

                                                                                2 024 102 272 frac14 89 090 80

                                                                                (5)1

                                                                                2 98 272 frac14 357 090 321

                                                                                Hfrac14 1345 MH frac14 2746

                                                                                (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                                                (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                                                XM frac14MV MH frac14 7790 kNm

                                                                                Lever arm of base resultant

                                                                                M

                                                                                Vfrac14 779

                                                                                488frac14 160

                                                                                Eccentricity of base resultant

                                                                                e frac14 200 160 frac14 040m

                                                                                39 m

                                                                                27 m

                                                                                40 m

                                                                                04 m

                                                                                04 m

                                                                                26 m

                                                                                (7)

                                                                                (9)

                                                                                (1)(2)

                                                                                (3)

                                                                                (4)

                                                                                (5)

                                                                                (8)(6)

                                                                                (10)

                                                                                WT

                                                                                10 kNm2

                                                                                Hydrostatic

                                                                                Figure Q64

                                                                                Lateral earth pressure 37

                                                                                Base pressures (Equation 627)

                                                                                p frac14 VB

                                                                                1 6e

                                                                                B

                                                                                frac14 488

                                                                                4eth1 060THORN

                                                                                frac14 195 kN=m2 and 49 kN=m2

                                                                                Factor of safety against sliding (Equation 628)

                                                                                F frac14 V tan

                                                                                Hfrac14 488 tan 25

                                                                                1345frac14 17

                                                                                (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                                                Hfrac14 1633 kN

                                                                                V frac14 4879 kN

                                                                                MH frac14 3453 kNm

                                                                                MV frac14 10536 kNm

                                                                                The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                                                65

                                                                                For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                                Kp

                                                                                Ffrac14 385

                                                                                2

                                                                                0 frac14 20 98 frac14 102 kN=m3

                                                                                The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                                                Force (kN) Arm (m) Moment (kN m)

                                                                                (1)1

                                                                                2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                                                (2) 026 17 45 d frac14 199d d2 995d2

                                                                                (3)1

                                                                                2 026 102 d2 frac14 133d2 d3 044d3

                                                                                (4)1

                                                                                2 385

                                                                                2 17 152 frac14 368 dthorn 05 368d 184

                                                                                (5)385

                                                                                2 17 15 d frac14 491d d2 2455d2

                                                                                (6)1

                                                                                2 385

                                                                                2 102 d2 frac14 982d2 d3 327d3

                                                                                38 Lateral earth pressure

                                                                                XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                                                d3 thorn 516d2 283d 1724 frac14 0

                                                                                d frac14 179m

                                                                                Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                                                Over additional 20 embedded depth

                                                                                pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                                                Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                                                66

                                                                                The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                                                Ka frac14 sin 69=sin 105

                                                                                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                                                ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                                                pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                                                26664

                                                                                37775

                                                                                2

                                                                                frac14 050

                                                                                The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                                                Pa frac14 1

                                                                                2 050 19 7502 frac14 267 kN=m

                                                                                Figure Q65

                                                                                Lateral earth pressure 39

                                                                                Horizontal component

                                                                                Ph frac14 267 cos 40 frac14 205 kN=m

                                                                                Vertical component

                                                                                Pv frac14 267 sin 40 frac14 172 kN=m

                                                                                Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                                Force (kN) Arm (m) Moment (kN m)

                                                                                (1)1

                                                                                2 175 650 235 frac14 1337 258 345

                                                                                (2) 050 650 235 frac14 764 175 134

                                                                                (3)1

                                                                                2 070 650 235 frac14 535 127 68

                                                                                (4) 100 400 235 frac14 940 200 188

                                                                                (5) 1

                                                                                2 080 050 235 frac14 47 027 1

                                                                                Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                                Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                                Lever arm of base resultant

                                                                                M

                                                                                Vfrac14 795

                                                                                525frac14 151m

                                                                                Eccentricity of base resultant

                                                                                e frac14 200 151 frac14 049m

                                                                                Figure Q66

                                                                                40 Lateral earth pressure

                                                                                Base pressures (Equation 627)

                                                                                p frac14 525

                                                                                41 6 049

                                                                                4

                                                                                frac14 228 kN=m2 and 35 kN=m2

                                                                                The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                                The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                                The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                                67

                                                                                For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                                Force (kN) Arm (m) Moment (kNm)

                                                                                (1)1

                                                                                2 027 17 52 frac14 574 183 1050

                                                                                (2) 027 17 5 3 frac14 689 500 3445

                                                                                (3)1

                                                                                2 027 102 32 frac14 124 550 682

                                                                                (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                                (5)1

                                                                                2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                                (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                                (7) 1

                                                                                2 267

                                                                                2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                                (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                                2 d frac14 163d d2thorn 650 82d2 1060d

                                                                                Tie rod force per m frac14 T 0 0

                                                                                XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                                d3 thorn 77d2 269d 1438 frac14 0

                                                                                d frac14 467m

                                                                                Depth of penetration frac14 12d frac14 560m

                                                                                Lateral earth pressure 41

                                                                                Algebraic sum of forces for d frac14 467m isX

                                                                                F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                                T frac14 905 kN=m

                                                                                Force in each tie rod frac14 25T frac14 226 kN

                                                                                68

                                                                                (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                                0 frac14 21 98 frac14 112 kN=m3

                                                                                The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                                uC frac14 150

                                                                                165 15 98 frac14 134 kN=m2

                                                                                The average seepage pressure is

                                                                                j frac14 15

                                                                                165 98 frac14 09 kN=m3

                                                                                Hence

                                                                                0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                                0 j frac14 112 09 frac14 103 kN=m3

                                                                                Figure Q67

                                                                                42 Lateral earth pressure

                                                                                Consider moments about the anchor point A (per m)

                                                                                Force (kN) Arm (m) Moment (kN m)

                                                                                (1) 10 026 150 frac14 390 60 2340

                                                                                (2)1

                                                                                2 026 18 452 frac14 474 15 711

                                                                                (3) 026 18 45 105 frac14 2211 825 18240

                                                                                (4)1

                                                                                2 026 121 1052 frac14 1734 100 17340

                                                                                (5)1

                                                                                2 134 15 frac14 101 40 404

                                                                                (6) 134 30 frac14 402 60 2412

                                                                                (7)1

                                                                                2 134 60 frac14 402 95 3819

                                                                                571 4527(8) Ppm

                                                                                115 115PPm

                                                                                XM frac14 0

                                                                                Ppm frac144527

                                                                                115frac14 394 kN=m

                                                                                Available passive resistance

                                                                                Pp frac14 1

                                                                                2 385 103 62 frac14 714 kN=m

                                                                                Factor of safety

                                                                                Fp frac14 Pp

                                                                                Ppm

                                                                                frac14 714

                                                                                394frac14 18

                                                                                Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                                Figure Q68

                                                                                Lateral earth pressure 43

                                                                                (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                                Consider moments (per m) about the tie point A

                                                                                Force (kN) Arm (m)

                                                                                (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                                (2)1

                                                                                2 033 18 452 frac14 601 15

                                                                                (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                                (4)1

                                                                                2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                                (5)1

                                                                                2 134 15 frac14 101 40

                                                                                (6) 134 30 frac14 402 60

                                                                                (7)1

                                                                                2 134 d frac14 67d d3thorn 75

                                                                                (8) 1

                                                                                2 30 103 d2 frac141545d2 2d3thorn 75

                                                                                Moment (kN m)

                                                                                (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                                XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                                d3 thorn 827d2 466d 1518 frac14 0

                                                                                By trial

                                                                                d frac14 544m

                                                                                The minimum depth of embedment required is 544m

                                                                                69

                                                                                For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                                0 frac14 20 98 frac14 102 kN=m3

                                                                                The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                                44 Lateral earth pressure

                                                                                uC frac14 147

                                                                                173 26 98 frac14 216 kN=m2

                                                                                and the average seepage pressure around the wall is

                                                                                j frac14 26

                                                                                173 98 frac14 15 kN=m3

                                                                                Consider moments about the prop (A) (per m)

                                                                                Force (kN) Arm (m) Moment (kN m)

                                                                                (1)1

                                                                                2 03 17 272 frac14 186 020 37

                                                                                (2) 03 17 27 53 frac14 730 335 2445

                                                                                (3)1

                                                                                2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                                (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                                (5)1

                                                                                2 216 26 frac14 281 243 684

                                                                                (6) 216 27 frac14 583 465 2712

                                                                                (7)1

                                                                                2 216 60 frac14 648 800 5184

                                                                                3055(8)

                                                                                1

                                                                                2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                                Factor of safety

                                                                                Fr frac14 6885

                                                                                3055frac14 225

                                                                                Figure Q69

                                                                                Lateral earth pressure 45

                                                                                610

                                                                                For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                                p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                                Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                                Using the recommendations of Twine and Roscoe

                                                                                p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                                Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                                611

                                                                                frac14 18 kN=m3 0 frac14 34

                                                                                H frac14 350m nH frac14 335m mH frac14 185m

                                                                                Consider a trial value of F frac14 20 Refer to Figure 635

                                                                                0m frac14 tan1tan 34

                                                                                20

                                                                                frac14 186

                                                                                Then

                                                                                frac14 45 thorn 0m2frac14 543

                                                                                W frac14 1

                                                                                2 18 3502 cot 543 frac14 792 kN=m

                                                                                Figure Q610

                                                                                46 Lateral earth pressure

                                                                                P frac14 1

                                                                                2 s 3352 frac14 561s kN=m

                                                                                U frac14 1

                                                                                2 98 1852 cosec 543 frac14 206 kN=m

                                                                                Equations 630 and 631 then become

                                                                                561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                                792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                                ie

                                                                                561s 0616N 405 frac14 0

                                                                                792 0857N thorn 563 frac14 0

                                                                                N frac14 848

                                                                                0857frac14 989 kN=m

                                                                                Then

                                                                                561s 609 405 frac14 0

                                                                                s frac14 649

                                                                                561frac14 116 kN=m3

                                                                                The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                                F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                                20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                                s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                                Figure Q611

                                                                                Lateral earth pressure 47

                                                                                612

                                                                                For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                45 thorn 0

                                                                                2frac14 63

                                                                                For the retained material between the surface and a depth of 36m

                                                                                Pa frac14 1

                                                                                2 030 18 362 frac14 350 kN=m

                                                                                Weight of reinforced fill between the surface and a depth of 36m is

                                                                                Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                Eccentricity of Rv

                                                                                e frac14 263 250 frac14 013m

                                                                                The average vertical stress at a depth of 36m is

                                                                                z frac14 Rv

                                                                                L 2efrac14 324

                                                                                474frac14 68 kN=m2

                                                                                (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                Tensile stress in the element frac14 138 103

                                                                                65 3frac14 71N=mm2

                                                                                Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                The weight of ABC is

                                                                                W frac14 1

                                                                                2 18 52 265 frac14 124 kN=m

                                                                                From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                48 Lateral earth pressure

                                                                                (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                Tp frac14 032 68 120 065 frac14 170 kN

                                                                                Tr frac14 213 420

                                                                                418frac14 214 kN

                                                                                Again the tensile failure and slipping limit states are satisfied for this element

                                                                                Figure Q612

                                                                                Lateral earth pressure 49

                                                                                Chapter 7

                                                                                Consolidation theory

                                                                                71

                                                                                Total change in thickness

                                                                                H frac14 782 602 frac14 180mm

                                                                                Average thickness frac14 1530thorn 180

                                                                                2frac14 1620mm

                                                                                Length of drainage path d frac14 1620

                                                                                2frac14 810mm

                                                                                Root time plot (Figure Q71a)

                                                                                ffiffiffiffiffiffit90p frac14 33

                                                                                t90 frac14 109min

                                                                                cv frac14 0848d2

                                                                                t90frac14 0848 8102

                                                                                109 1440 365

                                                                                106frac14 27m2=year

                                                                                r0 frac14 782 764

                                                                                782 602frac14 018

                                                                                180frac14 0100

                                                                                rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                10 119

                                                                                9 180frac14 0735

                                                                                rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                Log time plot (Figure Q71b)

                                                                                t50 frac14 26min

                                                                                cv frac14 0196d2

                                                                                t50frac14 0196 8102

                                                                                26 1440 365

                                                                                106frac14 26m2=year

                                                                                r0 frac14 782 763

                                                                                782 602frac14 019

                                                                                180frac14 0106

                                                                                rp frac14 763 623

                                                                                782 602frac14 140

                                                                                180frac14 0778

                                                                                rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                Figure Q71(a)

                                                                                Figure Q71(b)

                                                                                Final void ratio

                                                                                e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                e

                                                                                Hfrac14 1thorn e0

                                                                                H0frac14 1thorn e1 thorne

                                                                                H0

                                                                                ie

                                                                                e

                                                                                180frac14 1631thorne

                                                                                1710

                                                                                e frac14 2936

                                                                                1530frac14 0192

                                                                                Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                mv frac14 1

                                                                                1thorn e0 e0 e101 00

                                                                                frac14 1

                                                                                1823 0192

                                                                                0107frac14 098m2=MN

                                                                                k frac14 cvmvw frac14 265 098 98

                                                                                60 1440 365 103frac14 81 1010 m=s

                                                                                72

                                                                                Using Equation 77 (one-dimensional method)

                                                                                sc frac14 e0 e11thorn e0 H

                                                                                Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                Figure Q72

                                                                                52 Consolidation theory

                                                                                Settlement

                                                                                Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                318

                                                                                Notes 5 92y 460thorn 84

                                                                                Heave

                                                                                Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                38

                                                                                73

                                                                                U frac14 f ethTvTHORN frac14 f cvt

                                                                                d2

                                                                                Hence if cv is constant

                                                                                t1

                                                                                t2frac14 d

                                                                                21

                                                                                d22

                                                                                where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                d1 frac14 95mm and d2 frac14 2500mm

                                                                                for U frac14 050 t2 frac14 t1 d22

                                                                                d21

                                                                                frac14 20

                                                                                60 24 365 25002

                                                                                952frac14 263 years

                                                                                for U lt 060 Tv frac14

                                                                                4U2 (Equation 724(a))

                                                                                t030 frac14 t050 0302

                                                                                0502

                                                                                frac14 263 036 frac14 095 years

                                                                                Consolidation theory 53

                                                                                74

                                                                                The layer is open

                                                                                d frac14 8

                                                                                2frac14 4m

                                                                                Tv frac14 cvtd2frac14 24 3

                                                                                42frac14 0450

                                                                                ui frac14 frac14 84 kN=m2

                                                                                The excess pore water pressure is given by Equation 721

                                                                                ue frac14Xmfrac141mfrac140

                                                                                2ui

                                                                                Msin

                                                                                Mz

                                                                                d

                                                                                expethM2TvTHORN

                                                                                In this case z frac14 d

                                                                                sinMz

                                                                                d

                                                                                frac14 sinM

                                                                                where

                                                                                M frac14

                                                                                23

                                                                                25

                                                                                2

                                                                                M sin M M2Tv exp (M2Tv)

                                                                                2thorn1 1110 0329

                                                                                3

                                                                                21 9993 457 105

                                                                                ue frac14 2 84 2

                                                                                1 0329 ethother terms negligibleTHORN

                                                                                frac14 352 kN=m2

                                                                                75

                                                                                The layer is open

                                                                                d frac14 6

                                                                                2frac14 3m

                                                                                Tv frac14 cvtd2frac14 10 3

                                                                                32frac14 0333

                                                                                The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                54 Consolidation theory

                                                                                For an open layer

                                                                                Tv frac14 4n

                                                                                m2

                                                                                n frac14 0333 62

                                                                                4frac14 300

                                                                                The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                i j

                                                                                0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                The initial and 3-year isochrones are plotted in Figure Q75

                                                                                Area under initial isochrone frac14 180 units

                                                                                Area under 3-year isochrone frac14 63 units

                                                                                The average degree of consolidation is given by Equation 725Thus

                                                                                U frac14 1 63

                                                                                180frac14 065

                                                                                Figure Q75

                                                                                Consolidation theory 55

                                                                                76

                                                                                At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                The final consolidation settlement (one-dimensional method) is

                                                                                sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                Corrected time t frac14 2 1

                                                                                2

                                                                                40

                                                                                52

                                                                                frac14 1615 years

                                                                                Tv frac14 cvtd2frac14 44 1615

                                                                                42frac14 0444

                                                                                From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                77

                                                                                The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                Figure Q77

                                                                                56 Consolidation theory

                                                                                Point m n Ir (kNm2) sc (mm)

                                                                                13020frac14 15 20

                                                                                20frac14 10 0194 (4) 113 124

                                                                                260

                                                                                20frac14 30

                                                                                20

                                                                                20frac14 10 0204 (2) 59 65

                                                                                360

                                                                                20frac14 30

                                                                                40

                                                                                20frac14 20 0238 (1) 35 38

                                                                                430

                                                                                20frac14 15

                                                                                40

                                                                                20frac14 20 0224 (2) 65 72

                                                                                Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                78

                                                                                Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                (a) Immediate settlement

                                                                                H

                                                                                Bfrac14 30

                                                                                35frac14 086

                                                                                D

                                                                                Bfrac14 2

                                                                                35frac14 006

                                                                                Figure Q78

                                                                                Consolidation theory 57

                                                                                From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                si frac14 130131qB

                                                                                Eufrac14 10 032 105 35

                                                                                40frac14 30mm

                                                                                (b) Consolidation settlement

                                                                                Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                3150

                                                                                Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                Now

                                                                                H

                                                                                Bfrac14 30

                                                                                35frac14 086 and A frac14 065

                                                                                from Figure 712 13 frac14 079

                                                                                sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                Total settlement

                                                                                s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                79

                                                                                Without sand drains

                                                                                Uv frac14 025

                                                                                Tv frac14 0049 ethfrom Figure 718THORN

                                                                                t frac14 Tvd2

                                                                                cvfrac14 0049 82

                                                                                cvWith sand drains

                                                                                R frac14 0564S frac14 0564 3 frac14 169m

                                                                                n frac14 Rrfrac14 169

                                                                                015frac14 113

                                                                                Tr frac14 cht

                                                                                4R2frac14 ch

                                                                                4 1692 0049 82

                                                                                cvethand ch frac14 cvTHORN

                                                                                frac14 0275

                                                                                Ur frac14 073 (from Figure 730)

                                                                                58 Consolidation theory

                                                                                Using Equation 740

                                                                                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                U frac14 080

                                                                                710

                                                                                Without sand drains

                                                                                Uv frac14 090

                                                                                Tv frac14 0848

                                                                                t frac14 Tvd2

                                                                                cvfrac14 0848 102

                                                                                96frac14 88 years

                                                                                With sand drains

                                                                                R frac14 0564S frac14 0564 4 frac14 226m

                                                                                n frac14 Rrfrac14 226

                                                                                015frac14 15

                                                                                Tr

                                                                                Tvfrac14 chcv

                                                                                d2

                                                                                4R2ethsame tTHORN

                                                                                Tr

                                                                                Tvfrac14 140

                                                                                96 102

                                                                                4 2262frac14 714 eth1THORN

                                                                                Using Equation 740

                                                                                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                Thus

                                                                                Uv frac14 0295 and Ur frac14 086

                                                                                t frac14 88 00683

                                                                                0848frac14 07 years

                                                                                Consolidation theory 59

                                                                                Chapter 8

                                                                                Bearing capacity

                                                                                81

                                                                                (a) The ultimate bearing capacity is given by Equation 83

                                                                                qf frac14 cNc thorn DNq thorn 1

                                                                                2BN

                                                                                For u frac14 0

                                                                                Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                The net ultimate bearing capacity is

                                                                                qnf frac14 qf D frac14 540 kN=m2

                                                                                The net foundation pressure is

                                                                                qn frac14 q D frac14 425

                                                                                2 eth21 1THORN frac14 192 kN=m2

                                                                                The factor of safety (Equation 86) is

                                                                                F frac14 qnfqnfrac14 540

                                                                                192frac14 28

                                                                                (b) For 0 frac14 28

                                                                                Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                2 112 2 13

                                                                                frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                qnf frac14 574 112 frac14 563 kN=m2

                                                                                F frac14 563

                                                                                192frac14 29

                                                                                (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                82

                                                                                For 0 frac14 38

                                                                                Nq frac14 49 N frac14 67

                                                                                qnf frac14 DethNq 1THORN thorn 1

                                                                                2BN ethfrom Equation 83THORN

                                                                                frac14 eth18 075 48THORN thorn 1

                                                                                2 18 15 67

                                                                                frac14 648thorn 905 frac14 1553 kN=m2

                                                                                qn frac14 500

                                                                                15 eth18 075THORN frac14 320 kN=m2

                                                                                F frac14 qnfqnfrac14 1553

                                                                                320frac14 48

                                                                                0d frac14 tan1tan 38

                                                                                125

                                                                                frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                2 18 15 25

                                                                                frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                Design load (action) Vd frac14 500 kN=m

                                                                                The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                83

                                                                                D

                                                                                Bfrac14 350

                                                                                225frac14 155

                                                                                From Figure 85 for a square foundation

                                                                                Nc frac14 81

                                                                                Bearing capacity 61

                                                                                For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                Nc frac14 084thorn 016B

                                                                                L

                                                                                81 frac14 745

                                                                                Using Equation 810

                                                                                qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                For F frac14 3

                                                                                qn frac14 1006

                                                                                3frac14 335 kN=m2

                                                                                q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                Design load frac14 405 450 225 frac14 4100 kN

                                                                                Design undrained strength cud frac14 135

                                                                                14frac14 96 kN=m2

                                                                                Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                frac14 7241 kN

                                                                                Design load Vd frac14 4100 kN

                                                                                Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                84

                                                                                For 0 frac14 40

                                                                                Nq frac14 64 N frac14 95

                                                                                qnf frac14 DethNq 1THORN thorn 04BN

                                                                                (a) Water table 5m below ground level

                                                                                qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                qn frac14 400 17 frac14 383 kN=m2

                                                                                F frac14 2686

                                                                                383frac14 70

                                                                                (b) Water table 1m below ground level (ie at foundation level)

                                                                                0 frac14 20 98 frac14 102 kN=m3

                                                                                62 Bearing capacity

                                                                                qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                F frac14 2040

                                                                                383frac14 53

                                                                                (c) Water table at ground level with upward hydraulic gradient 02

                                                                                eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                F frac14 1296

                                                                                392frac14 33

                                                                                85

                                                                                The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                Design value of 0 frac14 tan1tan 39

                                                                                125

                                                                                frac14 33

                                                                                For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                86

                                                                                (a) Undrained shear for u frac14 0

                                                                                Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                qnf frac14 12cuNc

                                                                                frac14 12 100 514 frac14 617 kN=m2

                                                                                qn frac14 qnfFfrac14 617

                                                                                3frac14 206 kN=m2

                                                                                q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                Bearing capacity 63

                                                                                Drained shear for 0 frac14 32

                                                                                Nq frac14 23 N frac14 25

                                                                                0 frac14 21 98 frac14 112 kN=m3

                                                                                qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                frac14 694 kN=m2

                                                                                q frac14 694

                                                                                3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                Design load frac14 42 227 frac14 3632 kN

                                                                                (b) Design undrained strength cud frac14 100

                                                                                14frac14 71 kNm2

                                                                                Design bearing resistance Rd frac14 12cudNe area

                                                                                frac14 12 71 514 42

                                                                                frac14 7007 kN

                                                                                For drained shear 0d frac14 tan1tan 32

                                                                                125

                                                                                frac14 26

                                                                                Nq frac14 12 N frac14 10

                                                                                Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                1 2 100 0175 0700qn 0182qn

                                                                                2 6 033 0044 0176qn 0046qn

                                                                                3 10 020 0017 0068qn 0018qn

                                                                                0246qn

                                                                                Diameter of equivalent circle B frac14 45m

                                                                                H

                                                                                Bfrac14 12

                                                                                45frac14 27 and A frac14 042

                                                                                13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                64 Bearing capacity

                                                                                For sc frac14 30mm

                                                                                qn frac14 30

                                                                                0147frac14 204 kN=m2

                                                                                q frac14 204thorn 21 frac14 225 kN=m2

                                                                                Design load frac14 42 225 frac14 3600 kN

                                                                                The design load is 3600 kN settlement being the limiting criterion

                                                                                87

                                                                                D

                                                                                Bfrac14 8

                                                                                4frac14 20

                                                                                From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                F frac14 cuNc

                                                                                Dfrac14 40 71

                                                                                20 8frac14 18

                                                                                88

                                                                                Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                Design value of 0 frac14 tan1tan 38

                                                                                125

                                                                                frac14 32

                                                                                Figure Q86

                                                                                Bearing capacity 65

                                                                                For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                For B frac14 250m qn frac14 3750

                                                                                2502 17 frac14 583 kN=m2

                                                                                From Figure 510 m frac14 n frac14 126

                                                                                6frac14 021

                                                                                Ir frac14 0019

                                                                                Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                89

                                                                                Depth (m) N 0v (kNm2) CN N1

                                                                                070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                Cw frac14 05thorn 0530

                                                                                47

                                                                                frac14 082

                                                                                66 Bearing capacity

                                                                                Thus

                                                                                qa frac14 150 082 frac14 120 kN=m2

                                                                                (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                Thus

                                                                                qa frac14 90 15 frac14 135 kN=m2

                                                                                (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                Ic frac14 171

                                                                                1014frac14 0068

                                                                                From Equation 819(a) with s frac14 25mm

                                                                                q frac14 25

                                                                                3507 0068frac14 150 kN=m2

                                                                                810

                                                                                Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                Izp frac14 05thorn 01130

                                                                                16 27

                                                                                05

                                                                                frac14 067

                                                                                Refer to Figure Q810

                                                                                E frac14 25qc

                                                                                Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                Ez (mm3MN)

                                                                                1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                0203

                                                                                C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                130frac14 093

                                                                                C2 frac14 1 ethsayTHORN

                                                                                s frac14 C1C2qnX Iz

                                                                                Ez frac14 093 1 130 0203 frac14 25mm

                                                                                Bearing capacity 67

                                                                                811

                                                                                At pile base level

                                                                                cu frac14 220 kN=m2

                                                                                qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                Then

                                                                                Qf frac14 Abqb thorn Asqs

                                                                                frac14

                                                                                4 32 1980

                                                                                thorn eth 105 139 86THORN

                                                                                frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                0 01 02 03 04 05 06 07

                                                                                0 2 4 6 8 10 12 14

                                                                                1

                                                                                2

                                                                                3

                                                                                4

                                                                                5

                                                                                6

                                                                                7

                                                                                8

                                                                                (1)

                                                                                (2)

                                                                                (3)

                                                                                (4)

                                                                                (5)

                                                                                qc

                                                                                qc

                                                                                Iz

                                                                                Iz

                                                                                (MNm2)

                                                                                z (m)

                                                                                Figure Q810

                                                                                68 Bearing capacity

                                                                                Allowable load

                                                                                ethaTHORN Qf

                                                                                2frac14 17 937

                                                                                2frac14 8968 kN

                                                                                ethbTHORN Abqb

                                                                                3thorn Asqs frac14 13 996

                                                                                3thorn 3941 frac14 8606 kN

                                                                                ie allowable load frac14 8600 kN

                                                                                Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                According to the limit state method

                                                                                Characteristic undrained strength at base level cuk frac14 220

                                                                                150kN=m2

                                                                                Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                150frac14 1320 kN=m2

                                                                                Characteristic shaft resistance qsk frac14 00150

                                                                                frac14 86

                                                                                150frac14 57 kN=m2

                                                                                Characteristic base and shaft resistances

                                                                                Rbk frac14

                                                                                4 32 1320 frac14 9330 kN

                                                                                Rsk frac14 105 139 86

                                                                                150frac14 2629 kN

                                                                                For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                Design bearing resistance Rcd frac14 9330

                                                                                160thorn 2629

                                                                                130

                                                                                frac14 5831thorn 2022

                                                                                frac14 7850 kN

                                                                                Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                812

                                                                                ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                For a single pile

                                                                                Qf frac14 Abqb thorn Asqs

                                                                                frac14

                                                                                4 062 1305

                                                                                thorn eth 06 15 42THORN

                                                                                frac14 369thorn 1187 frac14 1556 kN

                                                                                Bearing capacity 69

                                                                                Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                qbkfrac14 9cuk frac14 9 220

                                                                                150frac14 1320 kN=m2

                                                                                qskfrac14cuk frac14 040 105

                                                                                150frac14 28 kN=m2

                                                                                Rbkfrac14

                                                                                4 0602 1320 frac14 373 kN

                                                                                Rskfrac14 060 15 28 frac14 791 kN

                                                                                Rcdfrac14 373

                                                                                160thorn 791

                                                                                130frac14 233thorn 608 frac14 841 kN

                                                                                Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                q frac14 21 000

                                                                                1762frac14 68 kN=m2

                                                                                Immediate settlement

                                                                                H

                                                                                Bfrac14 15

                                                                                176frac14 085

                                                                                D

                                                                                Bfrac14 13

                                                                                176frac14 074

                                                                                L

                                                                                Bfrac14 1

                                                                                Hence from Figure 515

                                                                                130 frac14 078 and 131 frac14 041

                                                                                70 Bearing capacity

                                                                                Thus using Equation 528

                                                                                si frac14 078 041 68 176

                                                                                65frac14 6mm

                                                                                Consolidation settlement

                                                                                Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                434 (sod)

                                                                                Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                sc frac14 056 434 frac14 24mm

                                                                                The total settlement is (6thorn 24) frac14 30mm

                                                                                813

                                                                                At base level N frac14 26 Then using Equation 830

                                                                                qb frac14 40NDb

                                                                                Bfrac14 40 26 2

                                                                                025frac14 8320 kN=m2

                                                                                ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                Figure Q812

                                                                                Bearing capacity 71

                                                                                Over the length embedded in sand

                                                                                N frac14 21 ie18thorn 24

                                                                                2

                                                                                Using Equation 831

                                                                                qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                For a single pile

                                                                                Qf frac14 Abqb thorn Asqs

                                                                                frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                For the pile group assuming a group efficiency of 12

                                                                                XQf frac14 12 9 604 frac14 6523 kN

                                                                                Then the load factor is

                                                                                F frac14 6523

                                                                                2000thorn 1000frac14 21

                                                                                (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                Characteristic base resistance per unit area qbk frac14 8320

                                                                                150frac14 5547 kNm2

                                                                                Characteristic shaft resistance per unit area qsk frac14 42

                                                                                150frac14 28 kNm2

                                                                                Characteristic base and shaft resistances for a single pile

                                                                                Rbk frac14 0252 5547 frac14 347 kN

                                                                                Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                For a driven pile the partial factors are b frac14 s frac14 130

                                                                                Design bearing resistance Rcd frac14 347

                                                                                130thorn 56

                                                                                130frac14 310 kN

                                                                                For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                72 Bearing capacity

                                                                                N frac14 24thorn 26thorn 34

                                                                                3frac14 28

                                                                                Ic frac14 171

                                                                                2814frac14 0016 ethEquation 818THORN

                                                                                s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                814

                                                                                Using Equation 841

                                                                                Tf frac14 DLcu thorn

                                                                                4ethD2 d2THORNcuNc

                                                                                frac14 eth 02 5 06 110THORN thorn

                                                                                4eth022 012THORN110 9

                                                                                frac14 207thorn 23 frac14 230 kN

                                                                                Figure Q813

                                                                                Bearing capacity 73

                                                                                Chapter 9

                                                                                Stability of slopes

                                                                                91

                                                                                Referring to Figure Q91

                                                                                W frac14 417 19 frac14 792 kN=m

                                                                                Q frac14 20 28 frac14 56 kN=m

                                                                                Arc lengthAB frac14

                                                                                180 73 90 frac14 115m

                                                                                Arc length BC frac14

                                                                                180 28 90 frac14 44m

                                                                                The factor of safety is given by

                                                                                F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                Depth of tension crack z0 frac14 2cu

                                                                                frac14 2 20

                                                                                19frac14 21m

                                                                                Arc length BD frac14

                                                                                180 13

                                                                                1

                                                                                2 90 frac14 21m

                                                                                F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                92

                                                                                u frac14 0

                                                                                Depth factor D frac14 11

                                                                                9frac14 122

                                                                                Using Equation 92 with F frac14 10

                                                                                Ns frac14 cu

                                                                                FHfrac14 30

                                                                                10 19 9frac14 0175

                                                                                Hence from Figure 93

                                                                                frac14 50

                                                                                For F frac14 12

                                                                                Ns frac14 30

                                                                                12 19 9frac14 0146

                                                                                frac14 27

                                                                                93

                                                                                Refer to Figure Q93

                                                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                74 m

                                                                                214 1deg

                                                                                213 1deg

                                                                                39 m

                                                                                WB

                                                                                D

                                                                                C

                                                                                28 m

                                                                                21 m

                                                                                A

                                                                                Q

                                                                                Soil (1)Soil (2)

                                                                                73deg

                                                                                Figure Q91

                                                                                Stability of slopes 75

                                                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                599 256 328 1372

                                                                                Figure Q93

                                                                                76 Stability of slopes

                                                                                XW cos frac14 b

                                                                                Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                W sin frac14 bX

                                                                                h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                Arc length La frac14

                                                                                180 57

                                                                                1

                                                                                2 326 frac14 327m

                                                                                The factor of safety is given by

                                                                                F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                W sin

                                                                                frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                frac14 091

                                                                                According to the limit state method

                                                                                0d frac14 tan1tan 32

                                                                                125

                                                                                frac14 265

                                                                                c0 frac14 8

                                                                                160frac14 5 kN=m2

                                                                                Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                Design disturbing moment frac14 1075 kN=m

                                                                                The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                94

                                                                                F frac14 1

                                                                                W sin

                                                                                Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                1thorn ethtan tan0=FTHORN

                                                                                c0 frac14 8 kN=m2

                                                                                0 frac14 32

                                                                                c0b frac14 8 2 frac14 16 kN=m

                                                                                W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                Try F frac14 100

                                                                                tan0

                                                                                Ffrac14 0625

                                                                                Stability of slopes 77

                                                                                Values of u are as obtained in Figure Q93

                                                                                SliceNo

                                                                                h(m)

                                                                                W frac14 bh(kNm)

                                                                                W sin(kNm)

                                                                                ub(kNm)

                                                                                c0bthorn (W ub) tan0(kNm)

                                                                                sec

                                                                                1thorn (tan tan0)FProduct(kNm)

                                                                                1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                23 33 30 1042 31

                                                                                3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                224 92 72 0931 67

                                                                                6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                12 58 112 90 0889 80

                                                                                8 60 252 1812

                                                                                80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                2154 88 116 0853 99

                                                                                14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                1074 1091

                                                                                F frac14 1091

                                                                                1074frac14 102 (assumed value 100)

                                                                                Thus

                                                                                F frac14 101

                                                                                95

                                                                                F frac14 1

                                                                                W sin

                                                                                XfWeth1 ruTHORN tan0g sec

                                                                                1thorn ethtan tan0THORN=F

                                                                                0 frac14 33

                                                                                ru frac14 020

                                                                                W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                Try F frac14 110

                                                                                tan 0

                                                                                Ffrac14 tan 33

                                                                                110frac14 0590

                                                                                78 Stability of slopes

                                                                                Referring to Figure Q95

                                                                                SliceNo

                                                                                h(m)

                                                                                W frac14 bh(kNm)

                                                                                W sin(kNm)

                                                                                W(1 ru) tan0(kNm)

                                                                                sec

                                                                                1thorn ( tan tan0)FProduct(kNm)

                                                                                1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                2120 234 0892 209

                                                                                4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                1185 1271

                                                                                Figure Q95

                                                                                Stability of slopes 79

                                                                                F frac14 1271

                                                                                1185frac14 107

                                                                                The trial value was 110 therefore take F to be 108

                                                                                96

                                                                                (a) Water table at surface the factor of safety is given by Equation 912

                                                                                F frac14 0

                                                                                sat

                                                                                tan0

                                                                                tan

                                                                                ptie 15 frac14 92

                                                                                19

                                                                                tan 36

                                                                                tan

                                                                                tan frac14 0234

                                                                                frac14 13

                                                                                Water table well below surface the factor of safety is given by Equation 911

                                                                                F frac14 tan0

                                                                                tan

                                                                                frac14 tan 36

                                                                                tan 13

                                                                                frac14 31

                                                                                (b) 0d frac14 tan1tan 36

                                                                                125

                                                                                frac14 30

                                                                                Depth of potential failure surface frac14 z

                                                                                Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                frac14 504z kN

                                                                                Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                frac14 19 z sin 13 cos 13

                                                                                frac14 416z kN

                                                                                Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                80 Stability of slopes

                                                                                • Book Cover
                                                                                • Title
                                                                                • Contents
                                                                                • Basic characteristics of soils
                                                                                • Seepage
                                                                                • Effective stress
                                                                                • Shear strength
                                                                                • Stresses and displacements
                                                                                • Lateral earth pressure
                                                                                • Consolidation theory
                                                                                • Bearing capacity
                                                                                • Stability of slopes

                                                                                  Distribution of active pressure (plotted in Figure Q62)

                                                                                  Depth (m) Soil Active pressure (kNm2)

                                                                                  3 1 0271 16 3 frac14 1305 1 (0271 16 3)thorn (0271 92 2) frac14 130thorn 50 frac14 1805 2 f(16 3)thorn (92 2)g 0375 (2 17 0613)frac14 249 209 frac14 408 2 40thorn (0375 102 3) frac14 40thorn 115 frac14 1558 3 f(16 3)thorn (92 2)thorn (102 3)g 0198 frac14 192

                                                                                  12 3 192thorn (0198 112 4) frac14 192thorn 89 frac14 281

                                                                                  At a depth of 12m the hydrostatic pressure frac14 98 9 frac14 882 kNm2 Calculation oftotal thrust and its point of application (forces are numbered as in Figure Q62 andmoments are taken about the top of the wall) per m

                                                                                  Total thrust frac14 571 kNm

                                                                                  Point of application is (4893571) m from the top of the wall ie 857m

                                                                                  Force (kN) Arm (m) Moment (kN m)

                                                                                  (1)1

                                                                                  2 0271 16 32 frac14 195 20 390

                                                                                  (2) 0271 16 3 2 frac14 260 40 1040

                                                                                  (3)1

                                                                                  2 0271 92 22 frac14 50 433 217

                                                                                  (4) [0375 f(16 3)thorn (92 2)g f2 17 0613g] 3 frac14 122 65 793

                                                                                  (5)1

                                                                                  2 0375 102 32 frac14 172 70 1204

                                                                                  (6) [0198 f(16 3)thorn (92 2)thorn (102 3)g] 4 frac14 768 100 7680

                                                                                  (7)1

                                                                                  2 0198 112 42 frac14 177 1067 1889

                                                                                  (8)1

                                                                                  2 98 92 frac14 3969 90 35721

                                                                                  5713 48934

                                                                                  Figure Q62

                                                                                  Lateral earth pressure 35

                                                                                  63

                                                                                  (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                                                  Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                                                  frac14 245

                                                                                  At the lower end of the piling

                                                                                  pa frac14 Kaqthorn Kasatz Kaccu

                                                                                  frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                                                  frac14 115 kN=m2

                                                                                  pp frac14 Kpsatzthorn Kpccu

                                                                                  frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                                                  frac14 202 kN=m2

                                                                                  (b) For 0 frac14 26 and frac14 1

                                                                                  20

                                                                                  Ka frac14 035

                                                                                  Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                                                  pfrac14 145 ethEquation 619THORN

                                                                                  Kp frac14 37

                                                                                  Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                                                  pfrac14 47 ethEquation 624THORN

                                                                                  At the lower end of the piling

                                                                                  pa frac14 Kaqthorn Ka0z Kacc

                                                                                  0

                                                                                  frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                                                  frac14 187 kN=m2

                                                                                  pp frac14 Kp0zthorn Kpcc

                                                                                  0

                                                                                  frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                                                  frac14 198 kN=m2

                                                                                  36 Lateral earth pressure

                                                                                  64

                                                                                  (a) For 0 frac14 38 Ka frac14 024

                                                                                  0 frac14 20 98 frac14 102 kN=m3

                                                                                  The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                                                  Force (kN) Arm (m) Moment (kN m)

                                                                                  (1) 024 10 66 frac14 159 33 525

                                                                                  (2)1

                                                                                  2 024 17 392 frac14 310 400 1240

                                                                                  (3) 024 17 39 27 frac14 430 135 580

                                                                                  (4)1

                                                                                  2 024 102 272 frac14 89 090 80

                                                                                  (5)1

                                                                                  2 98 272 frac14 357 090 321

                                                                                  Hfrac14 1345 MH frac14 2746

                                                                                  (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                                                  (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                                                  XM frac14MV MH frac14 7790 kNm

                                                                                  Lever arm of base resultant

                                                                                  M

                                                                                  Vfrac14 779

                                                                                  488frac14 160

                                                                                  Eccentricity of base resultant

                                                                                  e frac14 200 160 frac14 040m

                                                                                  39 m

                                                                                  27 m

                                                                                  40 m

                                                                                  04 m

                                                                                  04 m

                                                                                  26 m

                                                                                  (7)

                                                                                  (9)

                                                                                  (1)(2)

                                                                                  (3)

                                                                                  (4)

                                                                                  (5)

                                                                                  (8)(6)

                                                                                  (10)

                                                                                  WT

                                                                                  10 kNm2

                                                                                  Hydrostatic

                                                                                  Figure Q64

                                                                                  Lateral earth pressure 37

                                                                                  Base pressures (Equation 627)

                                                                                  p frac14 VB

                                                                                  1 6e

                                                                                  B

                                                                                  frac14 488

                                                                                  4eth1 060THORN

                                                                                  frac14 195 kN=m2 and 49 kN=m2

                                                                                  Factor of safety against sliding (Equation 628)

                                                                                  F frac14 V tan

                                                                                  Hfrac14 488 tan 25

                                                                                  1345frac14 17

                                                                                  (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                                                  Hfrac14 1633 kN

                                                                                  V frac14 4879 kN

                                                                                  MH frac14 3453 kNm

                                                                                  MV frac14 10536 kNm

                                                                                  The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                                                  65

                                                                                  For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                                  Kp

                                                                                  Ffrac14 385

                                                                                  2

                                                                                  0 frac14 20 98 frac14 102 kN=m3

                                                                                  The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                                                  Force (kN) Arm (m) Moment (kN m)

                                                                                  (1)1

                                                                                  2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                                                  (2) 026 17 45 d frac14 199d d2 995d2

                                                                                  (3)1

                                                                                  2 026 102 d2 frac14 133d2 d3 044d3

                                                                                  (4)1

                                                                                  2 385

                                                                                  2 17 152 frac14 368 dthorn 05 368d 184

                                                                                  (5)385

                                                                                  2 17 15 d frac14 491d d2 2455d2

                                                                                  (6)1

                                                                                  2 385

                                                                                  2 102 d2 frac14 982d2 d3 327d3

                                                                                  38 Lateral earth pressure

                                                                                  XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                                                  d3 thorn 516d2 283d 1724 frac14 0

                                                                                  d frac14 179m

                                                                                  Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                                                  Over additional 20 embedded depth

                                                                                  pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                                                  Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                                                  66

                                                                                  The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                                                  Ka frac14 sin 69=sin 105

                                                                                  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                                                  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                                                  pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                                                  26664

                                                                                  37775

                                                                                  2

                                                                                  frac14 050

                                                                                  The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                                                  Pa frac14 1

                                                                                  2 050 19 7502 frac14 267 kN=m

                                                                                  Figure Q65

                                                                                  Lateral earth pressure 39

                                                                                  Horizontal component

                                                                                  Ph frac14 267 cos 40 frac14 205 kN=m

                                                                                  Vertical component

                                                                                  Pv frac14 267 sin 40 frac14 172 kN=m

                                                                                  Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                                  Force (kN) Arm (m) Moment (kN m)

                                                                                  (1)1

                                                                                  2 175 650 235 frac14 1337 258 345

                                                                                  (2) 050 650 235 frac14 764 175 134

                                                                                  (3)1

                                                                                  2 070 650 235 frac14 535 127 68

                                                                                  (4) 100 400 235 frac14 940 200 188

                                                                                  (5) 1

                                                                                  2 080 050 235 frac14 47 027 1

                                                                                  Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                                  Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                                  Lever arm of base resultant

                                                                                  M

                                                                                  Vfrac14 795

                                                                                  525frac14 151m

                                                                                  Eccentricity of base resultant

                                                                                  e frac14 200 151 frac14 049m

                                                                                  Figure Q66

                                                                                  40 Lateral earth pressure

                                                                                  Base pressures (Equation 627)

                                                                                  p frac14 525

                                                                                  41 6 049

                                                                                  4

                                                                                  frac14 228 kN=m2 and 35 kN=m2

                                                                                  The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                                  The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                                  The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                                  67

                                                                                  For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                                  Force (kN) Arm (m) Moment (kNm)

                                                                                  (1)1

                                                                                  2 027 17 52 frac14 574 183 1050

                                                                                  (2) 027 17 5 3 frac14 689 500 3445

                                                                                  (3)1

                                                                                  2 027 102 32 frac14 124 550 682

                                                                                  (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                                  (5)1

                                                                                  2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                                  (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                                  (7) 1

                                                                                  2 267

                                                                                  2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                                  (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                                  2 d frac14 163d d2thorn 650 82d2 1060d

                                                                                  Tie rod force per m frac14 T 0 0

                                                                                  XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                                  d3 thorn 77d2 269d 1438 frac14 0

                                                                                  d frac14 467m

                                                                                  Depth of penetration frac14 12d frac14 560m

                                                                                  Lateral earth pressure 41

                                                                                  Algebraic sum of forces for d frac14 467m isX

                                                                                  F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                                  T frac14 905 kN=m

                                                                                  Force in each tie rod frac14 25T frac14 226 kN

                                                                                  68

                                                                                  (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                                  0 frac14 21 98 frac14 112 kN=m3

                                                                                  The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                                  uC frac14 150

                                                                                  165 15 98 frac14 134 kN=m2

                                                                                  The average seepage pressure is

                                                                                  j frac14 15

                                                                                  165 98 frac14 09 kN=m3

                                                                                  Hence

                                                                                  0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                                  0 j frac14 112 09 frac14 103 kN=m3

                                                                                  Figure Q67

                                                                                  42 Lateral earth pressure

                                                                                  Consider moments about the anchor point A (per m)

                                                                                  Force (kN) Arm (m) Moment (kN m)

                                                                                  (1) 10 026 150 frac14 390 60 2340

                                                                                  (2)1

                                                                                  2 026 18 452 frac14 474 15 711

                                                                                  (3) 026 18 45 105 frac14 2211 825 18240

                                                                                  (4)1

                                                                                  2 026 121 1052 frac14 1734 100 17340

                                                                                  (5)1

                                                                                  2 134 15 frac14 101 40 404

                                                                                  (6) 134 30 frac14 402 60 2412

                                                                                  (7)1

                                                                                  2 134 60 frac14 402 95 3819

                                                                                  571 4527(8) Ppm

                                                                                  115 115PPm

                                                                                  XM frac14 0

                                                                                  Ppm frac144527

                                                                                  115frac14 394 kN=m

                                                                                  Available passive resistance

                                                                                  Pp frac14 1

                                                                                  2 385 103 62 frac14 714 kN=m

                                                                                  Factor of safety

                                                                                  Fp frac14 Pp

                                                                                  Ppm

                                                                                  frac14 714

                                                                                  394frac14 18

                                                                                  Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                                  Figure Q68

                                                                                  Lateral earth pressure 43

                                                                                  (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                                  Consider moments (per m) about the tie point A

                                                                                  Force (kN) Arm (m)

                                                                                  (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                                  (2)1

                                                                                  2 033 18 452 frac14 601 15

                                                                                  (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                                  (4)1

                                                                                  2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                                  (5)1

                                                                                  2 134 15 frac14 101 40

                                                                                  (6) 134 30 frac14 402 60

                                                                                  (7)1

                                                                                  2 134 d frac14 67d d3thorn 75

                                                                                  (8) 1

                                                                                  2 30 103 d2 frac141545d2 2d3thorn 75

                                                                                  Moment (kN m)

                                                                                  (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                                  XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                                  d3 thorn 827d2 466d 1518 frac14 0

                                                                                  By trial

                                                                                  d frac14 544m

                                                                                  The minimum depth of embedment required is 544m

                                                                                  69

                                                                                  For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                                  0 frac14 20 98 frac14 102 kN=m3

                                                                                  The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                                  44 Lateral earth pressure

                                                                                  uC frac14 147

                                                                                  173 26 98 frac14 216 kN=m2

                                                                                  and the average seepage pressure around the wall is

                                                                                  j frac14 26

                                                                                  173 98 frac14 15 kN=m3

                                                                                  Consider moments about the prop (A) (per m)

                                                                                  Force (kN) Arm (m) Moment (kN m)

                                                                                  (1)1

                                                                                  2 03 17 272 frac14 186 020 37

                                                                                  (2) 03 17 27 53 frac14 730 335 2445

                                                                                  (3)1

                                                                                  2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                                  (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                                  (5)1

                                                                                  2 216 26 frac14 281 243 684

                                                                                  (6) 216 27 frac14 583 465 2712

                                                                                  (7)1

                                                                                  2 216 60 frac14 648 800 5184

                                                                                  3055(8)

                                                                                  1

                                                                                  2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                                  Factor of safety

                                                                                  Fr frac14 6885

                                                                                  3055frac14 225

                                                                                  Figure Q69

                                                                                  Lateral earth pressure 45

                                                                                  610

                                                                                  For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                                  p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                                  Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                                  Using the recommendations of Twine and Roscoe

                                                                                  p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                                  Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                                  611

                                                                                  frac14 18 kN=m3 0 frac14 34

                                                                                  H frac14 350m nH frac14 335m mH frac14 185m

                                                                                  Consider a trial value of F frac14 20 Refer to Figure 635

                                                                                  0m frac14 tan1tan 34

                                                                                  20

                                                                                  frac14 186

                                                                                  Then

                                                                                  frac14 45 thorn 0m2frac14 543

                                                                                  W frac14 1

                                                                                  2 18 3502 cot 543 frac14 792 kN=m

                                                                                  Figure Q610

                                                                                  46 Lateral earth pressure

                                                                                  P frac14 1

                                                                                  2 s 3352 frac14 561s kN=m

                                                                                  U frac14 1

                                                                                  2 98 1852 cosec 543 frac14 206 kN=m

                                                                                  Equations 630 and 631 then become

                                                                                  561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                                  792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                                  ie

                                                                                  561s 0616N 405 frac14 0

                                                                                  792 0857N thorn 563 frac14 0

                                                                                  N frac14 848

                                                                                  0857frac14 989 kN=m

                                                                                  Then

                                                                                  561s 609 405 frac14 0

                                                                                  s frac14 649

                                                                                  561frac14 116 kN=m3

                                                                                  The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                                  F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                                  20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                                  s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                                  Figure Q611

                                                                                  Lateral earth pressure 47

                                                                                  612

                                                                                  For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                  45 thorn 0

                                                                                  2frac14 63

                                                                                  For the retained material between the surface and a depth of 36m

                                                                                  Pa frac14 1

                                                                                  2 030 18 362 frac14 350 kN=m

                                                                                  Weight of reinforced fill between the surface and a depth of 36m is

                                                                                  Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                  eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                  Eccentricity of Rv

                                                                                  e frac14 263 250 frac14 013m

                                                                                  The average vertical stress at a depth of 36m is

                                                                                  z frac14 Rv

                                                                                  L 2efrac14 324

                                                                                  474frac14 68 kN=m2

                                                                                  (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                  Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                  Tensile stress in the element frac14 138 103

                                                                                  65 3frac14 71N=mm2

                                                                                  Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                  Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                  Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                  The weight of ABC is

                                                                                  W frac14 1

                                                                                  2 18 52 265 frac14 124 kN=m

                                                                                  From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                  48 Lateral earth pressure

                                                                                  (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                  Tp frac14 032 68 120 065 frac14 170 kN

                                                                                  Tr frac14 213 420

                                                                                  418frac14 214 kN

                                                                                  Again the tensile failure and slipping limit states are satisfied for this element

                                                                                  Figure Q612

                                                                                  Lateral earth pressure 49

                                                                                  Chapter 7

                                                                                  Consolidation theory

                                                                                  71

                                                                                  Total change in thickness

                                                                                  H frac14 782 602 frac14 180mm

                                                                                  Average thickness frac14 1530thorn 180

                                                                                  2frac14 1620mm

                                                                                  Length of drainage path d frac14 1620

                                                                                  2frac14 810mm

                                                                                  Root time plot (Figure Q71a)

                                                                                  ffiffiffiffiffiffit90p frac14 33

                                                                                  t90 frac14 109min

                                                                                  cv frac14 0848d2

                                                                                  t90frac14 0848 8102

                                                                                  109 1440 365

                                                                                  106frac14 27m2=year

                                                                                  r0 frac14 782 764

                                                                                  782 602frac14 018

                                                                                  180frac14 0100

                                                                                  rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                  10 119

                                                                                  9 180frac14 0735

                                                                                  rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                  Log time plot (Figure Q71b)

                                                                                  t50 frac14 26min

                                                                                  cv frac14 0196d2

                                                                                  t50frac14 0196 8102

                                                                                  26 1440 365

                                                                                  106frac14 26m2=year

                                                                                  r0 frac14 782 763

                                                                                  782 602frac14 019

                                                                                  180frac14 0106

                                                                                  rp frac14 763 623

                                                                                  782 602frac14 140

                                                                                  180frac14 0778

                                                                                  rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                  Figure Q71(a)

                                                                                  Figure Q71(b)

                                                                                  Final void ratio

                                                                                  e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                  e

                                                                                  Hfrac14 1thorn e0

                                                                                  H0frac14 1thorn e1 thorne

                                                                                  H0

                                                                                  ie

                                                                                  e

                                                                                  180frac14 1631thorne

                                                                                  1710

                                                                                  e frac14 2936

                                                                                  1530frac14 0192

                                                                                  Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                  mv frac14 1

                                                                                  1thorn e0 e0 e101 00

                                                                                  frac14 1

                                                                                  1823 0192

                                                                                  0107frac14 098m2=MN

                                                                                  k frac14 cvmvw frac14 265 098 98

                                                                                  60 1440 365 103frac14 81 1010 m=s

                                                                                  72

                                                                                  Using Equation 77 (one-dimensional method)

                                                                                  sc frac14 e0 e11thorn e0 H

                                                                                  Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                  Figure Q72

                                                                                  52 Consolidation theory

                                                                                  Settlement

                                                                                  Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                  1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                  318

                                                                                  Notes 5 92y 460thorn 84

                                                                                  Heave

                                                                                  Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                  1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                  38

                                                                                  73

                                                                                  U frac14 f ethTvTHORN frac14 f cvt

                                                                                  d2

                                                                                  Hence if cv is constant

                                                                                  t1

                                                                                  t2frac14 d

                                                                                  21

                                                                                  d22

                                                                                  where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                  d1 frac14 95mm and d2 frac14 2500mm

                                                                                  for U frac14 050 t2 frac14 t1 d22

                                                                                  d21

                                                                                  frac14 20

                                                                                  60 24 365 25002

                                                                                  952frac14 263 years

                                                                                  for U lt 060 Tv frac14

                                                                                  4U2 (Equation 724(a))

                                                                                  t030 frac14 t050 0302

                                                                                  0502

                                                                                  frac14 263 036 frac14 095 years

                                                                                  Consolidation theory 53

                                                                                  74

                                                                                  The layer is open

                                                                                  d frac14 8

                                                                                  2frac14 4m

                                                                                  Tv frac14 cvtd2frac14 24 3

                                                                                  42frac14 0450

                                                                                  ui frac14 frac14 84 kN=m2

                                                                                  The excess pore water pressure is given by Equation 721

                                                                                  ue frac14Xmfrac141mfrac140

                                                                                  2ui

                                                                                  Msin

                                                                                  Mz

                                                                                  d

                                                                                  expethM2TvTHORN

                                                                                  In this case z frac14 d

                                                                                  sinMz

                                                                                  d

                                                                                  frac14 sinM

                                                                                  where

                                                                                  M frac14

                                                                                  23

                                                                                  25

                                                                                  2

                                                                                  M sin M M2Tv exp (M2Tv)

                                                                                  2thorn1 1110 0329

                                                                                  3

                                                                                  21 9993 457 105

                                                                                  ue frac14 2 84 2

                                                                                  1 0329 ethother terms negligibleTHORN

                                                                                  frac14 352 kN=m2

                                                                                  75

                                                                                  The layer is open

                                                                                  d frac14 6

                                                                                  2frac14 3m

                                                                                  Tv frac14 cvtd2frac14 10 3

                                                                                  32frac14 0333

                                                                                  The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                  54 Consolidation theory

                                                                                  For an open layer

                                                                                  Tv frac14 4n

                                                                                  m2

                                                                                  n frac14 0333 62

                                                                                  4frac14 300

                                                                                  The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                  ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                  i j

                                                                                  0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                  0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                  The initial and 3-year isochrones are plotted in Figure Q75

                                                                                  Area under initial isochrone frac14 180 units

                                                                                  Area under 3-year isochrone frac14 63 units

                                                                                  The average degree of consolidation is given by Equation 725Thus

                                                                                  U frac14 1 63

                                                                                  180frac14 065

                                                                                  Figure Q75

                                                                                  Consolidation theory 55

                                                                                  76

                                                                                  At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                  0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                  The final consolidation settlement (one-dimensional method) is

                                                                                  sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                  Corrected time t frac14 2 1

                                                                                  2

                                                                                  40

                                                                                  52

                                                                                  frac14 1615 years

                                                                                  Tv frac14 cvtd2frac14 44 1615

                                                                                  42frac14 0444

                                                                                  From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                  77

                                                                                  The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                  Figure Q77

                                                                                  56 Consolidation theory

                                                                                  Point m n Ir (kNm2) sc (mm)

                                                                                  13020frac14 15 20

                                                                                  20frac14 10 0194 (4) 113 124

                                                                                  260

                                                                                  20frac14 30

                                                                                  20

                                                                                  20frac14 10 0204 (2) 59 65

                                                                                  360

                                                                                  20frac14 30

                                                                                  40

                                                                                  20frac14 20 0238 (1) 35 38

                                                                                  430

                                                                                  20frac14 15

                                                                                  40

                                                                                  20frac14 20 0224 (2) 65 72

                                                                                  Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                  78

                                                                                  Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                  (a) Immediate settlement

                                                                                  H

                                                                                  Bfrac14 30

                                                                                  35frac14 086

                                                                                  D

                                                                                  Bfrac14 2

                                                                                  35frac14 006

                                                                                  Figure Q78

                                                                                  Consolidation theory 57

                                                                                  From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                  si frac14 130131qB

                                                                                  Eufrac14 10 032 105 35

                                                                                  40frac14 30mm

                                                                                  (b) Consolidation settlement

                                                                                  Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                  1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                  3150

                                                                                  Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                  Now

                                                                                  H

                                                                                  Bfrac14 30

                                                                                  35frac14 086 and A frac14 065

                                                                                  from Figure 712 13 frac14 079

                                                                                  sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                  Total settlement

                                                                                  s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                  79

                                                                                  Without sand drains

                                                                                  Uv frac14 025

                                                                                  Tv frac14 0049 ethfrom Figure 718THORN

                                                                                  t frac14 Tvd2

                                                                                  cvfrac14 0049 82

                                                                                  cvWith sand drains

                                                                                  R frac14 0564S frac14 0564 3 frac14 169m

                                                                                  n frac14 Rrfrac14 169

                                                                                  015frac14 113

                                                                                  Tr frac14 cht

                                                                                  4R2frac14 ch

                                                                                  4 1692 0049 82

                                                                                  cvethand ch frac14 cvTHORN

                                                                                  frac14 0275

                                                                                  Ur frac14 073 (from Figure 730)

                                                                                  58 Consolidation theory

                                                                                  Using Equation 740

                                                                                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                  U frac14 080

                                                                                  710

                                                                                  Without sand drains

                                                                                  Uv frac14 090

                                                                                  Tv frac14 0848

                                                                                  t frac14 Tvd2

                                                                                  cvfrac14 0848 102

                                                                                  96frac14 88 years

                                                                                  With sand drains

                                                                                  R frac14 0564S frac14 0564 4 frac14 226m

                                                                                  n frac14 Rrfrac14 226

                                                                                  015frac14 15

                                                                                  Tr

                                                                                  Tvfrac14 chcv

                                                                                  d2

                                                                                  4R2ethsame tTHORN

                                                                                  Tr

                                                                                  Tvfrac14 140

                                                                                  96 102

                                                                                  4 2262frac14 714 eth1THORN

                                                                                  Using Equation 740

                                                                                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                  An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                  Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                  040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                  Thus

                                                                                  Uv frac14 0295 and Ur frac14 086

                                                                                  t frac14 88 00683

                                                                                  0848frac14 07 years

                                                                                  Consolidation theory 59

                                                                                  Chapter 8

                                                                                  Bearing capacity

                                                                                  81

                                                                                  (a) The ultimate bearing capacity is given by Equation 83

                                                                                  qf frac14 cNc thorn DNq thorn 1

                                                                                  2BN

                                                                                  For u frac14 0

                                                                                  Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                  qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                  The net ultimate bearing capacity is

                                                                                  qnf frac14 qf D frac14 540 kN=m2

                                                                                  The net foundation pressure is

                                                                                  qn frac14 q D frac14 425

                                                                                  2 eth21 1THORN frac14 192 kN=m2

                                                                                  The factor of safety (Equation 86) is

                                                                                  F frac14 qnfqnfrac14 540

                                                                                  192frac14 28

                                                                                  (b) For 0 frac14 28

                                                                                  Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                  qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                  2 112 2 13

                                                                                  frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                  qnf frac14 574 112 frac14 563 kN=m2

                                                                                  F frac14 563

                                                                                  192frac14 29

                                                                                  (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                  82

                                                                                  For 0 frac14 38

                                                                                  Nq frac14 49 N frac14 67

                                                                                  qnf frac14 DethNq 1THORN thorn 1

                                                                                  2BN ethfrom Equation 83THORN

                                                                                  frac14 eth18 075 48THORN thorn 1

                                                                                  2 18 15 67

                                                                                  frac14 648thorn 905 frac14 1553 kN=m2

                                                                                  qn frac14 500

                                                                                  15 eth18 075THORN frac14 320 kN=m2

                                                                                  F frac14 qnfqnfrac14 1553

                                                                                  320frac14 48

                                                                                  0d frac14 tan1tan 38

                                                                                  125

                                                                                  frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                  Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                  2 18 15 25

                                                                                  frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                  Design load (action) Vd frac14 500 kN=m

                                                                                  The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                  83

                                                                                  D

                                                                                  Bfrac14 350

                                                                                  225frac14 155

                                                                                  From Figure 85 for a square foundation

                                                                                  Nc frac14 81

                                                                                  Bearing capacity 61

                                                                                  For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                  Nc frac14 084thorn 016B

                                                                                  L

                                                                                  81 frac14 745

                                                                                  Using Equation 810

                                                                                  qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                  For F frac14 3

                                                                                  qn frac14 1006

                                                                                  3frac14 335 kN=m2

                                                                                  q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                  Design load frac14 405 450 225 frac14 4100 kN

                                                                                  Design undrained strength cud frac14 135

                                                                                  14frac14 96 kN=m2

                                                                                  Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                  frac14 7241 kN

                                                                                  Design load Vd frac14 4100 kN

                                                                                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                  84

                                                                                  For 0 frac14 40

                                                                                  Nq frac14 64 N frac14 95

                                                                                  qnf frac14 DethNq 1THORN thorn 04BN

                                                                                  (a) Water table 5m below ground level

                                                                                  qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                  qn frac14 400 17 frac14 383 kN=m2

                                                                                  F frac14 2686

                                                                                  383frac14 70

                                                                                  (b) Water table 1m below ground level (ie at foundation level)

                                                                                  0 frac14 20 98 frac14 102 kN=m3

                                                                                  62 Bearing capacity

                                                                                  qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                  F frac14 2040

                                                                                  383frac14 53

                                                                                  (c) Water table at ground level with upward hydraulic gradient 02

                                                                                  eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                  qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                  F frac14 1296

                                                                                  392frac14 33

                                                                                  85

                                                                                  The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                  Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                  Design value of 0 frac14 tan1tan 39

                                                                                  125

                                                                                  frac14 33

                                                                                  For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                  Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                  86

                                                                                  (a) Undrained shear for u frac14 0

                                                                                  Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                  qnf frac14 12cuNc

                                                                                  frac14 12 100 514 frac14 617 kN=m2

                                                                                  qn frac14 qnfFfrac14 617

                                                                                  3frac14 206 kN=m2

                                                                                  q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                  Bearing capacity 63

                                                                                  Drained shear for 0 frac14 32

                                                                                  Nq frac14 23 N frac14 25

                                                                                  0 frac14 21 98 frac14 112 kN=m3

                                                                                  qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                  frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                  frac14 694 kN=m2

                                                                                  q frac14 694

                                                                                  3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                  Design load frac14 42 227 frac14 3632 kN

                                                                                  (b) Design undrained strength cud frac14 100

                                                                                  14frac14 71 kNm2

                                                                                  Design bearing resistance Rd frac14 12cudNe area

                                                                                  frac14 12 71 514 42

                                                                                  frac14 7007 kN

                                                                                  For drained shear 0d frac14 tan1tan 32

                                                                                  125

                                                                                  frac14 26

                                                                                  Nq frac14 12 N frac14 10

                                                                                  Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                  (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                  Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                  1 2 100 0175 0700qn 0182qn

                                                                                  2 6 033 0044 0176qn 0046qn

                                                                                  3 10 020 0017 0068qn 0018qn

                                                                                  0246qn

                                                                                  Diameter of equivalent circle B frac14 45m

                                                                                  H

                                                                                  Bfrac14 12

                                                                                  45frac14 27 and A frac14 042

                                                                                  13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                  64 Bearing capacity

                                                                                  For sc frac14 30mm

                                                                                  qn frac14 30

                                                                                  0147frac14 204 kN=m2

                                                                                  q frac14 204thorn 21 frac14 225 kN=m2

                                                                                  Design load frac14 42 225 frac14 3600 kN

                                                                                  The design load is 3600 kN settlement being the limiting criterion

                                                                                  87

                                                                                  D

                                                                                  Bfrac14 8

                                                                                  4frac14 20

                                                                                  From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                  F frac14 cuNc

                                                                                  Dfrac14 40 71

                                                                                  20 8frac14 18

                                                                                  88

                                                                                  Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                  Design value of 0 frac14 tan1tan 38

                                                                                  125

                                                                                  frac14 32

                                                                                  Figure Q86

                                                                                  Bearing capacity 65

                                                                                  For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                  Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                  The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                  Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                  For B frac14 250m qn frac14 3750

                                                                                  2502 17 frac14 583 kN=m2

                                                                                  From Figure 510 m frac14 n frac14 126

                                                                                  6frac14 021

                                                                                  Ir frac14 0019

                                                                                  Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                  Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                  The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                  89

                                                                                  Depth (m) N 0v (kNm2) CN N1

                                                                                  070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                  Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                  (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                  Cw frac14 05thorn 0530

                                                                                  47

                                                                                  frac14 082

                                                                                  66 Bearing capacity

                                                                                  Thus

                                                                                  qa frac14 150 082 frac14 120 kN=m2

                                                                                  (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                  Thus

                                                                                  qa frac14 90 15 frac14 135 kN=m2

                                                                                  (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                  Ic frac14 171

                                                                                  1014frac14 0068

                                                                                  From Equation 819(a) with s frac14 25mm

                                                                                  q frac14 25

                                                                                  3507 0068frac14 150 kN=m2

                                                                                  810

                                                                                  Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                  Izp frac14 05thorn 01130

                                                                                  16 27

                                                                                  05

                                                                                  frac14 067

                                                                                  Refer to Figure Q810

                                                                                  E frac14 25qc

                                                                                  Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                  Ez (mm3MN)

                                                                                  1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                  0203

                                                                                  C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                  130frac14 093

                                                                                  C2 frac14 1 ethsayTHORN

                                                                                  s frac14 C1C2qnX Iz

                                                                                  Ez frac14 093 1 130 0203 frac14 25mm

                                                                                  Bearing capacity 67

                                                                                  811

                                                                                  At pile base level

                                                                                  cu frac14 220 kN=m2

                                                                                  qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                  Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                  00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                  qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                  Then

                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                  frac14

                                                                                  4 32 1980

                                                                                  thorn eth 105 139 86THORN

                                                                                  frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                  0 01 02 03 04 05 06 07

                                                                                  0 2 4 6 8 10 12 14

                                                                                  1

                                                                                  2

                                                                                  3

                                                                                  4

                                                                                  5

                                                                                  6

                                                                                  7

                                                                                  8

                                                                                  (1)

                                                                                  (2)

                                                                                  (3)

                                                                                  (4)

                                                                                  (5)

                                                                                  qc

                                                                                  qc

                                                                                  Iz

                                                                                  Iz

                                                                                  (MNm2)

                                                                                  z (m)

                                                                                  Figure Q810

                                                                                  68 Bearing capacity

                                                                                  Allowable load

                                                                                  ethaTHORN Qf

                                                                                  2frac14 17 937

                                                                                  2frac14 8968 kN

                                                                                  ethbTHORN Abqb

                                                                                  3thorn Asqs frac14 13 996

                                                                                  3thorn 3941 frac14 8606 kN

                                                                                  ie allowable load frac14 8600 kN

                                                                                  Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                  According to the limit state method

                                                                                  Characteristic undrained strength at base level cuk frac14 220

                                                                                  150kN=m2

                                                                                  Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                  150frac14 1320 kN=m2

                                                                                  Characteristic shaft resistance qsk frac14 00150

                                                                                  frac14 86

                                                                                  150frac14 57 kN=m2

                                                                                  Characteristic base and shaft resistances

                                                                                  Rbk frac14

                                                                                  4 32 1320 frac14 9330 kN

                                                                                  Rsk frac14 105 139 86

                                                                                  150frac14 2629 kN

                                                                                  For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                  Design bearing resistance Rcd frac14 9330

                                                                                  160thorn 2629

                                                                                  130

                                                                                  frac14 5831thorn 2022

                                                                                  frac14 7850 kN

                                                                                  Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                  812

                                                                                  ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                  qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                  For a single pile

                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                  frac14

                                                                                  4 062 1305

                                                                                  thorn eth 06 15 42THORN

                                                                                  frac14 369thorn 1187 frac14 1556 kN

                                                                                  Bearing capacity 69

                                                                                  Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                  eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                  (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                  (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                  qbkfrac14 9cuk frac14 9 220

                                                                                  150frac14 1320 kN=m2

                                                                                  qskfrac14cuk frac14 040 105

                                                                                  150frac14 28 kN=m2

                                                                                  Rbkfrac14

                                                                                  4 0602 1320 frac14 373 kN

                                                                                  Rskfrac14 060 15 28 frac14 791 kN

                                                                                  Rcdfrac14 373

                                                                                  160thorn 791

                                                                                  130frac14 233thorn 608 frac14 841 kN

                                                                                  Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                  Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                  (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                  q frac14 21 000

                                                                                  1762frac14 68 kN=m2

                                                                                  Immediate settlement

                                                                                  H

                                                                                  Bfrac14 15

                                                                                  176frac14 085

                                                                                  D

                                                                                  Bfrac14 13

                                                                                  176frac14 074

                                                                                  L

                                                                                  Bfrac14 1

                                                                                  Hence from Figure 515

                                                                                  130 frac14 078 and 131 frac14 041

                                                                                  70 Bearing capacity

                                                                                  Thus using Equation 528

                                                                                  si frac14 078 041 68 176

                                                                                  65frac14 6mm

                                                                                  Consolidation settlement

                                                                                  Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                  1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                  434 (sod)

                                                                                  Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                  sc frac14 056 434 frac14 24mm

                                                                                  The total settlement is (6thorn 24) frac14 30mm

                                                                                  813

                                                                                  At base level N frac14 26 Then using Equation 830

                                                                                  qb frac14 40NDb

                                                                                  Bfrac14 40 26 2

                                                                                  025frac14 8320 kN=m2

                                                                                  ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                  Figure Q812

                                                                                  Bearing capacity 71

                                                                                  Over the length embedded in sand

                                                                                  N frac14 21 ie18thorn 24

                                                                                  2

                                                                                  Using Equation 831

                                                                                  qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                  For a single pile

                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                  frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                  For the pile group assuming a group efficiency of 12

                                                                                  XQf frac14 12 9 604 frac14 6523 kN

                                                                                  Then the load factor is

                                                                                  F frac14 6523

                                                                                  2000thorn 1000frac14 21

                                                                                  (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                  Characteristic base resistance per unit area qbk frac14 8320

                                                                                  150frac14 5547 kNm2

                                                                                  Characteristic shaft resistance per unit area qsk frac14 42

                                                                                  150frac14 28 kNm2

                                                                                  Characteristic base and shaft resistances for a single pile

                                                                                  Rbk frac14 0252 5547 frac14 347 kN

                                                                                  Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                  For a driven pile the partial factors are b frac14 s frac14 130

                                                                                  Design bearing resistance Rcd frac14 347

                                                                                  130thorn 56

                                                                                  130frac14 310 kN

                                                                                  For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                  Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                  (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                  From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                  72 Bearing capacity

                                                                                  N frac14 24thorn 26thorn 34

                                                                                  3frac14 28

                                                                                  Ic frac14 171

                                                                                  2814frac14 0016 ethEquation 818THORN

                                                                                  s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                  814

                                                                                  Using Equation 841

                                                                                  Tf frac14 DLcu thorn

                                                                                  4ethD2 d2THORNcuNc

                                                                                  frac14 eth 02 5 06 110THORN thorn

                                                                                  4eth022 012THORN110 9

                                                                                  frac14 207thorn 23 frac14 230 kN

                                                                                  Figure Q813

                                                                                  Bearing capacity 73

                                                                                  Chapter 9

                                                                                  Stability of slopes

                                                                                  91

                                                                                  Referring to Figure Q91

                                                                                  W frac14 417 19 frac14 792 kN=m

                                                                                  Q frac14 20 28 frac14 56 kN=m

                                                                                  Arc lengthAB frac14

                                                                                  180 73 90 frac14 115m

                                                                                  Arc length BC frac14

                                                                                  180 28 90 frac14 44m

                                                                                  The factor of safety is given by

                                                                                  F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                  90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                  Depth of tension crack z0 frac14 2cu

                                                                                  frac14 2 20

                                                                                  19frac14 21m

                                                                                  Arc length BD frac14

                                                                                  180 13

                                                                                  1

                                                                                  2 90 frac14 21m

                                                                                  F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                  The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                  Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                  14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                  Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                  The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                  92

                                                                                  u frac14 0

                                                                                  Depth factor D frac14 11

                                                                                  9frac14 122

                                                                                  Using Equation 92 with F frac14 10

                                                                                  Ns frac14 cu

                                                                                  FHfrac14 30

                                                                                  10 19 9frac14 0175

                                                                                  Hence from Figure 93

                                                                                  frac14 50

                                                                                  For F frac14 12

                                                                                  Ns frac14 30

                                                                                  12 19 9frac14 0146

                                                                                  frac14 27

                                                                                  93

                                                                                  Refer to Figure Q93

                                                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                  1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                  74 m

                                                                                  214 1deg

                                                                                  213 1deg

                                                                                  39 m

                                                                                  WB

                                                                                  D

                                                                                  C

                                                                                  28 m

                                                                                  21 m

                                                                                  A

                                                                                  Q

                                                                                  Soil (1)Soil (2)

                                                                                  73deg

                                                                                  Figure Q91

                                                                                  Stability of slopes 75

                                                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                  9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                  599 256 328 1372

                                                                                  Figure Q93

                                                                                  76 Stability of slopes

                                                                                  XW cos frac14 b

                                                                                  Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                  W sin frac14 bX

                                                                                  h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                  Arc length La frac14

                                                                                  180 57

                                                                                  1

                                                                                  2 326 frac14 327m

                                                                                  The factor of safety is given by

                                                                                  F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                  W sin

                                                                                  frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                  frac14 091

                                                                                  According to the limit state method

                                                                                  0d frac14 tan1tan 32

                                                                                  125

                                                                                  frac14 265

                                                                                  c0 frac14 8

                                                                                  160frac14 5 kN=m2

                                                                                  Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                  Design disturbing moment frac14 1075 kN=m

                                                                                  The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                  94

                                                                                  F frac14 1

                                                                                  W sin

                                                                                  Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                  1thorn ethtan tan0=FTHORN

                                                                                  c0 frac14 8 kN=m2

                                                                                  0 frac14 32

                                                                                  c0b frac14 8 2 frac14 16 kN=m

                                                                                  W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                  Try F frac14 100

                                                                                  tan0

                                                                                  Ffrac14 0625

                                                                                  Stability of slopes 77

                                                                                  Values of u are as obtained in Figure Q93

                                                                                  SliceNo

                                                                                  h(m)

                                                                                  W frac14 bh(kNm)

                                                                                  W sin(kNm)

                                                                                  ub(kNm)

                                                                                  c0bthorn (W ub) tan0(kNm)

                                                                                  sec

                                                                                  1thorn (tan tan0)FProduct(kNm)

                                                                                  1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                  23 33 30 1042 31

                                                                                  3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                  224 92 72 0931 67

                                                                                  6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                  12 58 112 90 0889 80

                                                                                  8 60 252 1812

                                                                                  80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                  10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                  2154 88 116 0853 99

                                                                                  14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                  1074 1091

                                                                                  F frac14 1091

                                                                                  1074frac14 102 (assumed value 100)

                                                                                  Thus

                                                                                  F frac14 101

                                                                                  95

                                                                                  F frac14 1

                                                                                  W sin

                                                                                  XfWeth1 ruTHORN tan0g sec

                                                                                  1thorn ethtan tan0THORN=F

                                                                                  0 frac14 33

                                                                                  ru frac14 020

                                                                                  W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                  eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                  Try F frac14 110

                                                                                  tan 0

                                                                                  Ffrac14 tan 33

                                                                                  110frac14 0590

                                                                                  78 Stability of slopes

                                                                                  Referring to Figure Q95

                                                                                  SliceNo

                                                                                  h(m)

                                                                                  W frac14 bh(kNm)

                                                                                  W sin(kNm)

                                                                                  W(1 ru) tan0(kNm)

                                                                                  sec

                                                                                  1thorn ( tan tan0)FProduct(kNm)

                                                                                  1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                  2120 234 0892 209

                                                                                  4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                  1185 1271

                                                                                  Figure Q95

                                                                                  Stability of slopes 79

                                                                                  F frac14 1271

                                                                                  1185frac14 107

                                                                                  The trial value was 110 therefore take F to be 108

                                                                                  96

                                                                                  (a) Water table at surface the factor of safety is given by Equation 912

                                                                                  F frac14 0

                                                                                  sat

                                                                                  tan0

                                                                                  tan

                                                                                  ptie 15 frac14 92

                                                                                  19

                                                                                  tan 36

                                                                                  tan

                                                                                  tan frac14 0234

                                                                                  frac14 13

                                                                                  Water table well below surface the factor of safety is given by Equation 911

                                                                                  F frac14 tan0

                                                                                  tan

                                                                                  frac14 tan 36

                                                                                  tan 13

                                                                                  frac14 31

                                                                                  (b) 0d frac14 tan1tan 36

                                                                                  125

                                                                                  frac14 30

                                                                                  Depth of potential failure surface frac14 z

                                                                                  Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                  frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                  frac14 504z kN

                                                                                  Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                  frac14 19 z sin 13 cos 13

                                                                                  frac14 416z kN

                                                                                  Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                  80 Stability of slopes

                                                                                  • Book Cover
                                                                                  • Title
                                                                                  • Contents
                                                                                  • Basic characteristics of soils
                                                                                  • Seepage
                                                                                  • Effective stress
                                                                                  • Shear strength
                                                                                  • Stresses and displacements
                                                                                  • Lateral earth pressure
                                                                                  • Consolidation theory
                                                                                  • Bearing capacity
                                                                                  • Stability of slopes

                                                                                    63

                                                                                    (a) For u frac14 0 Ka frac14 Kp frac14 1

                                                                                    Kac frac14 Kpc frac14 2ffiffiffiffiffiffiffi15p

                                                                                    frac14 245

                                                                                    At the lower end of the piling

                                                                                    pa frac14 Kaqthorn Kasatz Kaccu

                                                                                    frac14 eth1 18 3THORN thorn eth1 20 4THORN eth245 50THORNfrac14 54thorn 80 1225

                                                                                    frac14 115 kN=m2

                                                                                    pp frac14 Kpsatzthorn Kpccu

                                                                                    frac14 eth1 20 4THORN thorn eth245 50THORNfrac14 80thorn 1225

                                                                                    frac14 202 kN=m2

                                                                                    (b) For 0 frac14 26 and frac14 1

                                                                                    20

                                                                                    Ka frac14 035

                                                                                    Kac frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth035 15THORN

                                                                                    pfrac14 145 ethEquation 619THORN

                                                                                    Kp frac14 37

                                                                                    Kpc frac14 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffieth37 15THORN

                                                                                    pfrac14 47 ethEquation 624THORN

                                                                                    At the lower end of the piling

                                                                                    pa frac14 Kaqthorn Ka0z Kacc

                                                                                    0

                                                                                    frac14 eth035 18 3THORN thorn eth035 102 4THORN eth145 10THORNfrac14 189thorn 143 145

                                                                                    frac14 187 kN=m2

                                                                                    pp frac14 Kp0zthorn Kpcc

                                                                                    0

                                                                                    frac14 eth37 102 4THORN thorn eth47 10THORNfrac14 151thorn 47

                                                                                    frac14 198 kN=m2

                                                                                    36 Lateral earth pressure

                                                                                    64

                                                                                    (a) For 0 frac14 38 Ka frac14 024

                                                                                    0 frac14 20 98 frac14 102 kN=m3

                                                                                    The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                                                    Force (kN) Arm (m) Moment (kN m)

                                                                                    (1) 024 10 66 frac14 159 33 525

                                                                                    (2)1

                                                                                    2 024 17 392 frac14 310 400 1240

                                                                                    (3) 024 17 39 27 frac14 430 135 580

                                                                                    (4)1

                                                                                    2 024 102 272 frac14 89 090 80

                                                                                    (5)1

                                                                                    2 98 272 frac14 357 090 321

                                                                                    Hfrac14 1345 MH frac14 2746

                                                                                    (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                                                    (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                                                    XM frac14MV MH frac14 7790 kNm

                                                                                    Lever arm of base resultant

                                                                                    M

                                                                                    Vfrac14 779

                                                                                    488frac14 160

                                                                                    Eccentricity of base resultant

                                                                                    e frac14 200 160 frac14 040m

                                                                                    39 m

                                                                                    27 m

                                                                                    40 m

                                                                                    04 m

                                                                                    04 m

                                                                                    26 m

                                                                                    (7)

                                                                                    (9)

                                                                                    (1)(2)

                                                                                    (3)

                                                                                    (4)

                                                                                    (5)

                                                                                    (8)(6)

                                                                                    (10)

                                                                                    WT

                                                                                    10 kNm2

                                                                                    Hydrostatic

                                                                                    Figure Q64

                                                                                    Lateral earth pressure 37

                                                                                    Base pressures (Equation 627)

                                                                                    p frac14 VB

                                                                                    1 6e

                                                                                    B

                                                                                    frac14 488

                                                                                    4eth1 060THORN

                                                                                    frac14 195 kN=m2 and 49 kN=m2

                                                                                    Factor of safety against sliding (Equation 628)

                                                                                    F frac14 V tan

                                                                                    Hfrac14 488 tan 25

                                                                                    1345frac14 17

                                                                                    (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                                                    Hfrac14 1633 kN

                                                                                    V frac14 4879 kN

                                                                                    MH frac14 3453 kNm

                                                                                    MV frac14 10536 kNm

                                                                                    The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                                                    65

                                                                                    For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                                    Kp

                                                                                    Ffrac14 385

                                                                                    2

                                                                                    0 frac14 20 98 frac14 102 kN=m3

                                                                                    The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                                                    Force (kN) Arm (m) Moment (kN m)

                                                                                    (1)1

                                                                                    2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                                                    (2) 026 17 45 d frac14 199d d2 995d2

                                                                                    (3)1

                                                                                    2 026 102 d2 frac14 133d2 d3 044d3

                                                                                    (4)1

                                                                                    2 385

                                                                                    2 17 152 frac14 368 dthorn 05 368d 184

                                                                                    (5)385

                                                                                    2 17 15 d frac14 491d d2 2455d2

                                                                                    (6)1

                                                                                    2 385

                                                                                    2 102 d2 frac14 982d2 d3 327d3

                                                                                    38 Lateral earth pressure

                                                                                    XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                                                    d3 thorn 516d2 283d 1724 frac14 0

                                                                                    d frac14 179m

                                                                                    Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                                                    Over additional 20 embedded depth

                                                                                    pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                                                    Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                                                    66

                                                                                    The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                                                    Ka frac14 sin 69=sin 105

                                                                                    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                                                    ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                                                    pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                                                    26664

                                                                                    37775

                                                                                    2

                                                                                    frac14 050

                                                                                    The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                                                    Pa frac14 1

                                                                                    2 050 19 7502 frac14 267 kN=m

                                                                                    Figure Q65

                                                                                    Lateral earth pressure 39

                                                                                    Horizontal component

                                                                                    Ph frac14 267 cos 40 frac14 205 kN=m

                                                                                    Vertical component

                                                                                    Pv frac14 267 sin 40 frac14 172 kN=m

                                                                                    Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                                    Force (kN) Arm (m) Moment (kN m)

                                                                                    (1)1

                                                                                    2 175 650 235 frac14 1337 258 345

                                                                                    (2) 050 650 235 frac14 764 175 134

                                                                                    (3)1

                                                                                    2 070 650 235 frac14 535 127 68

                                                                                    (4) 100 400 235 frac14 940 200 188

                                                                                    (5) 1

                                                                                    2 080 050 235 frac14 47 027 1

                                                                                    Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                                    Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                                    Lever arm of base resultant

                                                                                    M

                                                                                    Vfrac14 795

                                                                                    525frac14 151m

                                                                                    Eccentricity of base resultant

                                                                                    e frac14 200 151 frac14 049m

                                                                                    Figure Q66

                                                                                    40 Lateral earth pressure

                                                                                    Base pressures (Equation 627)

                                                                                    p frac14 525

                                                                                    41 6 049

                                                                                    4

                                                                                    frac14 228 kN=m2 and 35 kN=m2

                                                                                    The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                                    The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                                    The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                                    67

                                                                                    For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                                    Force (kN) Arm (m) Moment (kNm)

                                                                                    (1)1

                                                                                    2 027 17 52 frac14 574 183 1050

                                                                                    (2) 027 17 5 3 frac14 689 500 3445

                                                                                    (3)1

                                                                                    2 027 102 32 frac14 124 550 682

                                                                                    (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                                    (5)1

                                                                                    2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                                    (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                                    (7) 1

                                                                                    2 267

                                                                                    2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                                    (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                                    2 d frac14 163d d2thorn 650 82d2 1060d

                                                                                    Tie rod force per m frac14 T 0 0

                                                                                    XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                                    d3 thorn 77d2 269d 1438 frac14 0

                                                                                    d frac14 467m

                                                                                    Depth of penetration frac14 12d frac14 560m

                                                                                    Lateral earth pressure 41

                                                                                    Algebraic sum of forces for d frac14 467m isX

                                                                                    F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                                    T frac14 905 kN=m

                                                                                    Force in each tie rod frac14 25T frac14 226 kN

                                                                                    68

                                                                                    (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                                    0 frac14 21 98 frac14 112 kN=m3

                                                                                    The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                                    uC frac14 150

                                                                                    165 15 98 frac14 134 kN=m2

                                                                                    The average seepage pressure is

                                                                                    j frac14 15

                                                                                    165 98 frac14 09 kN=m3

                                                                                    Hence

                                                                                    0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                                    0 j frac14 112 09 frac14 103 kN=m3

                                                                                    Figure Q67

                                                                                    42 Lateral earth pressure

                                                                                    Consider moments about the anchor point A (per m)

                                                                                    Force (kN) Arm (m) Moment (kN m)

                                                                                    (1) 10 026 150 frac14 390 60 2340

                                                                                    (2)1

                                                                                    2 026 18 452 frac14 474 15 711

                                                                                    (3) 026 18 45 105 frac14 2211 825 18240

                                                                                    (4)1

                                                                                    2 026 121 1052 frac14 1734 100 17340

                                                                                    (5)1

                                                                                    2 134 15 frac14 101 40 404

                                                                                    (6) 134 30 frac14 402 60 2412

                                                                                    (7)1

                                                                                    2 134 60 frac14 402 95 3819

                                                                                    571 4527(8) Ppm

                                                                                    115 115PPm

                                                                                    XM frac14 0

                                                                                    Ppm frac144527

                                                                                    115frac14 394 kN=m

                                                                                    Available passive resistance

                                                                                    Pp frac14 1

                                                                                    2 385 103 62 frac14 714 kN=m

                                                                                    Factor of safety

                                                                                    Fp frac14 Pp

                                                                                    Ppm

                                                                                    frac14 714

                                                                                    394frac14 18

                                                                                    Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                                    Figure Q68

                                                                                    Lateral earth pressure 43

                                                                                    (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                                    Consider moments (per m) about the tie point A

                                                                                    Force (kN) Arm (m)

                                                                                    (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                                    (2)1

                                                                                    2 033 18 452 frac14 601 15

                                                                                    (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                                    (4)1

                                                                                    2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                                    (5)1

                                                                                    2 134 15 frac14 101 40

                                                                                    (6) 134 30 frac14 402 60

                                                                                    (7)1

                                                                                    2 134 d frac14 67d d3thorn 75

                                                                                    (8) 1

                                                                                    2 30 103 d2 frac141545d2 2d3thorn 75

                                                                                    Moment (kN m)

                                                                                    (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                                    XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                                    d3 thorn 827d2 466d 1518 frac14 0

                                                                                    By trial

                                                                                    d frac14 544m

                                                                                    The minimum depth of embedment required is 544m

                                                                                    69

                                                                                    For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                                    0 frac14 20 98 frac14 102 kN=m3

                                                                                    The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                                    44 Lateral earth pressure

                                                                                    uC frac14 147

                                                                                    173 26 98 frac14 216 kN=m2

                                                                                    and the average seepage pressure around the wall is

                                                                                    j frac14 26

                                                                                    173 98 frac14 15 kN=m3

                                                                                    Consider moments about the prop (A) (per m)

                                                                                    Force (kN) Arm (m) Moment (kN m)

                                                                                    (1)1

                                                                                    2 03 17 272 frac14 186 020 37

                                                                                    (2) 03 17 27 53 frac14 730 335 2445

                                                                                    (3)1

                                                                                    2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                                    (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                                    (5)1

                                                                                    2 216 26 frac14 281 243 684

                                                                                    (6) 216 27 frac14 583 465 2712

                                                                                    (7)1

                                                                                    2 216 60 frac14 648 800 5184

                                                                                    3055(8)

                                                                                    1

                                                                                    2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                                    Factor of safety

                                                                                    Fr frac14 6885

                                                                                    3055frac14 225

                                                                                    Figure Q69

                                                                                    Lateral earth pressure 45

                                                                                    610

                                                                                    For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                                    p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                                    Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                                    Using the recommendations of Twine and Roscoe

                                                                                    p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                                    Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                                    611

                                                                                    frac14 18 kN=m3 0 frac14 34

                                                                                    H frac14 350m nH frac14 335m mH frac14 185m

                                                                                    Consider a trial value of F frac14 20 Refer to Figure 635

                                                                                    0m frac14 tan1tan 34

                                                                                    20

                                                                                    frac14 186

                                                                                    Then

                                                                                    frac14 45 thorn 0m2frac14 543

                                                                                    W frac14 1

                                                                                    2 18 3502 cot 543 frac14 792 kN=m

                                                                                    Figure Q610

                                                                                    46 Lateral earth pressure

                                                                                    P frac14 1

                                                                                    2 s 3352 frac14 561s kN=m

                                                                                    U frac14 1

                                                                                    2 98 1852 cosec 543 frac14 206 kN=m

                                                                                    Equations 630 and 631 then become

                                                                                    561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                                    792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                                    ie

                                                                                    561s 0616N 405 frac14 0

                                                                                    792 0857N thorn 563 frac14 0

                                                                                    N frac14 848

                                                                                    0857frac14 989 kN=m

                                                                                    Then

                                                                                    561s 609 405 frac14 0

                                                                                    s frac14 649

                                                                                    561frac14 116 kN=m3

                                                                                    The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                                    F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                                    20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                                    s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                                    Figure Q611

                                                                                    Lateral earth pressure 47

                                                                                    612

                                                                                    For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                    45 thorn 0

                                                                                    2frac14 63

                                                                                    For the retained material between the surface and a depth of 36m

                                                                                    Pa frac14 1

                                                                                    2 030 18 362 frac14 350 kN=m

                                                                                    Weight of reinforced fill between the surface and a depth of 36m is

                                                                                    Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                    eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                    Eccentricity of Rv

                                                                                    e frac14 263 250 frac14 013m

                                                                                    The average vertical stress at a depth of 36m is

                                                                                    z frac14 Rv

                                                                                    L 2efrac14 324

                                                                                    474frac14 68 kN=m2

                                                                                    (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                    Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                    Tensile stress in the element frac14 138 103

                                                                                    65 3frac14 71N=mm2

                                                                                    Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                    Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                    Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                    The weight of ABC is

                                                                                    W frac14 1

                                                                                    2 18 52 265 frac14 124 kN=m

                                                                                    From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                    48 Lateral earth pressure

                                                                                    (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                    Tp frac14 032 68 120 065 frac14 170 kN

                                                                                    Tr frac14 213 420

                                                                                    418frac14 214 kN

                                                                                    Again the tensile failure and slipping limit states are satisfied for this element

                                                                                    Figure Q612

                                                                                    Lateral earth pressure 49

                                                                                    Chapter 7

                                                                                    Consolidation theory

                                                                                    71

                                                                                    Total change in thickness

                                                                                    H frac14 782 602 frac14 180mm

                                                                                    Average thickness frac14 1530thorn 180

                                                                                    2frac14 1620mm

                                                                                    Length of drainage path d frac14 1620

                                                                                    2frac14 810mm

                                                                                    Root time plot (Figure Q71a)

                                                                                    ffiffiffiffiffiffit90p frac14 33

                                                                                    t90 frac14 109min

                                                                                    cv frac14 0848d2

                                                                                    t90frac14 0848 8102

                                                                                    109 1440 365

                                                                                    106frac14 27m2=year

                                                                                    r0 frac14 782 764

                                                                                    782 602frac14 018

                                                                                    180frac14 0100

                                                                                    rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                    10 119

                                                                                    9 180frac14 0735

                                                                                    rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                    Log time plot (Figure Q71b)

                                                                                    t50 frac14 26min

                                                                                    cv frac14 0196d2

                                                                                    t50frac14 0196 8102

                                                                                    26 1440 365

                                                                                    106frac14 26m2=year

                                                                                    r0 frac14 782 763

                                                                                    782 602frac14 019

                                                                                    180frac14 0106

                                                                                    rp frac14 763 623

                                                                                    782 602frac14 140

                                                                                    180frac14 0778

                                                                                    rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                    Figure Q71(a)

                                                                                    Figure Q71(b)

                                                                                    Final void ratio

                                                                                    e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                    e

                                                                                    Hfrac14 1thorn e0

                                                                                    H0frac14 1thorn e1 thorne

                                                                                    H0

                                                                                    ie

                                                                                    e

                                                                                    180frac14 1631thorne

                                                                                    1710

                                                                                    e frac14 2936

                                                                                    1530frac14 0192

                                                                                    Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                    mv frac14 1

                                                                                    1thorn e0 e0 e101 00

                                                                                    frac14 1

                                                                                    1823 0192

                                                                                    0107frac14 098m2=MN

                                                                                    k frac14 cvmvw frac14 265 098 98

                                                                                    60 1440 365 103frac14 81 1010 m=s

                                                                                    72

                                                                                    Using Equation 77 (one-dimensional method)

                                                                                    sc frac14 e0 e11thorn e0 H

                                                                                    Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                    Figure Q72

                                                                                    52 Consolidation theory

                                                                                    Settlement

                                                                                    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                    1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                    318

                                                                                    Notes 5 92y 460thorn 84

                                                                                    Heave

                                                                                    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                    1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                    38

                                                                                    73

                                                                                    U frac14 f ethTvTHORN frac14 f cvt

                                                                                    d2

                                                                                    Hence if cv is constant

                                                                                    t1

                                                                                    t2frac14 d

                                                                                    21

                                                                                    d22

                                                                                    where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                    d1 frac14 95mm and d2 frac14 2500mm

                                                                                    for U frac14 050 t2 frac14 t1 d22

                                                                                    d21

                                                                                    frac14 20

                                                                                    60 24 365 25002

                                                                                    952frac14 263 years

                                                                                    for U lt 060 Tv frac14

                                                                                    4U2 (Equation 724(a))

                                                                                    t030 frac14 t050 0302

                                                                                    0502

                                                                                    frac14 263 036 frac14 095 years

                                                                                    Consolidation theory 53

                                                                                    74

                                                                                    The layer is open

                                                                                    d frac14 8

                                                                                    2frac14 4m

                                                                                    Tv frac14 cvtd2frac14 24 3

                                                                                    42frac14 0450

                                                                                    ui frac14 frac14 84 kN=m2

                                                                                    The excess pore water pressure is given by Equation 721

                                                                                    ue frac14Xmfrac141mfrac140

                                                                                    2ui

                                                                                    Msin

                                                                                    Mz

                                                                                    d

                                                                                    expethM2TvTHORN

                                                                                    In this case z frac14 d

                                                                                    sinMz

                                                                                    d

                                                                                    frac14 sinM

                                                                                    where

                                                                                    M frac14

                                                                                    23

                                                                                    25

                                                                                    2

                                                                                    M sin M M2Tv exp (M2Tv)

                                                                                    2thorn1 1110 0329

                                                                                    3

                                                                                    21 9993 457 105

                                                                                    ue frac14 2 84 2

                                                                                    1 0329 ethother terms negligibleTHORN

                                                                                    frac14 352 kN=m2

                                                                                    75

                                                                                    The layer is open

                                                                                    d frac14 6

                                                                                    2frac14 3m

                                                                                    Tv frac14 cvtd2frac14 10 3

                                                                                    32frac14 0333

                                                                                    The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                    54 Consolidation theory

                                                                                    For an open layer

                                                                                    Tv frac14 4n

                                                                                    m2

                                                                                    n frac14 0333 62

                                                                                    4frac14 300

                                                                                    The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                    ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                    i j

                                                                                    0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                    0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                    The initial and 3-year isochrones are plotted in Figure Q75

                                                                                    Area under initial isochrone frac14 180 units

                                                                                    Area under 3-year isochrone frac14 63 units

                                                                                    The average degree of consolidation is given by Equation 725Thus

                                                                                    U frac14 1 63

                                                                                    180frac14 065

                                                                                    Figure Q75

                                                                                    Consolidation theory 55

                                                                                    76

                                                                                    At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                    0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                    The final consolidation settlement (one-dimensional method) is

                                                                                    sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                    Corrected time t frac14 2 1

                                                                                    2

                                                                                    40

                                                                                    52

                                                                                    frac14 1615 years

                                                                                    Tv frac14 cvtd2frac14 44 1615

                                                                                    42frac14 0444

                                                                                    From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                    77

                                                                                    The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                    Figure Q77

                                                                                    56 Consolidation theory

                                                                                    Point m n Ir (kNm2) sc (mm)

                                                                                    13020frac14 15 20

                                                                                    20frac14 10 0194 (4) 113 124

                                                                                    260

                                                                                    20frac14 30

                                                                                    20

                                                                                    20frac14 10 0204 (2) 59 65

                                                                                    360

                                                                                    20frac14 30

                                                                                    40

                                                                                    20frac14 20 0238 (1) 35 38

                                                                                    430

                                                                                    20frac14 15

                                                                                    40

                                                                                    20frac14 20 0224 (2) 65 72

                                                                                    Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                    78

                                                                                    Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                    (a) Immediate settlement

                                                                                    H

                                                                                    Bfrac14 30

                                                                                    35frac14 086

                                                                                    D

                                                                                    Bfrac14 2

                                                                                    35frac14 006

                                                                                    Figure Q78

                                                                                    Consolidation theory 57

                                                                                    From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                    si frac14 130131qB

                                                                                    Eufrac14 10 032 105 35

                                                                                    40frac14 30mm

                                                                                    (b) Consolidation settlement

                                                                                    Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                    1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                    3150

                                                                                    Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                    Now

                                                                                    H

                                                                                    Bfrac14 30

                                                                                    35frac14 086 and A frac14 065

                                                                                    from Figure 712 13 frac14 079

                                                                                    sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                    Total settlement

                                                                                    s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                    79

                                                                                    Without sand drains

                                                                                    Uv frac14 025

                                                                                    Tv frac14 0049 ethfrom Figure 718THORN

                                                                                    t frac14 Tvd2

                                                                                    cvfrac14 0049 82

                                                                                    cvWith sand drains

                                                                                    R frac14 0564S frac14 0564 3 frac14 169m

                                                                                    n frac14 Rrfrac14 169

                                                                                    015frac14 113

                                                                                    Tr frac14 cht

                                                                                    4R2frac14 ch

                                                                                    4 1692 0049 82

                                                                                    cvethand ch frac14 cvTHORN

                                                                                    frac14 0275

                                                                                    Ur frac14 073 (from Figure 730)

                                                                                    58 Consolidation theory

                                                                                    Using Equation 740

                                                                                    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                    U frac14 080

                                                                                    710

                                                                                    Without sand drains

                                                                                    Uv frac14 090

                                                                                    Tv frac14 0848

                                                                                    t frac14 Tvd2

                                                                                    cvfrac14 0848 102

                                                                                    96frac14 88 years

                                                                                    With sand drains

                                                                                    R frac14 0564S frac14 0564 4 frac14 226m

                                                                                    n frac14 Rrfrac14 226

                                                                                    015frac14 15

                                                                                    Tr

                                                                                    Tvfrac14 chcv

                                                                                    d2

                                                                                    4R2ethsame tTHORN

                                                                                    Tr

                                                                                    Tvfrac14 140

                                                                                    96 102

                                                                                    4 2262frac14 714 eth1THORN

                                                                                    Using Equation 740

                                                                                    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                    An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                    Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                    040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                    Thus

                                                                                    Uv frac14 0295 and Ur frac14 086

                                                                                    t frac14 88 00683

                                                                                    0848frac14 07 years

                                                                                    Consolidation theory 59

                                                                                    Chapter 8

                                                                                    Bearing capacity

                                                                                    81

                                                                                    (a) The ultimate bearing capacity is given by Equation 83

                                                                                    qf frac14 cNc thorn DNq thorn 1

                                                                                    2BN

                                                                                    For u frac14 0

                                                                                    Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                    qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                    The net ultimate bearing capacity is

                                                                                    qnf frac14 qf D frac14 540 kN=m2

                                                                                    The net foundation pressure is

                                                                                    qn frac14 q D frac14 425

                                                                                    2 eth21 1THORN frac14 192 kN=m2

                                                                                    The factor of safety (Equation 86) is

                                                                                    F frac14 qnfqnfrac14 540

                                                                                    192frac14 28

                                                                                    (b) For 0 frac14 28

                                                                                    Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                    qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                    2 112 2 13

                                                                                    frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                    qnf frac14 574 112 frac14 563 kN=m2

                                                                                    F frac14 563

                                                                                    192frac14 29

                                                                                    (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                    82

                                                                                    For 0 frac14 38

                                                                                    Nq frac14 49 N frac14 67

                                                                                    qnf frac14 DethNq 1THORN thorn 1

                                                                                    2BN ethfrom Equation 83THORN

                                                                                    frac14 eth18 075 48THORN thorn 1

                                                                                    2 18 15 67

                                                                                    frac14 648thorn 905 frac14 1553 kN=m2

                                                                                    qn frac14 500

                                                                                    15 eth18 075THORN frac14 320 kN=m2

                                                                                    F frac14 qnfqnfrac14 1553

                                                                                    320frac14 48

                                                                                    0d frac14 tan1tan 38

                                                                                    125

                                                                                    frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                    Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                    2 18 15 25

                                                                                    frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                    Design load (action) Vd frac14 500 kN=m

                                                                                    The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                    83

                                                                                    D

                                                                                    Bfrac14 350

                                                                                    225frac14 155

                                                                                    From Figure 85 for a square foundation

                                                                                    Nc frac14 81

                                                                                    Bearing capacity 61

                                                                                    For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                    Nc frac14 084thorn 016B

                                                                                    L

                                                                                    81 frac14 745

                                                                                    Using Equation 810

                                                                                    qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                    For F frac14 3

                                                                                    qn frac14 1006

                                                                                    3frac14 335 kN=m2

                                                                                    q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                    Design load frac14 405 450 225 frac14 4100 kN

                                                                                    Design undrained strength cud frac14 135

                                                                                    14frac14 96 kN=m2

                                                                                    Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                    frac14 7241 kN

                                                                                    Design load Vd frac14 4100 kN

                                                                                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                    84

                                                                                    For 0 frac14 40

                                                                                    Nq frac14 64 N frac14 95

                                                                                    qnf frac14 DethNq 1THORN thorn 04BN

                                                                                    (a) Water table 5m below ground level

                                                                                    qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                    qn frac14 400 17 frac14 383 kN=m2

                                                                                    F frac14 2686

                                                                                    383frac14 70

                                                                                    (b) Water table 1m below ground level (ie at foundation level)

                                                                                    0 frac14 20 98 frac14 102 kN=m3

                                                                                    62 Bearing capacity

                                                                                    qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                    F frac14 2040

                                                                                    383frac14 53

                                                                                    (c) Water table at ground level with upward hydraulic gradient 02

                                                                                    eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                    qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                    F frac14 1296

                                                                                    392frac14 33

                                                                                    85

                                                                                    The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                    Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                    Design value of 0 frac14 tan1tan 39

                                                                                    125

                                                                                    frac14 33

                                                                                    For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                    Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                    86

                                                                                    (a) Undrained shear for u frac14 0

                                                                                    Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                    qnf frac14 12cuNc

                                                                                    frac14 12 100 514 frac14 617 kN=m2

                                                                                    qn frac14 qnfFfrac14 617

                                                                                    3frac14 206 kN=m2

                                                                                    q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                    Bearing capacity 63

                                                                                    Drained shear for 0 frac14 32

                                                                                    Nq frac14 23 N frac14 25

                                                                                    0 frac14 21 98 frac14 112 kN=m3

                                                                                    qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                    frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                    frac14 694 kN=m2

                                                                                    q frac14 694

                                                                                    3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                    Design load frac14 42 227 frac14 3632 kN

                                                                                    (b) Design undrained strength cud frac14 100

                                                                                    14frac14 71 kNm2

                                                                                    Design bearing resistance Rd frac14 12cudNe area

                                                                                    frac14 12 71 514 42

                                                                                    frac14 7007 kN

                                                                                    For drained shear 0d frac14 tan1tan 32

                                                                                    125

                                                                                    frac14 26

                                                                                    Nq frac14 12 N frac14 10

                                                                                    Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                    (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                    Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                    1 2 100 0175 0700qn 0182qn

                                                                                    2 6 033 0044 0176qn 0046qn

                                                                                    3 10 020 0017 0068qn 0018qn

                                                                                    0246qn

                                                                                    Diameter of equivalent circle B frac14 45m

                                                                                    H

                                                                                    Bfrac14 12

                                                                                    45frac14 27 and A frac14 042

                                                                                    13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                    64 Bearing capacity

                                                                                    For sc frac14 30mm

                                                                                    qn frac14 30

                                                                                    0147frac14 204 kN=m2

                                                                                    q frac14 204thorn 21 frac14 225 kN=m2

                                                                                    Design load frac14 42 225 frac14 3600 kN

                                                                                    The design load is 3600 kN settlement being the limiting criterion

                                                                                    87

                                                                                    D

                                                                                    Bfrac14 8

                                                                                    4frac14 20

                                                                                    From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                    F frac14 cuNc

                                                                                    Dfrac14 40 71

                                                                                    20 8frac14 18

                                                                                    88

                                                                                    Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                    Design value of 0 frac14 tan1tan 38

                                                                                    125

                                                                                    frac14 32

                                                                                    Figure Q86

                                                                                    Bearing capacity 65

                                                                                    For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                    Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                    The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                    Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                    For B frac14 250m qn frac14 3750

                                                                                    2502 17 frac14 583 kN=m2

                                                                                    From Figure 510 m frac14 n frac14 126

                                                                                    6frac14 021

                                                                                    Ir frac14 0019

                                                                                    Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                    Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                    The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                    89

                                                                                    Depth (m) N 0v (kNm2) CN N1

                                                                                    070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                    Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                    (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                    Cw frac14 05thorn 0530

                                                                                    47

                                                                                    frac14 082

                                                                                    66 Bearing capacity

                                                                                    Thus

                                                                                    qa frac14 150 082 frac14 120 kN=m2

                                                                                    (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                    Thus

                                                                                    qa frac14 90 15 frac14 135 kN=m2

                                                                                    (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                    Ic frac14 171

                                                                                    1014frac14 0068

                                                                                    From Equation 819(a) with s frac14 25mm

                                                                                    q frac14 25

                                                                                    3507 0068frac14 150 kN=m2

                                                                                    810

                                                                                    Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                    Izp frac14 05thorn 01130

                                                                                    16 27

                                                                                    05

                                                                                    frac14 067

                                                                                    Refer to Figure Q810

                                                                                    E frac14 25qc

                                                                                    Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                    Ez (mm3MN)

                                                                                    1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                    0203

                                                                                    C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                    130frac14 093

                                                                                    C2 frac14 1 ethsayTHORN

                                                                                    s frac14 C1C2qnX Iz

                                                                                    Ez frac14 093 1 130 0203 frac14 25mm

                                                                                    Bearing capacity 67

                                                                                    811

                                                                                    At pile base level

                                                                                    cu frac14 220 kN=m2

                                                                                    qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                    Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                    00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                    qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                    Then

                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                    frac14

                                                                                    4 32 1980

                                                                                    thorn eth 105 139 86THORN

                                                                                    frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                    0 01 02 03 04 05 06 07

                                                                                    0 2 4 6 8 10 12 14

                                                                                    1

                                                                                    2

                                                                                    3

                                                                                    4

                                                                                    5

                                                                                    6

                                                                                    7

                                                                                    8

                                                                                    (1)

                                                                                    (2)

                                                                                    (3)

                                                                                    (4)

                                                                                    (5)

                                                                                    qc

                                                                                    qc

                                                                                    Iz

                                                                                    Iz

                                                                                    (MNm2)

                                                                                    z (m)

                                                                                    Figure Q810

                                                                                    68 Bearing capacity

                                                                                    Allowable load

                                                                                    ethaTHORN Qf

                                                                                    2frac14 17 937

                                                                                    2frac14 8968 kN

                                                                                    ethbTHORN Abqb

                                                                                    3thorn Asqs frac14 13 996

                                                                                    3thorn 3941 frac14 8606 kN

                                                                                    ie allowable load frac14 8600 kN

                                                                                    Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                    According to the limit state method

                                                                                    Characteristic undrained strength at base level cuk frac14 220

                                                                                    150kN=m2

                                                                                    Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                    150frac14 1320 kN=m2

                                                                                    Characteristic shaft resistance qsk frac14 00150

                                                                                    frac14 86

                                                                                    150frac14 57 kN=m2

                                                                                    Characteristic base and shaft resistances

                                                                                    Rbk frac14

                                                                                    4 32 1320 frac14 9330 kN

                                                                                    Rsk frac14 105 139 86

                                                                                    150frac14 2629 kN

                                                                                    For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                    Design bearing resistance Rcd frac14 9330

                                                                                    160thorn 2629

                                                                                    130

                                                                                    frac14 5831thorn 2022

                                                                                    frac14 7850 kN

                                                                                    Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                    812

                                                                                    ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                    qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                    For a single pile

                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                    frac14

                                                                                    4 062 1305

                                                                                    thorn eth 06 15 42THORN

                                                                                    frac14 369thorn 1187 frac14 1556 kN

                                                                                    Bearing capacity 69

                                                                                    Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                    eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                    (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                    (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                    qbkfrac14 9cuk frac14 9 220

                                                                                    150frac14 1320 kN=m2

                                                                                    qskfrac14cuk frac14 040 105

                                                                                    150frac14 28 kN=m2

                                                                                    Rbkfrac14

                                                                                    4 0602 1320 frac14 373 kN

                                                                                    Rskfrac14 060 15 28 frac14 791 kN

                                                                                    Rcdfrac14 373

                                                                                    160thorn 791

                                                                                    130frac14 233thorn 608 frac14 841 kN

                                                                                    Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                    Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                    (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                    q frac14 21 000

                                                                                    1762frac14 68 kN=m2

                                                                                    Immediate settlement

                                                                                    H

                                                                                    Bfrac14 15

                                                                                    176frac14 085

                                                                                    D

                                                                                    Bfrac14 13

                                                                                    176frac14 074

                                                                                    L

                                                                                    Bfrac14 1

                                                                                    Hence from Figure 515

                                                                                    130 frac14 078 and 131 frac14 041

                                                                                    70 Bearing capacity

                                                                                    Thus using Equation 528

                                                                                    si frac14 078 041 68 176

                                                                                    65frac14 6mm

                                                                                    Consolidation settlement

                                                                                    Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                    1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                    434 (sod)

                                                                                    Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                    sc frac14 056 434 frac14 24mm

                                                                                    The total settlement is (6thorn 24) frac14 30mm

                                                                                    813

                                                                                    At base level N frac14 26 Then using Equation 830

                                                                                    qb frac14 40NDb

                                                                                    Bfrac14 40 26 2

                                                                                    025frac14 8320 kN=m2

                                                                                    ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                    Figure Q812

                                                                                    Bearing capacity 71

                                                                                    Over the length embedded in sand

                                                                                    N frac14 21 ie18thorn 24

                                                                                    2

                                                                                    Using Equation 831

                                                                                    qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                    For a single pile

                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                    frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                    For the pile group assuming a group efficiency of 12

                                                                                    XQf frac14 12 9 604 frac14 6523 kN

                                                                                    Then the load factor is

                                                                                    F frac14 6523

                                                                                    2000thorn 1000frac14 21

                                                                                    (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                    Characteristic base resistance per unit area qbk frac14 8320

                                                                                    150frac14 5547 kNm2

                                                                                    Characteristic shaft resistance per unit area qsk frac14 42

                                                                                    150frac14 28 kNm2

                                                                                    Characteristic base and shaft resistances for a single pile

                                                                                    Rbk frac14 0252 5547 frac14 347 kN

                                                                                    Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                    For a driven pile the partial factors are b frac14 s frac14 130

                                                                                    Design bearing resistance Rcd frac14 347

                                                                                    130thorn 56

                                                                                    130frac14 310 kN

                                                                                    For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                    Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                    (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                    From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                    72 Bearing capacity

                                                                                    N frac14 24thorn 26thorn 34

                                                                                    3frac14 28

                                                                                    Ic frac14 171

                                                                                    2814frac14 0016 ethEquation 818THORN

                                                                                    s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                    814

                                                                                    Using Equation 841

                                                                                    Tf frac14 DLcu thorn

                                                                                    4ethD2 d2THORNcuNc

                                                                                    frac14 eth 02 5 06 110THORN thorn

                                                                                    4eth022 012THORN110 9

                                                                                    frac14 207thorn 23 frac14 230 kN

                                                                                    Figure Q813

                                                                                    Bearing capacity 73

                                                                                    Chapter 9

                                                                                    Stability of slopes

                                                                                    91

                                                                                    Referring to Figure Q91

                                                                                    W frac14 417 19 frac14 792 kN=m

                                                                                    Q frac14 20 28 frac14 56 kN=m

                                                                                    Arc lengthAB frac14

                                                                                    180 73 90 frac14 115m

                                                                                    Arc length BC frac14

                                                                                    180 28 90 frac14 44m

                                                                                    The factor of safety is given by

                                                                                    F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                    90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                    Depth of tension crack z0 frac14 2cu

                                                                                    frac14 2 20

                                                                                    19frac14 21m

                                                                                    Arc length BD frac14

                                                                                    180 13

                                                                                    1

                                                                                    2 90 frac14 21m

                                                                                    F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                    The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                    Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                    14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                    Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                    The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                    92

                                                                                    u frac14 0

                                                                                    Depth factor D frac14 11

                                                                                    9frac14 122

                                                                                    Using Equation 92 with F frac14 10

                                                                                    Ns frac14 cu

                                                                                    FHfrac14 30

                                                                                    10 19 9frac14 0175

                                                                                    Hence from Figure 93

                                                                                    frac14 50

                                                                                    For F frac14 12

                                                                                    Ns frac14 30

                                                                                    12 19 9frac14 0146

                                                                                    frac14 27

                                                                                    93

                                                                                    Refer to Figure Q93

                                                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                    1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                    74 m

                                                                                    214 1deg

                                                                                    213 1deg

                                                                                    39 m

                                                                                    WB

                                                                                    D

                                                                                    C

                                                                                    28 m

                                                                                    21 m

                                                                                    A

                                                                                    Q

                                                                                    Soil (1)Soil (2)

                                                                                    73deg

                                                                                    Figure Q91

                                                                                    Stability of slopes 75

                                                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                    9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                    599 256 328 1372

                                                                                    Figure Q93

                                                                                    76 Stability of slopes

                                                                                    XW cos frac14 b

                                                                                    Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                    W sin frac14 bX

                                                                                    h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                    Arc length La frac14

                                                                                    180 57

                                                                                    1

                                                                                    2 326 frac14 327m

                                                                                    The factor of safety is given by

                                                                                    F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                    W sin

                                                                                    frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                    frac14 091

                                                                                    According to the limit state method

                                                                                    0d frac14 tan1tan 32

                                                                                    125

                                                                                    frac14 265

                                                                                    c0 frac14 8

                                                                                    160frac14 5 kN=m2

                                                                                    Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                    Design disturbing moment frac14 1075 kN=m

                                                                                    The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                    94

                                                                                    F frac14 1

                                                                                    W sin

                                                                                    Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                    1thorn ethtan tan0=FTHORN

                                                                                    c0 frac14 8 kN=m2

                                                                                    0 frac14 32

                                                                                    c0b frac14 8 2 frac14 16 kN=m

                                                                                    W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                    Try F frac14 100

                                                                                    tan0

                                                                                    Ffrac14 0625

                                                                                    Stability of slopes 77

                                                                                    Values of u are as obtained in Figure Q93

                                                                                    SliceNo

                                                                                    h(m)

                                                                                    W frac14 bh(kNm)

                                                                                    W sin(kNm)

                                                                                    ub(kNm)

                                                                                    c0bthorn (W ub) tan0(kNm)

                                                                                    sec

                                                                                    1thorn (tan tan0)FProduct(kNm)

                                                                                    1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                    23 33 30 1042 31

                                                                                    3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                    224 92 72 0931 67

                                                                                    6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                    12 58 112 90 0889 80

                                                                                    8 60 252 1812

                                                                                    80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                    10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                    2154 88 116 0853 99

                                                                                    14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                    1074 1091

                                                                                    F frac14 1091

                                                                                    1074frac14 102 (assumed value 100)

                                                                                    Thus

                                                                                    F frac14 101

                                                                                    95

                                                                                    F frac14 1

                                                                                    W sin

                                                                                    XfWeth1 ruTHORN tan0g sec

                                                                                    1thorn ethtan tan0THORN=F

                                                                                    0 frac14 33

                                                                                    ru frac14 020

                                                                                    W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                    eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                    Try F frac14 110

                                                                                    tan 0

                                                                                    Ffrac14 tan 33

                                                                                    110frac14 0590

                                                                                    78 Stability of slopes

                                                                                    Referring to Figure Q95

                                                                                    SliceNo

                                                                                    h(m)

                                                                                    W frac14 bh(kNm)

                                                                                    W sin(kNm)

                                                                                    W(1 ru) tan0(kNm)

                                                                                    sec

                                                                                    1thorn ( tan tan0)FProduct(kNm)

                                                                                    1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                    2120 234 0892 209

                                                                                    4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                    1185 1271

                                                                                    Figure Q95

                                                                                    Stability of slopes 79

                                                                                    F frac14 1271

                                                                                    1185frac14 107

                                                                                    The trial value was 110 therefore take F to be 108

                                                                                    96

                                                                                    (a) Water table at surface the factor of safety is given by Equation 912

                                                                                    F frac14 0

                                                                                    sat

                                                                                    tan0

                                                                                    tan

                                                                                    ptie 15 frac14 92

                                                                                    19

                                                                                    tan 36

                                                                                    tan

                                                                                    tan frac14 0234

                                                                                    frac14 13

                                                                                    Water table well below surface the factor of safety is given by Equation 911

                                                                                    F frac14 tan0

                                                                                    tan

                                                                                    frac14 tan 36

                                                                                    tan 13

                                                                                    frac14 31

                                                                                    (b) 0d frac14 tan1tan 36

                                                                                    125

                                                                                    frac14 30

                                                                                    Depth of potential failure surface frac14 z

                                                                                    Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                    frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                    frac14 504z kN

                                                                                    Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                    frac14 19 z sin 13 cos 13

                                                                                    frac14 416z kN

                                                                                    Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                    80 Stability of slopes

                                                                                    • Book Cover
                                                                                    • Title
                                                                                    • Contents
                                                                                    • Basic characteristics of soils
                                                                                    • Seepage
                                                                                    • Effective stress
                                                                                    • Shear strength
                                                                                    • Stresses and displacements
                                                                                    • Lateral earth pressure
                                                                                    • Consolidation theory
                                                                                    • Bearing capacity
                                                                                    • Stability of slopes

                                                                                      64

                                                                                      (a) For 0 frac14 38 Ka frac14 024

                                                                                      0 frac14 20 98 frac14 102 kN=m3

                                                                                      The pressure distribution is shown in Figure Q64 Consider moments (per m length ofwall) about the toe

                                                                                      Force (kN) Arm (m) Moment (kN m)

                                                                                      (1) 024 10 66 frac14 159 33 525

                                                                                      (2)1

                                                                                      2 024 17 392 frac14 310 400 1240

                                                                                      (3) 024 17 39 27 frac14 430 135 580

                                                                                      (4)1

                                                                                      2 024 102 272 frac14 89 090 80

                                                                                      (5)1

                                                                                      2 98 272 frac14 357 090 321

                                                                                      Hfrac14 1345 MH frac14 2746

                                                                                      (6) 62 04 235 frac14 583 120 700(7) 40 04 235 frac14 376 200 752(8) 39 26 17 frac14 1724 270 4655(9) 23 26 20 frac14 1196 270 3229

                                                                                      (10) 100 frac14 1000 120 1200Vfrac14 4879 MV frac14 10536

                                                                                      XM frac14MV MH frac14 7790 kNm

                                                                                      Lever arm of base resultant

                                                                                      M

                                                                                      Vfrac14 779

                                                                                      488frac14 160

                                                                                      Eccentricity of base resultant

                                                                                      e frac14 200 160 frac14 040m

                                                                                      39 m

                                                                                      27 m

                                                                                      40 m

                                                                                      04 m

                                                                                      04 m

                                                                                      26 m

                                                                                      (7)

                                                                                      (9)

                                                                                      (1)(2)

                                                                                      (3)

                                                                                      (4)

                                                                                      (5)

                                                                                      (8)(6)

                                                                                      (10)

                                                                                      WT

                                                                                      10 kNm2

                                                                                      Hydrostatic

                                                                                      Figure Q64

                                                                                      Lateral earth pressure 37

                                                                                      Base pressures (Equation 627)

                                                                                      p frac14 VB

                                                                                      1 6e

                                                                                      B

                                                                                      frac14 488

                                                                                      4eth1 060THORN

                                                                                      frac14 195 kN=m2 and 49 kN=m2

                                                                                      Factor of safety against sliding (Equation 628)

                                                                                      F frac14 V tan

                                                                                      Hfrac14 488 tan 25

                                                                                      1345frac14 17

                                                                                      (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                                                      Hfrac14 1633 kN

                                                                                      V frac14 4879 kN

                                                                                      MH frac14 3453 kNm

                                                                                      MV frac14 10536 kNm

                                                                                      The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                                                      65

                                                                                      For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                                      Kp

                                                                                      Ffrac14 385

                                                                                      2

                                                                                      0 frac14 20 98 frac14 102 kN=m3

                                                                                      The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                                                      Force (kN) Arm (m) Moment (kN m)

                                                                                      (1)1

                                                                                      2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                                                      (2) 026 17 45 d frac14 199d d2 995d2

                                                                                      (3)1

                                                                                      2 026 102 d2 frac14 133d2 d3 044d3

                                                                                      (4)1

                                                                                      2 385

                                                                                      2 17 152 frac14 368 dthorn 05 368d 184

                                                                                      (5)385

                                                                                      2 17 15 d frac14 491d d2 2455d2

                                                                                      (6)1

                                                                                      2 385

                                                                                      2 102 d2 frac14 982d2 d3 327d3

                                                                                      38 Lateral earth pressure

                                                                                      XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                                                      d3 thorn 516d2 283d 1724 frac14 0

                                                                                      d frac14 179m

                                                                                      Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                                                      Over additional 20 embedded depth

                                                                                      pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                                                      Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                                                      66

                                                                                      The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                                                      Ka frac14 sin 69=sin 105

                                                                                      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                                                      ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                                                      pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                                                      26664

                                                                                      37775

                                                                                      2

                                                                                      frac14 050

                                                                                      The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                                                      Pa frac14 1

                                                                                      2 050 19 7502 frac14 267 kN=m

                                                                                      Figure Q65

                                                                                      Lateral earth pressure 39

                                                                                      Horizontal component

                                                                                      Ph frac14 267 cos 40 frac14 205 kN=m

                                                                                      Vertical component

                                                                                      Pv frac14 267 sin 40 frac14 172 kN=m

                                                                                      Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                                      Force (kN) Arm (m) Moment (kN m)

                                                                                      (1)1

                                                                                      2 175 650 235 frac14 1337 258 345

                                                                                      (2) 050 650 235 frac14 764 175 134

                                                                                      (3)1

                                                                                      2 070 650 235 frac14 535 127 68

                                                                                      (4) 100 400 235 frac14 940 200 188

                                                                                      (5) 1

                                                                                      2 080 050 235 frac14 47 027 1

                                                                                      Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                                      Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                                      Lever arm of base resultant

                                                                                      M

                                                                                      Vfrac14 795

                                                                                      525frac14 151m

                                                                                      Eccentricity of base resultant

                                                                                      e frac14 200 151 frac14 049m

                                                                                      Figure Q66

                                                                                      40 Lateral earth pressure

                                                                                      Base pressures (Equation 627)

                                                                                      p frac14 525

                                                                                      41 6 049

                                                                                      4

                                                                                      frac14 228 kN=m2 and 35 kN=m2

                                                                                      The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                                      The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                                      The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                                      67

                                                                                      For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                                      Force (kN) Arm (m) Moment (kNm)

                                                                                      (1)1

                                                                                      2 027 17 52 frac14 574 183 1050

                                                                                      (2) 027 17 5 3 frac14 689 500 3445

                                                                                      (3)1

                                                                                      2 027 102 32 frac14 124 550 682

                                                                                      (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                                      (5)1

                                                                                      2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                                      (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                                      (7) 1

                                                                                      2 267

                                                                                      2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                                      (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                                      2 d frac14 163d d2thorn 650 82d2 1060d

                                                                                      Tie rod force per m frac14 T 0 0

                                                                                      XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                                      d3 thorn 77d2 269d 1438 frac14 0

                                                                                      d frac14 467m

                                                                                      Depth of penetration frac14 12d frac14 560m

                                                                                      Lateral earth pressure 41

                                                                                      Algebraic sum of forces for d frac14 467m isX

                                                                                      F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                                      T frac14 905 kN=m

                                                                                      Force in each tie rod frac14 25T frac14 226 kN

                                                                                      68

                                                                                      (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                                      0 frac14 21 98 frac14 112 kN=m3

                                                                                      The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                                      uC frac14 150

                                                                                      165 15 98 frac14 134 kN=m2

                                                                                      The average seepage pressure is

                                                                                      j frac14 15

                                                                                      165 98 frac14 09 kN=m3

                                                                                      Hence

                                                                                      0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                                      0 j frac14 112 09 frac14 103 kN=m3

                                                                                      Figure Q67

                                                                                      42 Lateral earth pressure

                                                                                      Consider moments about the anchor point A (per m)

                                                                                      Force (kN) Arm (m) Moment (kN m)

                                                                                      (1) 10 026 150 frac14 390 60 2340

                                                                                      (2)1

                                                                                      2 026 18 452 frac14 474 15 711

                                                                                      (3) 026 18 45 105 frac14 2211 825 18240

                                                                                      (4)1

                                                                                      2 026 121 1052 frac14 1734 100 17340

                                                                                      (5)1

                                                                                      2 134 15 frac14 101 40 404

                                                                                      (6) 134 30 frac14 402 60 2412

                                                                                      (7)1

                                                                                      2 134 60 frac14 402 95 3819

                                                                                      571 4527(8) Ppm

                                                                                      115 115PPm

                                                                                      XM frac14 0

                                                                                      Ppm frac144527

                                                                                      115frac14 394 kN=m

                                                                                      Available passive resistance

                                                                                      Pp frac14 1

                                                                                      2 385 103 62 frac14 714 kN=m

                                                                                      Factor of safety

                                                                                      Fp frac14 Pp

                                                                                      Ppm

                                                                                      frac14 714

                                                                                      394frac14 18

                                                                                      Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                                      Figure Q68

                                                                                      Lateral earth pressure 43

                                                                                      (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                                      Consider moments (per m) about the tie point A

                                                                                      Force (kN) Arm (m)

                                                                                      (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                                      (2)1

                                                                                      2 033 18 452 frac14 601 15

                                                                                      (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                                      (4)1

                                                                                      2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                                      (5)1

                                                                                      2 134 15 frac14 101 40

                                                                                      (6) 134 30 frac14 402 60

                                                                                      (7)1

                                                                                      2 134 d frac14 67d d3thorn 75

                                                                                      (8) 1

                                                                                      2 30 103 d2 frac141545d2 2d3thorn 75

                                                                                      Moment (kN m)

                                                                                      (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                                      XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                                      d3 thorn 827d2 466d 1518 frac14 0

                                                                                      By trial

                                                                                      d frac14 544m

                                                                                      The minimum depth of embedment required is 544m

                                                                                      69

                                                                                      For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                                      0 frac14 20 98 frac14 102 kN=m3

                                                                                      The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                                      44 Lateral earth pressure

                                                                                      uC frac14 147

                                                                                      173 26 98 frac14 216 kN=m2

                                                                                      and the average seepage pressure around the wall is

                                                                                      j frac14 26

                                                                                      173 98 frac14 15 kN=m3

                                                                                      Consider moments about the prop (A) (per m)

                                                                                      Force (kN) Arm (m) Moment (kN m)

                                                                                      (1)1

                                                                                      2 03 17 272 frac14 186 020 37

                                                                                      (2) 03 17 27 53 frac14 730 335 2445

                                                                                      (3)1

                                                                                      2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                                      (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                                      (5)1

                                                                                      2 216 26 frac14 281 243 684

                                                                                      (6) 216 27 frac14 583 465 2712

                                                                                      (7)1

                                                                                      2 216 60 frac14 648 800 5184

                                                                                      3055(8)

                                                                                      1

                                                                                      2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                                      Factor of safety

                                                                                      Fr frac14 6885

                                                                                      3055frac14 225

                                                                                      Figure Q69

                                                                                      Lateral earth pressure 45

                                                                                      610

                                                                                      For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                                      p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                                      Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                                      Using the recommendations of Twine and Roscoe

                                                                                      p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                                      Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                                      611

                                                                                      frac14 18 kN=m3 0 frac14 34

                                                                                      H frac14 350m nH frac14 335m mH frac14 185m

                                                                                      Consider a trial value of F frac14 20 Refer to Figure 635

                                                                                      0m frac14 tan1tan 34

                                                                                      20

                                                                                      frac14 186

                                                                                      Then

                                                                                      frac14 45 thorn 0m2frac14 543

                                                                                      W frac14 1

                                                                                      2 18 3502 cot 543 frac14 792 kN=m

                                                                                      Figure Q610

                                                                                      46 Lateral earth pressure

                                                                                      P frac14 1

                                                                                      2 s 3352 frac14 561s kN=m

                                                                                      U frac14 1

                                                                                      2 98 1852 cosec 543 frac14 206 kN=m

                                                                                      Equations 630 and 631 then become

                                                                                      561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                                      792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                                      ie

                                                                                      561s 0616N 405 frac14 0

                                                                                      792 0857N thorn 563 frac14 0

                                                                                      N frac14 848

                                                                                      0857frac14 989 kN=m

                                                                                      Then

                                                                                      561s 609 405 frac14 0

                                                                                      s frac14 649

                                                                                      561frac14 116 kN=m3

                                                                                      The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                                      F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                                      20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                                      s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                                      Figure Q611

                                                                                      Lateral earth pressure 47

                                                                                      612

                                                                                      For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                      45 thorn 0

                                                                                      2frac14 63

                                                                                      For the retained material between the surface and a depth of 36m

                                                                                      Pa frac14 1

                                                                                      2 030 18 362 frac14 350 kN=m

                                                                                      Weight of reinforced fill between the surface and a depth of 36m is

                                                                                      Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                      eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                      Eccentricity of Rv

                                                                                      e frac14 263 250 frac14 013m

                                                                                      The average vertical stress at a depth of 36m is

                                                                                      z frac14 Rv

                                                                                      L 2efrac14 324

                                                                                      474frac14 68 kN=m2

                                                                                      (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                      Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                      Tensile stress in the element frac14 138 103

                                                                                      65 3frac14 71N=mm2

                                                                                      Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                      Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                      Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                      The weight of ABC is

                                                                                      W frac14 1

                                                                                      2 18 52 265 frac14 124 kN=m

                                                                                      From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                      48 Lateral earth pressure

                                                                                      (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                      Tp frac14 032 68 120 065 frac14 170 kN

                                                                                      Tr frac14 213 420

                                                                                      418frac14 214 kN

                                                                                      Again the tensile failure and slipping limit states are satisfied for this element

                                                                                      Figure Q612

                                                                                      Lateral earth pressure 49

                                                                                      Chapter 7

                                                                                      Consolidation theory

                                                                                      71

                                                                                      Total change in thickness

                                                                                      H frac14 782 602 frac14 180mm

                                                                                      Average thickness frac14 1530thorn 180

                                                                                      2frac14 1620mm

                                                                                      Length of drainage path d frac14 1620

                                                                                      2frac14 810mm

                                                                                      Root time plot (Figure Q71a)

                                                                                      ffiffiffiffiffiffit90p frac14 33

                                                                                      t90 frac14 109min

                                                                                      cv frac14 0848d2

                                                                                      t90frac14 0848 8102

                                                                                      109 1440 365

                                                                                      106frac14 27m2=year

                                                                                      r0 frac14 782 764

                                                                                      782 602frac14 018

                                                                                      180frac14 0100

                                                                                      rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                      10 119

                                                                                      9 180frac14 0735

                                                                                      rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                      Log time plot (Figure Q71b)

                                                                                      t50 frac14 26min

                                                                                      cv frac14 0196d2

                                                                                      t50frac14 0196 8102

                                                                                      26 1440 365

                                                                                      106frac14 26m2=year

                                                                                      r0 frac14 782 763

                                                                                      782 602frac14 019

                                                                                      180frac14 0106

                                                                                      rp frac14 763 623

                                                                                      782 602frac14 140

                                                                                      180frac14 0778

                                                                                      rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                      Figure Q71(a)

                                                                                      Figure Q71(b)

                                                                                      Final void ratio

                                                                                      e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                      e

                                                                                      Hfrac14 1thorn e0

                                                                                      H0frac14 1thorn e1 thorne

                                                                                      H0

                                                                                      ie

                                                                                      e

                                                                                      180frac14 1631thorne

                                                                                      1710

                                                                                      e frac14 2936

                                                                                      1530frac14 0192

                                                                                      Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                      mv frac14 1

                                                                                      1thorn e0 e0 e101 00

                                                                                      frac14 1

                                                                                      1823 0192

                                                                                      0107frac14 098m2=MN

                                                                                      k frac14 cvmvw frac14 265 098 98

                                                                                      60 1440 365 103frac14 81 1010 m=s

                                                                                      72

                                                                                      Using Equation 77 (one-dimensional method)

                                                                                      sc frac14 e0 e11thorn e0 H

                                                                                      Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                      Figure Q72

                                                                                      52 Consolidation theory

                                                                                      Settlement

                                                                                      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                      1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                      318

                                                                                      Notes 5 92y 460thorn 84

                                                                                      Heave

                                                                                      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                      1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                      38

                                                                                      73

                                                                                      U frac14 f ethTvTHORN frac14 f cvt

                                                                                      d2

                                                                                      Hence if cv is constant

                                                                                      t1

                                                                                      t2frac14 d

                                                                                      21

                                                                                      d22

                                                                                      where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                      d1 frac14 95mm and d2 frac14 2500mm

                                                                                      for U frac14 050 t2 frac14 t1 d22

                                                                                      d21

                                                                                      frac14 20

                                                                                      60 24 365 25002

                                                                                      952frac14 263 years

                                                                                      for U lt 060 Tv frac14

                                                                                      4U2 (Equation 724(a))

                                                                                      t030 frac14 t050 0302

                                                                                      0502

                                                                                      frac14 263 036 frac14 095 years

                                                                                      Consolidation theory 53

                                                                                      74

                                                                                      The layer is open

                                                                                      d frac14 8

                                                                                      2frac14 4m

                                                                                      Tv frac14 cvtd2frac14 24 3

                                                                                      42frac14 0450

                                                                                      ui frac14 frac14 84 kN=m2

                                                                                      The excess pore water pressure is given by Equation 721

                                                                                      ue frac14Xmfrac141mfrac140

                                                                                      2ui

                                                                                      Msin

                                                                                      Mz

                                                                                      d

                                                                                      expethM2TvTHORN

                                                                                      In this case z frac14 d

                                                                                      sinMz

                                                                                      d

                                                                                      frac14 sinM

                                                                                      where

                                                                                      M frac14

                                                                                      23

                                                                                      25

                                                                                      2

                                                                                      M sin M M2Tv exp (M2Tv)

                                                                                      2thorn1 1110 0329

                                                                                      3

                                                                                      21 9993 457 105

                                                                                      ue frac14 2 84 2

                                                                                      1 0329 ethother terms negligibleTHORN

                                                                                      frac14 352 kN=m2

                                                                                      75

                                                                                      The layer is open

                                                                                      d frac14 6

                                                                                      2frac14 3m

                                                                                      Tv frac14 cvtd2frac14 10 3

                                                                                      32frac14 0333

                                                                                      The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                      54 Consolidation theory

                                                                                      For an open layer

                                                                                      Tv frac14 4n

                                                                                      m2

                                                                                      n frac14 0333 62

                                                                                      4frac14 300

                                                                                      The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                      ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                      i j

                                                                                      0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                      0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                      The initial and 3-year isochrones are plotted in Figure Q75

                                                                                      Area under initial isochrone frac14 180 units

                                                                                      Area under 3-year isochrone frac14 63 units

                                                                                      The average degree of consolidation is given by Equation 725Thus

                                                                                      U frac14 1 63

                                                                                      180frac14 065

                                                                                      Figure Q75

                                                                                      Consolidation theory 55

                                                                                      76

                                                                                      At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                      0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                      The final consolidation settlement (one-dimensional method) is

                                                                                      sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                      Corrected time t frac14 2 1

                                                                                      2

                                                                                      40

                                                                                      52

                                                                                      frac14 1615 years

                                                                                      Tv frac14 cvtd2frac14 44 1615

                                                                                      42frac14 0444

                                                                                      From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                      77

                                                                                      The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                      Figure Q77

                                                                                      56 Consolidation theory

                                                                                      Point m n Ir (kNm2) sc (mm)

                                                                                      13020frac14 15 20

                                                                                      20frac14 10 0194 (4) 113 124

                                                                                      260

                                                                                      20frac14 30

                                                                                      20

                                                                                      20frac14 10 0204 (2) 59 65

                                                                                      360

                                                                                      20frac14 30

                                                                                      40

                                                                                      20frac14 20 0238 (1) 35 38

                                                                                      430

                                                                                      20frac14 15

                                                                                      40

                                                                                      20frac14 20 0224 (2) 65 72

                                                                                      Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                      78

                                                                                      Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                      (a) Immediate settlement

                                                                                      H

                                                                                      Bfrac14 30

                                                                                      35frac14 086

                                                                                      D

                                                                                      Bfrac14 2

                                                                                      35frac14 006

                                                                                      Figure Q78

                                                                                      Consolidation theory 57

                                                                                      From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                      si frac14 130131qB

                                                                                      Eufrac14 10 032 105 35

                                                                                      40frac14 30mm

                                                                                      (b) Consolidation settlement

                                                                                      Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                      1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                      3150

                                                                                      Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                      Now

                                                                                      H

                                                                                      Bfrac14 30

                                                                                      35frac14 086 and A frac14 065

                                                                                      from Figure 712 13 frac14 079

                                                                                      sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                      Total settlement

                                                                                      s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                      79

                                                                                      Without sand drains

                                                                                      Uv frac14 025

                                                                                      Tv frac14 0049 ethfrom Figure 718THORN

                                                                                      t frac14 Tvd2

                                                                                      cvfrac14 0049 82

                                                                                      cvWith sand drains

                                                                                      R frac14 0564S frac14 0564 3 frac14 169m

                                                                                      n frac14 Rrfrac14 169

                                                                                      015frac14 113

                                                                                      Tr frac14 cht

                                                                                      4R2frac14 ch

                                                                                      4 1692 0049 82

                                                                                      cvethand ch frac14 cvTHORN

                                                                                      frac14 0275

                                                                                      Ur frac14 073 (from Figure 730)

                                                                                      58 Consolidation theory

                                                                                      Using Equation 740

                                                                                      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                      U frac14 080

                                                                                      710

                                                                                      Without sand drains

                                                                                      Uv frac14 090

                                                                                      Tv frac14 0848

                                                                                      t frac14 Tvd2

                                                                                      cvfrac14 0848 102

                                                                                      96frac14 88 years

                                                                                      With sand drains

                                                                                      R frac14 0564S frac14 0564 4 frac14 226m

                                                                                      n frac14 Rrfrac14 226

                                                                                      015frac14 15

                                                                                      Tr

                                                                                      Tvfrac14 chcv

                                                                                      d2

                                                                                      4R2ethsame tTHORN

                                                                                      Tr

                                                                                      Tvfrac14 140

                                                                                      96 102

                                                                                      4 2262frac14 714 eth1THORN

                                                                                      Using Equation 740

                                                                                      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                      An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                      Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                      040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                      Thus

                                                                                      Uv frac14 0295 and Ur frac14 086

                                                                                      t frac14 88 00683

                                                                                      0848frac14 07 years

                                                                                      Consolidation theory 59

                                                                                      Chapter 8

                                                                                      Bearing capacity

                                                                                      81

                                                                                      (a) The ultimate bearing capacity is given by Equation 83

                                                                                      qf frac14 cNc thorn DNq thorn 1

                                                                                      2BN

                                                                                      For u frac14 0

                                                                                      Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                      qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                      The net ultimate bearing capacity is

                                                                                      qnf frac14 qf D frac14 540 kN=m2

                                                                                      The net foundation pressure is

                                                                                      qn frac14 q D frac14 425

                                                                                      2 eth21 1THORN frac14 192 kN=m2

                                                                                      The factor of safety (Equation 86) is

                                                                                      F frac14 qnfqnfrac14 540

                                                                                      192frac14 28

                                                                                      (b) For 0 frac14 28

                                                                                      Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                      qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                      2 112 2 13

                                                                                      frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                      qnf frac14 574 112 frac14 563 kN=m2

                                                                                      F frac14 563

                                                                                      192frac14 29

                                                                                      (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                      82

                                                                                      For 0 frac14 38

                                                                                      Nq frac14 49 N frac14 67

                                                                                      qnf frac14 DethNq 1THORN thorn 1

                                                                                      2BN ethfrom Equation 83THORN

                                                                                      frac14 eth18 075 48THORN thorn 1

                                                                                      2 18 15 67

                                                                                      frac14 648thorn 905 frac14 1553 kN=m2

                                                                                      qn frac14 500

                                                                                      15 eth18 075THORN frac14 320 kN=m2

                                                                                      F frac14 qnfqnfrac14 1553

                                                                                      320frac14 48

                                                                                      0d frac14 tan1tan 38

                                                                                      125

                                                                                      frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                      Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                      2 18 15 25

                                                                                      frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                      Design load (action) Vd frac14 500 kN=m

                                                                                      The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                      83

                                                                                      D

                                                                                      Bfrac14 350

                                                                                      225frac14 155

                                                                                      From Figure 85 for a square foundation

                                                                                      Nc frac14 81

                                                                                      Bearing capacity 61

                                                                                      For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                      Nc frac14 084thorn 016B

                                                                                      L

                                                                                      81 frac14 745

                                                                                      Using Equation 810

                                                                                      qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                      For F frac14 3

                                                                                      qn frac14 1006

                                                                                      3frac14 335 kN=m2

                                                                                      q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                      Design load frac14 405 450 225 frac14 4100 kN

                                                                                      Design undrained strength cud frac14 135

                                                                                      14frac14 96 kN=m2

                                                                                      Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                      frac14 7241 kN

                                                                                      Design load Vd frac14 4100 kN

                                                                                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                      84

                                                                                      For 0 frac14 40

                                                                                      Nq frac14 64 N frac14 95

                                                                                      qnf frac14 DethNq 1THORN thorn 04BN

                                                                                      (a) Water table 5m below ground level

                                                                                      qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                      qn frac14 400 17 frac14 383 kN=m2

                                                                                      F frac14 2686

                                                                                      383frac14 70

                                                                                      (b) Water table 1m below ground level (ie at foundation level)

                                                                                      0 frac14 20 98 frac14 102 kN=m3

                                                                                      62 Bearing capacity

                                                                                      qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                      F frac14 2040

                                                                                      383frac14 53

                                                                                      (c) Water table at ground level with upward hydraulic gradient 02

                                                                                      eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                      qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                      F frac14 1296

                                                                                      392frac14 33

                                                                                      85

                                                                                      The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                      Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                      Design value of 0 frac14 tan1tan 39

                                                                                      125

                                                                                      frac14 33

                                                                                      For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                      Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                      86

                                                                                      (a) Undrained shear for u frac14 0

                                                                                      Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                      qnf frac14 12cuNc

                                                                                      frac14 12 100 514 frac14 617 kN=m2

                                                                                      qn frac14 qnfFfrac14 617

                                                                                      3frac14 206 kN=m2

                                                                                      q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                      Bearing capacity 63

                                                                                      Drained shear for 0 frac14 32

                                                                                      Nq frac14 23 N frac14 25

                                                                                      0 frac14 21 98 frac14 112 kN=m3

                                                                                      qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                      frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                      frac14 694 kN=m2

                                                                                      q frac14 694

                                                                                      3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                      Design load frac14 42 227 frac14 3632 kN

                                                                                      (b) Design undrained strength cud frac14 100

                                                                                      14frac14 71 kNm2

                                                                                      Design bearing resistance Rd frac14 12cudNe area

                                                                                      frac14 12 71 514 42

                                                                                      frac14 7007 kN

                                                                                      For drained shear 0d frac14 tan1tan 32

                                                                                      125

                                                                                      frac14 26

                                                                                      Nq frac14 12 N frac14 10

                                                                                      Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                      (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                      Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                      1 2 100 0175 0700qn 0182qn

                                                                                      2 6 033 0044 0176qn 0046qn

                                                                                      3 10 020 0017 0068qn 0018qn

                                                                                      0246qn

                                                                                      Diameter of equivalent circle B frac14 45m

                                                                                      H

                                                                                      Bfrac14 12

                                                                                      45frac14 27 and A frac14 042

                                                                                      13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                      64 Bearing capacity

                                                                                      For sc frac14 30mm

                                                                                      qn frac14 30

                                                                                      0147frac14 204 kN=m2

                                                                                      q frac14 204thorn 21 frac14 225 kN=m2

                                                                                      Design load frac14 42 225 frac14 3600 kN

                                                                                      The design load is 3600 kN settlement being the limiting criterion

                                                                                      87

                                                                                      D

                                                                                      Bfrac14 8

                                                                                      4frac14 20

                                                                                      From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                      F frac14 cuNc

                                                                                      Dfrac14 40 71

                                                                                      20 8frac14 18

                                                                                      88

                                                                                      Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                      Design value of 0 frac14 tan1tan 38

                                                                                      125

                                                                                      frac14 32

                                                                                      Figure Q86

                                                                                      Bearing capacity 65

                                                                                      For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                      Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                      The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                      Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                      For B frac14 250m qn frac14 3750

                                                                                      2502 17 frac14 583 kN=m2

                                                                                      From Figure 510 m frac14 n frac14 126

                                                                                      6frac14 021

                                                                                      Ir frac14 0019

                                                                                      Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                      Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                      The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                      89

                                                                                      Depth (m) N 0v (kNm2) CN N1

                                                                                      070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                      Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                      (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                      Cw frac14 05thorn 0530

                                                                                      47

                                                                                      frac14 082

                                                                                      66 Bearing capacity

                                                                                      Thus

                                                                                      qa frac14 150 082 frac14 120 kN=m2

                                                                                      (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                      Thus

                                                                                      qa frac14 90 15 frac14 135 kN=m2

                                                                                      (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                      Ic frac14 171

                                                                                      1014frac14 0068

                                                                                      From Equation 819(a) with s frac14 25mm

                                                                                      q frac14 25

                                                                                      3507 0068frac14 150 kN=m2

                                                                                      810

                                                                                      Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                      Izp frac14 05thorn 01130

                                                                                      16 27

                                                                                      05

                                                                                      frac14 067

                                                                                      Refer to Figure Q810

                                                                                      E frac14 25qc

                                                                                      Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                      Ez (mm3MN)

                                                                                      1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                      0203

                                                                                      C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                      130frac14 093

                                                                                      C2 frac14 1 ethsayTHORN

                                                                                      s frac14 C1C2qnX Iz

                                                                                      Ez frac14 093 1 130 0203 frac14 25mm

                                                                                      Bearing capacity 67

                                                                                      811

                                                                                      At pile base level

                                                                                      cu frac14 220 kN=m2

                                                                                      qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                      Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                      00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                      qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                      Then

                                                                                      Qf frac14 Abqb thorn Asqs

                                                                                      frac14

                                                                                      4 32 1980

                                                                                      thorn eth 105 139 86THORN

                                                                                      frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                      0 01 02 03 04 05 06 07

                                                                                      0 2 4 6 8 10 12 14

                                                                                      1

                                                                                      2

                                                                                      3

                                                                                      4

                                                                                      5

                                                                                      6

                                                                                      7

                                                                                      8

                                                                                      (1)

                                                                                      (2)

                                                                                      (3)

                                                                                      (4)

                                                                                      (5)

                                                                                      qc

                                                                                      qc

                                                                                      Iz

                                                                                      Iz

                                                                                      (MNm2)

                                                                                      z (m)

                                                                                      Figure Q810

                                                                                      68 Bearing capacity

                                                                                      Allowable load

                                                                                      ethaTHORN Qf

                                                                                      2frac14 17 937

                                                                                      2frac14 8968 kN

                                                                                      ethbTHORN Abqb

                                                                                      3thorn Asqs frac14 13 996

                                                                                      3thorn 3941 frac14 8606 kN

                                                                                      ie allowable load frac14 8600 kN

                                                                                      Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                      According to the limit state method

                                                                                      Characteristic undrained strength at base level cuk frac14 220

                                                                                      150kN=m2

                                                                                      Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                      150frac14 1320 kN=m2

                                                                                      Characteristic shaft resistance qsk frac14 00150

                                                                                      frac14 86

                                                                                      150frac14 57 kN=m2

                                                                                      Characteristic base and shaft resistances

                                                                                      Rbk frac14

                                                                                      4 32 1320 frac14 9330 kN

                                                                                      Rsk frac14 105 139 86

                                                                                      150frac14 2629 kN

                                                                                      For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                      Design bearing resistance Rcd frac14 9330

                                                                                      160thorn 2629

                                                                                      130

                                                                                      frac14 5831thorn 2022

                                                                                      frac14 7850 kN

                                                                                      Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                      812

                                                                                      ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                      qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                      For a single pile

                                                                                      Qf frac14 Abqb thorn Asqs

                                                                                      frac14

                                                                                      4 062 1305

                                                                                      thorn eth 06 15 42THORN

                                                                                      frac14 369thorn 1187 frac14 1556 kN

                                                                                      Bearing capacity 69

                                                                                      Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                      eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                      (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                      (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                      qbkfrac14 9cuk frac14 9 220

                                                                                      150frac14 1320 kN=m2

                                                                                      qskfrac14cuk frac14 040 105

                                                                                      150frac14 28 kN=m2

                                                                                      Rbkfrac14

                                                                                      4 0602 1320 frac14 373 kN

                                                                                      Rskfrac14 060 15 28 frac14 791 kN

                                                                                      Rcdfrac14 373

                                                                                      160thorn 791

                                                                                      130frac14 233thorn 608 frac14 841 kN

                                                                                      Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                      Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                      (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                      q frac14 21 000

                                                                                      1762frac14 68 kN=m2

                                                                                      Immediate settlement

                                                                                      H

                                                                                      Bfrac14 15

                                                                                      176frac14 085

                                                                                      D

                                                                                      Bfrac14 13

                                                                                      176frac14 074

                                                                                      L

                                                                                      Bfrac14 1

                                                                                      Hence from Figure 515

                                                                                      130 frac14 078 and 131 frac14 041

                                                                                      70 Bearing capacity

                                                                                      Thus using Equation 528

                                                                                      si frac14 078 041 68 176

                                                                                      65frac14 6mm

                                                                                      Consolidation settlement

                                                                                      Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                      1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                      434 (sod)

                                                                                      Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                      sc frac14 056 434 frac14 24mm

                                                                                      The total settlement is (6thorn 24) frac14 30mm

                                                                                      813

                                                                                      At base level N frac14 26 Then using Equation 830

                                                                                      qb frac14 40NDb

                                                                                      Bfrac14 40 26 2

                                                                                      025frac14 8320 kN=m2

                                                                                      ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                      Figure Q812

                                                                                      Bearing capacity 71

                                                                                      Over the length embedded in sand

                                                                                      N frac14 21 ie18thorn 24

                                                                                      2

                                                                                      Using Equation 831

                                                                                      qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                      For a single pile

                                                                                      Qf frac14 Abqb thorn Asqs

                                                                                      frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                      For the pile group assuming a group efficiency of 12

                                                                                      XQf frac14 12 9 604 frac14 6523 kN

                                                                                      Then the load factor is

                                                                                      F frac14 6523

                                                                                      2000thorn 1000frac14 21

                                                                                      (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                      Characteristic base resistance per unit area qbk frac14 8320

                                                                                      150frac14 5547 kNm2

                                                                                      Characteristic shaft resistance per unit area qsk frac14 42

                                                                                      150frac14 28 kNm2

                                                                                      Characteristic base and shaft resistances for a single pile

                                                                                      Rbk frac14 0252 5547 frac14 347 kN

                                                                                      Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                      For a driven pile the partial factors are b frac14 s frac14 130

                                                                                      Design bearing resistance Rcd frac14 347

                                                                                      130thorn 56

                                                                                      130frac14 310 kN

                                                                                      For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                      Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                      (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                      From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                      72 Bearing capacity

                                                                                      N frac14 24thorn 26thorn 34

                                                                                      3frac14 28

                                                                                      Ic frac14 171

                                                                                      2814frac14 0016 ethEquation 818THORN

                                                                                      s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                      814

                                                                                      Using Equation 841

                                                                                      Tf frac14 DLcu thorn

                                                                                      4ethD2 d2THORNcuNc

                                                                                      frac14 eth 02 5 06 110THORN thorn

                                                                                      4eth022 012THORN110 9

                                                                                      frac14 207thorn 23 frac14 230 kN

                                                                                      Figure Q813

                                                                                      Bearing capacity 73

                                                                                      Chapter 9

                                                                                      Stability of slopes

                                                                                      91

                                                                                      Referring to Figure Q91

                                                                                      W frac14 417 19 frac14 792 kN=m

                                                                                      Q frac14 20 28 frac14 56 kN=m

                                                                                      Arc lengthAB frac14

                                                                                      180 73 90 frac14 115m

                                                                                      Arc length BC frac14

                                                                                      180 28 90 frac14 44m

                                                                                      The factor of safety is given by

                                                                                      F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                      90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                      Depth of tension crack z0 frac14 2cu

                                                                                      frac14 2 20

                                                                                      19frac14 21m

                                                                                      Arc length BD frac14

                                                                                      180 13

                                                                                      1

                                                                                      2 90 frac14 21m

                                                                                      F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                      The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                      Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                      14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                      Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                      The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                      92

                                                                                      u frac14 0

                                                                                      Depth factor D frac14 11

                                                                                      9frac14 122

                                                                                      Using Equation 92 with F frac14 10

                                                                                      Ns frac14 cu

                                                                                      FHfrac14 30

                                                                                      10 19 9frac14 0175

                                                                                      Hence from Figure 93

                                                                                      frac14 50

                                                                                      For F frac14 12

                                                                                      Ns frac14 30

                                                                                      12 19 9frac14 0146

                                                                                      frac14 27

                                                                                      93

                                                                                      Refer to Figure Q93

                                                                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                      1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                      74 m

                                                                                      214 1deg

                                                                                      213 1deg

                                                                                      39 m

                                                                                      WB

                                                                                      D

                                                                                      C

                                                                                      28 m

                                                                                      21 m

                                                                                      A

                                                                                      Q

                                                                                      Soil (1)Soil (2)

                                                                                      73deg

                                                                                      Figure Q91

                                                                                      Stability of slopes 75

                                                                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                      9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                      599 256 328 1372

                                                                                      Figure Q93

                                                                                      76 Stability of slopes

                                                                                      XW cos frac14 b

                                                                                      Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                      W sin frac14 bX

                                                                                      h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                      Arc length La frac14

                                                                                      180 57

                                                                                      1

                                                                                      2 326 frac14 327m

                                                                                      The factor of safety is given by

                                                                                      F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                      W sin

                                                                                      frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                      frac14 091

                                                                                      According to the limit state method

                                                                                      0d frac14 tan1tan 32

                                                                                      125

                                                                                      frac14 265

                                                                                      c0 frac14 8

                                                                                      160frac14 5 kN=m2

                                                                                      Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                      Design disturbing moment frac14 1075 kN=m

                                                                                      The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                      94

                                                                                      F frac14 1

                                                                                      W sin

                                                                                      Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                      1thorn ethtan tan0=FTHORN

                                                                                      c0 frac14 8 kN=m2

                                                                                      0 frac14 32

                                                                                      c0b frac14 8 2 frac14 16 kN=m

                                                                                      W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                      Try F frac14 100

                                                                                      tan0

                                                                                      Ffrac14 0625

                                                                                      Stability of slopes 77

                                                                                      Values of u are as obtained in Figure Q93

                                                                                      SliceNo

                                                                                      h(m)

                                                                                      W frac14 bh(kNm)

                                                                                      W sin(kNm)

                                                                                      ub(kNm)

                                                                                      c0bthorn (W ub) tan0(kNm)

                                                                                      sec

                                                                                      1thorn (tan tan0)FProduct(kNm)

                                                                                      1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                      23 33 30 1042 31

                                                                                      3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                      224 92 72 0931 67

                                                                                      6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                      12 58 112 90 0889 80

                                                                                      8 60 252 1812

                                                                                      80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                      10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                      2154 88 116 0853 99

                                                                                      14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                      1074 1091

                                                                                      F frac14 1091

                                                                                      1074frac14 102 (assumed value 100)

                                                                                      Thus

                                                                                      F frac14 101

                                                                                      95

                                                                                      F frac14 1

                                                                                      W sin

                                                                                      XfWeth1 ruTHORN tan0g sec

                                                                                      1thorn ethtan tan0THORN=F

                                                                                      0 frac14 33

                                                                                      ru frac14 020

                                                                                      W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                      eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                      Try F frac14 110

                                                                                      tan 0

                                                                                      Ffrac14 tan 33

                                                                                      110frac14 0590

                                                                                      78 Stability of slopes

                                                                                      Referring to Figure Q95

                                                                                      SliceNo

                                                                                      h(m)

                                                                                      W frac14 bh(kNm)

                                                                                      W sin(kNm)

                                                                                      W(1 ru) tan0(kNm)

                                                                                      sec

                                                                                      1thorn ( tan tan0)FProduct(kNm)

                                                                                      1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                      2120 234 0892 209

                                                                                      4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                      1185 1271

                                                                                      Figure Q95

                                                                                      Stability of slopes 79

                                                                                      F frac14 1271

                                                                                      1185frac14 107

                                                                                      The trial value was 110 therefore take F to be 108

                                                                                      96

                                                                                      (a) Water table at surface the factor of safety is given by Equation 912

                                                                                      F frac14 0

                                                                                      sat

                                                                                      tan0

                                                                                      tan

                                                                                      ptie 15 frac14 92

                                                                                      19

                                                                                      tan 36

                                                                                      tan

                                                                                      tan frac14 0234

                                                                                      frac14 13

                                                                                      Water table well below surface the factor of safety is given by Equation 911

                                                                                      F frac14 tan0

                                                                                      tan

                                                                                      frac14 tan 36

                                                                                      tan 13

                                                                                      frac14 31

                                                                                      (b) 0d frac14 tan1tan 36

                                                                                      125

                                                                                      frac14 30

                                                                                      Depth of potential failure surface frac14 z

                                                                                      Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                      frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                      frac14 504z kN

                                                                                      Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                      frac14 19 z sin 13 cos 13

                                                                                      frac14 416z kN

                                                                                      Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                      80 Stability of slopes

                                                                                      • Book Cover
                                                                                      • Title
                                                                                      • Contents
                                                                                      • Basic characteristics of soils
                                                                                      • Seepage
                                                                                      • Effective stress
                                                                                      • Shear strength
                                                                                      • Stresses and displacements
                                                                                      • Lateral earth pressure
                                                                                      • Consolidation theory
                                                                                      • Bearing capacity
                                                                                      • Stability of slopes

                                                                                        Base pressures (Equation 627)

                                                                                        p frac14 VB

                                                                                        1 6e

                                                                                        B

                                                                                        frac14 488

                                                                                        4eth1 060THORN

                                                                                        frac14 195 kN=m2 and 49 kN=m2

                                                                                        Factor of safety against sliding (Equation 628)

                                                                                        F frac14 V tan

                                                                                        Hfrac14 488 tan 25

                                                                                        1345frac14 17

                                                                                        (b) Using a partial factor of 125 the design value of 0 is tan1 ( tan 38125) frac14 32Therefore Ka frac14 031 and the forces and moments are

                                                                                        Hfrac14 1633 kN

                                                                                        V frac14 4879 kN

                                                                                        MH frac14 3453 kNm

                                                                                        MV frac14 10536 kNm

                                                                                        The overturning limit state is satisfied the restoring moment (MV ) being greater thanthe overturning moment (MH) The sliding limit state is satisfied the resisting force(V tan frac14 2275 kN) being greater than the disturbing force (H )

                                                                                        65

                                                                                        For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                                        Kp

                                                                                        Ffrac14 385

                                                                                        2

                                                                                        0 frac14 20 98 frac14 102 kN=m3

                                                                                        The pressure distribution is shown in Figure Q65 hydrostatic pressure on the twosides of the wall balances Consider moments about X (per m) assuming d gt 0

                                                                                        Force (kN) Arm (m) Moment (kN m)

                                                                                        (1)1

                                                                                        2 026 17 452 frac14 448 dthorn 15 448dthorn 672

                                                                                        (2) 026 17 45 d frac14 199d d2 995d2

                                                                                        (3)1

                                                                                        2 026 102 d2 frac14 133d2 d3 044d3

                                                                                        (4)1

                                                                                        2 385

                                                                                        2 17 152 frac14 368 dthorn 05 368d 184

                                                                                        (5)385

                                                                                        2 17 15 d frac14 491d d2 2455d2

                                                                                        (6)1

                                                                                        2 385

                                                                                        2 102 d2 frac14 982d2 d3 327d3

                                                                                        38 Lateral earth pressure

                                                                                        XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                                                        d3 thorn 516d2 283d 1724 frac14 0

                                                                                        d frac14 179m

                                                                                        Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                                                        Over additional 20 embedded depth

                                                                                        pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                                                        Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                                                        66

                                                                                        The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                                                        Ka frac14 sin 69=sin 105

                                                                                        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                                                        ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                                                        pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                                                        26664

                                                                                        37775

                                                                                        2

                                                                                        frac14 050

                                                                                        The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                                                        Pa frac14 1

                                                                                        2 050 19 7502 frac14 267 kN=m

                                                                                        Figure Q65

                                                                                        Lateral earth pressure 39

                                                                                        Horizontal component

                                                                                        Ph frac14 267 cos 40 frac14 205 kN=m

                                                                                        Vertical component

                                                                                        Pv frac14 267 sin 40 frac14 172 kN=m

                                                                                        Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                                        Force (kN) Arm (m) Moment (kN m)

                                                                                        (1)1

                                                                                        2 175 650 235 frac14 1337 258 345

                                                                                        (2) 050 650 235 frac14 764 175 134

                                                                                        (3)1

                                                                                        2 070 650 235 frac14 535 127 68

                                                                                        (4) 100 400 235 frac14 940 200 188

                                                                                        (5) 1

                                                                                        2 080 050 235 frac14 47 027 1

                                                                                        Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                                        Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                                        Lever arm of base resultant

                                                                                        M

                                                                                        Vfrac14 795

                                                                                        525frac14 151m

                                                                                        Eccentricity of base resultant

                                                                                        e frac14 200 151 frac14 049m

                                                                                        Figure Q66

                                                                                        40 Lateral earth pressure

                                                                                        Base pressures (Equation 627)

                                                                                        p frac14 525

                                                                                        41 6 049

                                                                                        4

                                                                                        frac14 228 kN=m2 and 35 kN=m2

                                                                                        The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                                        The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                                        The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                                        67

                                                                                        For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                                        Force (kN) Arm (m) Moment (kNm)

                                                                                        (1)1

                                                                                        2 027 17 52 frac14 574 183 1050

                                                                                        (2) 027 17 5 3 frac14 689 500 3445

                                                                                        (3)1

                                                                                        2 027 102 32 frac14 124 550 682

                                                                                        (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                                        (5)1

                                                                                        2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                                        (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                                        (7) 1

                                                                                        2 267

                                                                                        2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                                        (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                                        2 d frac14 163d d2thorn 650 82d2 1060d

                                                                                        Tie rod force per m frac14 T 0 0

                                                                                        XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                                        d3 thorn 77d2 269d 1438 frac14 0

                                                                                        d frac14 467m

                                                                                        Depth of penetration frac14 12d frac14 560m

                                                                                        Lateral earth pressure 41

                                                                                        Algebraic sum of forces for d frac14 467m isX

                                                                                        F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                                        T frac14 905 kN=m

                                                                                        Force in each tie rod frac14 25T frac14 226 kN

                                                                                        68

                                                                                        (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                                        0 frac14 21 98 frac14 112 kN=m3

                                                                                        The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                                        uC frac14 150

                                                                                        165 15 98 frac14 134 kN=m2

                                                                                        The average seepage pressure is

                                                                                        j frac14 15

                                                                                        165 98 frac14 09 kN=m3

                                                                                        Hence

                                                                                        0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                                        0 j frac14 112 09 frac14 103 kN=m3

                                                                                        Figure Q67

                                                                                        42 Lateral earth pressure

                                                                                        Consider moments about the anchor point A (per m)

                                                                                        Force (kN) Arm (m) Moment (kN m)

                                                                                        (1) 10 026 150 frac14 390 60 2340

                                                                                        (2)1

                                                                                        2 026 18 452 frac14 474 15 711

                                                                                        (3) 026 18 45 105 frac14 2211 825 18240

                                                                                        (4)1

                                                                                        2 026 121 1052 frac14 1734 100 17340

                                                                                        (5)1

                                                                                        2 134 15 frac14 101 40 404

                                                                                        (6) 134 30 frac14 402 60 2412

                                                                                        (7)1

                                                                                        2 134 60 frac14 402 95 3819

                                                                                        571 4527(8) Ppm

                                                                                        115 115PPm

                                                                                        XM frac14 0

                                                                                        Ppm frac144527

                                                                                        115frac14 394 kN=m

                                                                                        Available passive resistance

                                                                                        Pp frac14 1

                                                                                        2 385 103 62 frac14 714 kN=m

                                                                                        Factor of safety

                                                                                        Fp frac14 Pp

                                                                                        Ppm

                                                                                        frac14 714

                                                                                        394frac14 18

                                                                                        Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                                        Figure Q68

                                                                                        Lateral earth pressure 43

                                                                                        (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                                        Consider moments (per m) about the tie point A

                                                                                        Force (kN) Arm (m)

                                                                                        (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                                        (2)1

                                                                                        2 033 18 452 frac14 601 15

                                                                                        (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                                        (4)1

                                                                                        2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                                        (5)1

                                                                                        2 134 15 frac14 101 40

                                                                                        (6) 134 30 frac14 402 60

                                                                                        (7)1

                                                                                        2 134 d frac14 67d d3thorn 75

                                                                                        (8) 1

                                                                                        2 30 103 d2 frac141545d2 2d3thorn 75

                                                                                        Moment (kN m)

                                                                                        (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                                        XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                                        d3 thorn 827d2 466d 1518 frac14 0

                                                                                        By trial

                                                                                        d frac14 544m

                                                                                        The minimum depth of embedment required is 544m

                                                                                        69

                                                                                        For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                                        0 frac14 20 98 frac14 102 kN=m3

                                                                                        The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                                        44 Lateral earth pressure

                                                                                        uC frac14 147

                                                                                        173 26 98 frac14 216 kN=m2

                                                                                        and the average seepage pressure around the wall is

                                                                                        j frac14 26

                                                                                        173 98 frac14 15 kN=m3

                                                                                        Consider moments about the prop (A) (per m)

                                                                                        Force (kN) Arm (m) Moment (kN m)

                                                                                        (1)1

                                                                                        2 03 17 272 frac14 186 020 37

                                                                                        (2) 03 17 27 53 frac14 730 335 2445

                                                                                        (3)1

                                                                                        2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                                        (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                                        (5)1

                                                                                        2 216 26 frac14 281 243 684

                                                                                        (6) 216 27 frac14 583 465 2712

                                                                                        (7)1

                                                                                        2 216 60 frac14 648 800 5184

                                                                                        3055(8)

                                                                                        1

                                                                                        2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                                        Factor of safety

                                                                                        Fr frac14 6885

                                                                                        3055frac14 225

                                                                                        Figure Q69

                                                                                        Lateral earth pressure 45

                                                                                        610

                                                                                        For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                                        p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                                        Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                                        Using the recommendations of Twine and Roscoe

                                                                                        p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                                        Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                                        611

                                                                                        frac14 18 kN=m3 0 frac14 34

                                                                                        H frac14 350m nH frac14 335m mH frac14 185m

                                                                                        Consider a trial value of F frac14 20 Refer to Figure 635

                                                                                        0m frac14 tan1tan 34

                                                                                        20

                                                                                        frac14 186

                                                                                        Then

                                                                                        frac14 45 thorn 0m2frac14 543

                                                                                        W frac14 1

                                                                                        2 18 3502 cot 543 frac14 792 kN=m

                                                                                        Figure Q610

                                                                                        46 Lateral earth pressure

                                                                                        P frac14 1

                                                                                        2 s 3352 frac14 561s kN=m

                                                                                        U frac14 1

                                                                                        2 98 1852 cosec 543 frac14 206 kN=m

                                                                                        Equations 630 and 631 then become

                                                                                        561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                                        792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                                        ie

                                                                                        561s 0616N 405 frac14 0

                                                                                        792 0857N thorn 563 frac14 0

                                                                                        N frac14 848

                                                                                        0857frac14 989 kN=m

                                                                                        Then

                                                                                        561s 609 405 frac14 0

                                                                                        s frac14 649

                                                                                        561frac14 116 kN=m3

                                                                                        The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                                        F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                                        20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                                        s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                                        Figure Q611

                                                                                        Lateral earth pressure 47

                                                                                        612

                                                                                        For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                        45 thorn 0

                                                                                        2frac14 63

                                                                                        For the retained material between the surface and a depth of 36m

                                                                                        Pa frac14 1

                                                                                        2 030 18 362 frac14 350 kN=m

                                                                                        Weight of reinforced fill between the surface and a depth of 36m is

                                                                                        Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                        eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                        Eccentricity of Rv

                                                                                        e frac14 263 250 frac14 013m

                                                                                        The average vertical stress at a depth of 36m is

                                                                                        z frac14 Rv

                                                                                        L 2efrac14 324

                                                                                        474frac14 68 kN=m2

                                                                                        (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                        Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                        Tensile stress in the element frac14 138 103

                                                                                        65 3frac14 71N=mm2

                                                                                        Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                        Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                        Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                        The weight of ABC is

                                                                                        W frac14 1

                                                                                        2 18 52 265 frac14 124 kN=m

                                                                                        From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                        48 Lateral earth pressure

                                                                                        (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                        Tp frac14 032 68 120 065 frac14 170 kN

                                                                                        Tr frac14 213 420

                                                                                        418frac14 214 kN

                                                                                        Again the tensile failure and slipping limit states are satisfied for this element

                                                                                        Figure Q612

                                                                                        Lateral earth pressure 49

                                                                                        Chapter 7

                                                                                        Consolidation theory

                                                                                        71

                                                                                        Total change in thickness

                                                                                        H frac14 782 602 frac14 180mm

                                                                                        Average thickness frac14 1530thorn 180

                                                                                        2frac14 1620mm

                                                                                        Length of drainage path d frac14 1620

                                                                                        2frac14 810mm

                                                                                        Root time plot (Figure Q71a)

                                                                                        ffiffiffiffiffiffit90p frac14 33

                                                                                        t90 frac14 109min

                                                                                        cv frac14 0848d2

                                                                                        t90frac14 0848 8102

                                                                                        109 1440 365

                                                                                        106frac14 27m2=year

                                                                                        r0 frac14 782 764

                                                                                        782 602frac14 018

                                                                                        180frac14 0100

                                                                                        rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                        10 119

                                                                                        9 180frac14 0735

                                                                                        rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                        Log time plot (Figure Q71b)

                                                                                        t50 frac14 26min

                                                                                        cv frac14 0196d2

                                                                                        t50frac14 0196 8102

                                                                                        26 1440 365

                                                                                        106frac14 26m2=year

                                                                                        r0 frac14 782 763

                                                                                        782 602frac14 019

                                                                                        180frac14 0106

                                                                                        rp frac14 763 623

                                                                                        782 602frac14 140

                                                                                        180frac14 0778

                                                                                        rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                        Figure Q71(a)

                                                                                        Figure Q71(b)

                                                                                        Final void ratio

                                                                                        e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                        e

                                                                                        Hfrac14 1thorn e0

                                                                                        H0frac14 1thorn e1 thorne

                                                                                        H0

                                                                                        ie

                                                                                        e

                                                                                        180frac14 1631thorne

                                                                                        1710

                                                                                        e frac14 2936

                                                                                        1530frac14 0192

                                                                                        Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                        mv frac14 1

                                                                                        1thorn e0 e0 e101 00

                                                                                        frac14 1

                                                                                        1823 0192

                                                                                        0107frac14 098m2=MN

                                                                                        k frac14 cvmvw frac14 265 098 98

                                                                                        60 1440 365 103frac14 81 1010 m=s

                                                                                        72

                                                                                        Using Equation 77 (one-dimensional method)

                                                                                        sc frac14 e0 e11thorn e0 H

                                                                                        Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                        Figure Q72

                                                                                        52 Consolidation theory

                                                                                        Settlement

                                                                                        Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                        1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                        318

                                                                                        Notes 5 92y 460thorn 84

                                                                                        Heave

                                                                                        Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                        1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                        38

                                                                                        73

                                                                                        U frac14 f ethTvTHORN frac14 f cvt

                                                                                        d2

                                                                                        Hence if cv is constant

                                                                                        t1

                                                                                        t2frac14 d

                                                                                        21

                                                                                        d22

                                                                                        where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                        d1 frac14 95mm and d2 frac14 2500mm

                                                                                        for U frac14 050 t2 frac14 t1 d22

                                                                                        d21

                                                                                        frac14 20

                                                                                        60 24 365 25002

                                                                                        952frac14 263 years

                                                                                        for U lt 060 Tv frac14

                                                                                        4U2 (Equation 724(a))

                                                                                        t030 frac14 t050 0302

                                                                                        0502

                                                                                        frac14 263 036 frac14 095 years

                                                                                        Consolidation theory 53

                                                                                        74

                                                                                        The layer is open

                                                                                        d frac14 8

                                                                                        2frac14 4m

                                                                                        Tv frac14 cvtd2frac14 24 3

                                                                                        42frac14 0450

                                                                                        ui frac14 frac14 84 kN=m2

                                                                                        The excess pore water pressure is given by Equation 721

                                                                                        ue frac14Xmfrac141mfrac140

                                                                                        2ui

                                                                                        Msin

                                                                                        Mz

                                                                                        d

                                                                                        expethM2TvTHORN

                                                                                        In this case z frac14 d

                                                                                        sinMz

                                                                                        d

                                                                                        frac14 sinM

                                                                                        where

                                                                                        M frac14

                                                                                        23

                                                                                        25

                                                                                        2

                                                                                        M sin M M2Tv exp (M2Tv)

                                                                                        2thorn1 1110 0329

                                                                                        3

                                                                                        21 9993 457 105

                                                                                        ue frac14 2 84 2

                                                                                        1 0329 ethother terms negligibleTHORN

                                                                                        frac14 352 kN=m2

                                                                                        75

                                                                                        The layer is open

                                                                                        d frac14 6

                                                                                        2frac14 3m

                                                                                        Tv frac14 cvtd2frac14 10 3

                                                                                        32frac14 0333

                                                                                        The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                        54 Consolidation theory

                                                                                        For an open layer

                                                                                        Tv frac14 4n

                                                                                        m2

                                                                                        n frac14 0333 62

                                                                                        4frac14 300

                                                                                        The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                        ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                        i j

                                                                                        0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                        0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                        The initial and 3-year isochrones are plotted in Figure Q75

                                                                                        Area under initial isochrone frac14 180 units

                                                                                        Area under 3-year isochrone frac14 63 units

                                                                                        The average degree of consolidation is given by Equation 725Thus

                                                                                        U frac14 1 63

                                                                                        180frac14 065

                                                                                        Figure Q75

                                                                                        Consolidation theory 55

                                                                                        76

                                                                                        At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                        0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                        The final consolidation settlement (one-dimensional method) is

                                                                                        sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                        Corrected time t frac14 2 1

                                                                                        2

                                                                                        40

                                                                                        52

                                                                                        frac14 1615 years

                                                                                        Tv frac14 cvtd2frac14 44 1615

                                                                                        42frac14 0444

                                                                                        From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                        77

                                                                                        The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                        Figure Q77

                                                                                        56 Consolidation theory

                                                                                        Point m n Ir (kNm2) sc (mm)

                                                                                        13020frac14 15 20

                                                                                        20frac14 10 0194 (4) 113 124

                                                                                        260

                                                                                        20frac14 30

                                                                                        20

                                                                                        20frac14 10 0204 (2) 59 65

                                                                                        360

                                                                                        20frac14 30

                                                                                        40

                                                                                        20frac14 20 0238 (1) 35 38

                                                                                        430

                                                                                        20frac14 15

                                                                                        40

                                                                                        20frac14 20 0224 (2) 65 72

                                                                                        Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                        78

                                                                                        Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                        (a) Immediate settlement

                                                                                        H

                                                                                        Bfrac14 30

                                                                                        35frac14 086

                                                                                        D

                                                                                        Bfrac14 2

                                                                                        35frac14 006

                                                                                        Figure Q78

                                                                                        Consolidation theory 57

                                                                                        From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                        si frac14 130131qB

                                                                                        Eufrac14 10 032 105 35

                                                                                        40frac14 30mm

                                                                                        (b) Consolidation settlement

                                                                                        Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                        1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                        3150

                                                                                        Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                        Now

                                                                                        H

                                                                                        Bfrac14 30

                                                                                        35frac14 086 and A frac14 065

                                                                                        from Figure 712 13 frac14 079

                                                                                        sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                        Total settlement

                                                                                        s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                        79

                                                                                        Without sand drains

                                                                                        Uv frac14 025

                                                                                        Tv frac14 0049 ethfrom Figure 718THORN

                                                                                        t frac14 Tvd2

                                                                                        cvfrac14 0049 82

                                                                                        cvWith sand drains

                                                                                        R frac14 0564S frac14 0564 3 frac14 169m

                                                                                        n frac14 Rrfrac14 169

                                                                                        015frac14 113

                                                                                        Tr frac14 cht

                                                                                        4R2frac14 ch

                                                                                        4 1692 0049 82

                                                                                        cvethand ch frac14 cvTHORN

                                                                                        frac14 0275

                                                                                        Ur frac14 073 (from Figure 730)

                                                                                        58 Consolidation theory

                                                                                        Using Equation 740

                                                                                        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                        U frac14 080

                                                                                        710

                                                                                        Without sand drains

                                                                                        Uv frac14 090

                                                                                        Tv frac14 0848

                                                                                        t frac14 Tvd2

                                                                                        cvfrac14 0848 102

                                                                                        96frac14 88 years

                                                                                        With sand drains

                                                                                        R frac14 0564S frac14 0564 4 frac14 226m

                                                                                        n frac14 Rrfrac14 226

                                                                                        015frac14 15

                                                                                        Tr

                                                                                        Tvfrac14 chcv

                                                                                        d2

                                                                                        4R2ethsame tTHORN

                                                                                        Tr

                                                                                        Tvfrac14 140

                                                                                        96 102

                                                                                        4 2262frac14 714 eth1THORN

                                                                                        Using Equation 740

                                                                                        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                        An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                        Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                        040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                        Thus

                                                                                        Uv frac14 0295 and Ur frac14 086

                                                                                        t frac14 88 00683

                                                                                        0848frac14 07 years

                                                                                        Consolidation theory 59

                                                                                        Chapter 8

                                                                                        Bearing capacity

                                                                                        81

                                                                                        (a) The ultimate bearing capacity is given by Equation 83

                                                                                        qf frac14 cNc thorn DNq thorn 1

                                                                                        2BN

                                                                                        For u frac14 0

                                                                                        Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                        qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                        The net ultimate bearing capacity is

                                                                                        qnf frac14 qf D frac14 540 kN=m2

                                                                                        The net foundation pressure is

                                                                                        qn frac14 q D frac14 425

                                                                                        2 eth21 1THORN frac14 192 kN=m2

                                                                                        The factor of safety (Equation 86) is

                                                                                        F frac14 qnfqnfrac14 540

                                                                                        192frac14 28

                                                                                        (b) For 0 frac14 28

                                                                                        Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                        qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                        2 112 2 13

                                                                                        frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                        qnf frac14 574 112 frac14 563 kN=m2

                                                                                        F frac14 563

                                                                                        192frac14 29

                                                                                        (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                        82

                                                                                        For 0 frac14 38

                                                                                        Nq frac14 49 N frac14 67

                                                                                        qnf frac14 DethNq 1THORN thorn 1

                                                                                        2BN ethfrom Equation 83THORN

                                                                                        frac14 eth18 075 48THORN thorn 1

                                                                                        2 18 15 67

                                                                                        frac14 648thorn 905 frac14 1553 kN=m2

                                                                                        qn frac14 500

                                                                                        15 eth18 075THORN frac14 320 kN=m2

                                                                                        F frac14 qnfqnfrac14 1553

                                                                                        320frac14 48

                                                                                        0d frac14 tan1tan 38

                                                                                        125

                                                                                        frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                        Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                        2 18 15 25

                                                                                        frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                        Design load (action) Vd frac14 500 kN=m

                                                                                        The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                        83

                                                                                        D

                                                                                        Bfrac14 350

                                                                                        225frac14 155

                                                                                        From Figure 85 for a square foundation

                                                                                        Nc frac14 81

                                                                                        Bearing capacity 61

                                                                                        For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                        Nc frac14 084thorn 016B

                                                                                        L

                                                                                        81 frac14 745

                                                                                        Using Equation 810

                                                                                        qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                        For F frac14 3

                                                                                        qn frac14 1006

                                                                                        3frac14 335 kN=m2

                                                                                        q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                        Design load frac14 405 450 225 frac14 4100 kN

                                                                                        Design undrained strength cud frac14 135

                                                                                        14frac14 96 kN=m2

                                                                                        Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                        frac14 7241 kN

                                                                                        Design load Vd frac14 4100 kN

                                                                                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                        84

                                                                                        For 0 frac14 40

                                                                                        Nq frac14 64 N frac14 95

                                                                                        qnf frac14 DethNq 1THORN thorn 04BN

                                                                                        (a) Water table 5m below ground level

                                                                                        qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                        qn frac14 400 17 frac14 383 kN=m2

                                                                                        F frac14 2686

                                                                                        383frac14 70

                                                                                        (b) Water table 1m below ground level (ie at foundation level)

                                                                                        0 frac14 20 98 frac14 102 kN=m3

                                                                                        62 Bearing capacity

                                                                                        qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                        F frac14 2040

                                                                                        383frac14 53

                                                                                        (c) Water table at ground level with upward hydraulic gradient 02

                                                                                        eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                        qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                        F frac14 1296

                                                                                        392frac14 33

                                                                                        85

                                                                                        The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                        Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                        Design value of 0 frac14 tan1tan 39

                                                                                        125

                                                                                        frac14 33

                                                                                        For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                        Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                        86

                                                                                        (a) Undrained shear for u frac14 0

                                                                                        Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                        qnf frac14 12cuNc

                                                                                        frac14 12 100 514 frac14 617 kN=m2

                                                                                        qn frac14 qnfFfrac14 617

                                                                                        3frac14 206 kN=m2

                                                                                        q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                        Bearing capacity 63

                                                                                        Drained shear for 0 frac14 32

                                                                                        Nq frac14 23 N frac14 25

                                                                                        0 frac14 21 98 frac14 112 kN=m3

                                                                                        qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                        frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                        frac14 694 kN=m2

                                                                                        q frac14 694

                                                                                        3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                        Design load frac14 42 227 frac14 3632 kN

                                                                                        (b) Design undrained strength cud frac14 100

                                                                                        14frac14 71 kNm2

                                                                                        Design bearing resistance Rd frac14 12cudNe area

                                                                                        frac14 12 71 514 42

                                                                                        frac14 7007 kN

                                                                                        For drained shear 0d frac14 tan1tan 32

                                                                                        125

                                                                                        frac14 26

                                                                                        Nq frac14 12 N frac14 10

                                                                                        Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                        (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                        Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                        1 2 100 0175 0700qn 0182qn

                                                                                        2 6 033 0044 0176qn 0046qn

                                                                                        3 10 020 0017 0068qn 0018qn

                                                                                        0246qn

                                                                                        Diameter of equivalent circle B frac14 45m

                                                                                        H

                                                                                        Bfrac14 12

                                                                                        45frac14 27 and A frac14 042

                                                                                        13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                        64 Bearing capacity

                                                                                        For sc frac14 30mm

                                                                                        qn frac14 30

                                                                                        0147frac14 204 kN=m2

                                                                                        q frac14 204thorn 21 frac14 225 kN=m2

                                                                                        Design load frac14 42 225 frac14 3600 kN

                                                                                        The design load is 3600 kN settlement being the limiting criterion

                                                                                        87

                                                                                        D

                                                                                        Bfrac14 8

                                                                                        4frac14 20

                                                                                        From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                        F frac14 cuNc

                                                                                        Dfrac14 40 71

                                                                                        20 8frac14 18

                                                                                        88

                                                                                        Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                        Design value of 0 frac14 tan1tan 38

                                                                                        125

                                                                                        frac14 32

                                                                                        Figure Q86

                                                                                        Bearing capacity 65

                                                                                        For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                        Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                        The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                        Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                        For B frac14 250m qn frac14 3750

                                                                                        2502 17 frac14 583 kN=m2

                                                                                        From Figure 510 m frac14 n frac14 126

                                                                                        6frac14 021

                                                                                        Ir frac14 0019

                                                                                        Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                        Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                        The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                        89

                                                                                        Depth (m) N 0v (kNm2) CN N1

                                                                                        070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                        Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                        (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                        Cw frac14 05thorn 0530

                                                                                        47

                                                                                        frac14 082

                                                                                        66 Bearing capacity

                                                                                        Thus

                                                                                        qa frac14 150 082 frac14 120 kN=m2

                                                                                        (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                        Thus

                                                                                        qa frac14 90 15 frac14 135 kN=m2

                                                                                        (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                        Ic frac14 171

                                                                                        1014frac14 0068

                                                                                        From Equation 819(a) with s frac14 25mm

                                                                                        q frac14 25

                                                                                        3507 0068frac14 150 kN=m2

                                                                                        810

                                                                                        Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                        Izp frac14 05thorn 01130

                                                                                        16 27

                                                                                        05

                                                                                        frac14 067

                                                                                        Refer to Figure Q810

                                                                                        E frac14 25qc

                                                                                        Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                        Ez (mm3MN)

                                                                                        1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                        0203

                                                                                        C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                        130frac14 093

                                                                                        C2 frac14 1 ethsayTHORN

                                                                                        s frac14 C1C2qnX Iz

                                                                                        Ez frac14 093 1 130 0203 frac14 25mm

                                                                                        Bearing capacity 67

                                                                                        811

                                                                                        At pile base level

                                                                                        cu frac14 220 kN=m2

                                                                                        qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                        Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                        00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                        qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                        Then

                                                                                        Qf frac14 Abqb thorn Asqs

                                                                                        frac14

                                                                                        4 32 1980

                                                                                        thorn eth 105 139 86THORN

                                                                                        frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                        0 01 02 03 04 05 06 07

                                                                                        0 2 4 6 8 10 12 14

                                                                                        1

                                                                                        2

                                                                                        3

                                                                                        4

                                                                                        5

                                                                                        6

                                                                                        7

                                                                                        8

                                                                                        (1)

                                                                                        (2)

                                                                                        (3)

                                                                                        (4)

                                                                                        (5)

                                                                                        qc

                                                                                        qc

                                                                                        Iz

                                                                                        Iz

                                                                                        (MNm2)

                                                                                        z (m)

                                                                                        Figure Q810

                                                                                        68 Bearing capacity

                                                                                        Allowable load

                                                                                        ethaTHORN Qf

                                                                                        2frac14 17 937

                                                                                        2frac14 8968 kN

                                                                                        ethbTHORN Abqb

                                                                                        3thorn Asqs frac14 13 996

                                                                                        3thorn 3941 frac14 8606 kN

                                                                                        ie allowable load frac14 8600 kN

                                                                                        Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                        According to the limit state method

                                                                                        Characteristic undrained strength at base level cuk frac14 220

                                                                                        150kN=m2

                                                                                        Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                        150frac14 1320 kN=m2

                                                                                        Characteristic shaft resistance qsk frac14 00150

                                                                                        frac14 86

                                                                                        150frac14 57 kN=m2

                                                                                        Characteristic base and shaft resistances

                                                                                        Rbk frac14

                                                                                        4 32 1320 frac14 9330 kN

                                                                                        Rsk frac14 105 139 86

                                                                                        150frac14 2629 kN

                                                                                        For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                        Design bearing resistance Rcd frac14 9330

                                                                                        160thorn 2629

                                                                                        130

                                                                                        frac14 5831thorn 2022

                                                                                        frac14 7850 kN

                                                                                        Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                        812

                                                                                        ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                        qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                        For a single pile

                                                                                        Qf frac14 Abqb thorn Asqs

                                                                                        frac14

                                                                                        4 062 1305

                                                                                        thorn eth 06 15 42THORN

                                                                                        frac14 369thorn 1187 frac14 1556 kN

                                                                                        Bearing capacity 69

                                                                                        Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                        eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                        (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                        (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                        qbkfrac14 9cuk frac14 9 220

                                                                                        150frac14 1320 kN=m2

                                                                                        qskfrac14cuk frac14 040 105

                                                                                        150frac14 28 kN=m2

                                                                                        Rbkfrac14

                                                                                        4 0602 1320 frac14 373 kN

                                                                                        Rskfrac14 060 15 28 frac14 791 kN

                                                                                        Rcdfrac14 373

                                                                                        160thorn 791

                                                                                        130frac14 233thorn 608 frac14 841 kN

                                                                                        Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                        Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                        (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                        q frac14 21 000

                                                                                        1762frac14 68 kN=m2

                                                                                        Immediate settlement

                                                                                        H

                                                                                        Bfrac14 15

                                                                                        176frac14 085

                                                                                        D

                                                                                        Bfrac14 13

                                                                                        176frac14 074

                                                                                        L

                                                                                        Bfrac14 1

                                                                                        Hence from Figure 515

                                                                                        130 frac14 078 and 131 frac14 041

                                                                                        70 Bearing capacity

                                                                                        Thus using Equation 528

                                                                                        si frac14 078 041 68 176

                                                                                        65frac14 6mm

                                                                                        Consolidation settlement

                                                                                        Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                        1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                        434 (sod)

                                                                                        Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                        sc frac14 056 434 frac14 24mm

                                                                                        The total settlement is (6thorn 24) frac14 30mm

                                                                                        813

                                                                                        At base level N frac14 26 Then using Equation 830

                                                                                        qb frac14 40NDb

                                                                                        Bfrac14 40 26 2

                                                                                        025frac14 8320 kN=m2

                                                                                        ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                        Figure Q812

                                                                                        Bearing capacity 71

                                                                                        Over the length embedded in sand

                                                                                        N frac14 21 ie18thorn 24

                                                                                        2

                                                                                        Using Equation 831

                                                                                        qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                        For a single pile

                                                                                        Qf frac14 Abqb thorn Asqs

                                                                                        frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                        For the pile group assuming a group efficiency of 12

                                                                                        XQf frac14 12 9 604 frac14 6523 kN

                                                                                        Then the load factor is

                                                                                        F frac14 6523

                                                                                        2000thorn 1000frac14 21

                                                                                        (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                        Characteristic base resistance per unit area qbk frac14 8320

                                                                                        150frac14 5547 kNm2

                                                                                        Characteristic shaft resistance per unit area qsk frac14 42

                                                                                        150frac14 28 kNm2

                                                                                        Characteristic base and shaft resistances for a single pile

                                                                                        Rbk frac14 0252 5547 frac14 347 kN

                                                                                        Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                        For a driven pile the partial factors are b frac14 s frac14 130

                                                                                        Design bearing resistance Rcd frac14 347

                                                                                        130thorn 56

                                                                                        130frac14 310 kN

                                                                                        For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                        Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                        (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                        From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                        72 Bearing capacity

                                                                                        N frac14 24thorn 26thorn 34

                                                                                        3frac14 28

                                                                                        Ic frac14 171

                                                                                        2814frac14 0016 ethEquation 818THORN

                                                                                        s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                        814

                                                                                        Using Equation 841

                                                                                        Tf frac14 DLcu thorn

                                                                                        4ethD2 d2THORNcuNc

                                                                                        frac14 eth 02 5 06 110THORN thorn

                                                                                        4eth022 012THORN110 9

                                                                                        frac14 207thorn 23 frac14 230 kN

                                                                                        Figure Q813

                                                                                        Bearing capacity 73

                                                                                        Chapter 9

                                                                                        Stability of slopes

                                                                                        91

                                                                                        Referring to Figure Q91

                                                                                        W frac14 417 19 frac14 792 kN=m

                                                                                        Q frac14 20 28 frac14 56 kN=m

                                                                                        Arc lengthAB frac14

                                                                                        180 73 90 frac14 115m

                                                                                        Arc length BC frac14

                                                                                        180 28 90 frac14 44m

                                                                                        The factor of safety is given by

                                                                                        F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                        90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                        Depth of tension crack z0 frac14 2cu

                                                                                        frac14 2 20

                                                                                        19frac14 21m

                                                                                        Arc length BD frac14

                                                                                        180 13

                                                                                        1

                                                                                        2 90 frac14 21m

                                                                                        F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                        The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                        Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                        14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                        Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                        The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                        92

                                                                                        u frac14 0

                                                                                        Depth factor D frac14 11

                                                                                        9frac14 122

                                                                                        Using Equation 92 with F frac14 10

                                                                                        Ns frac14 cu

                                                                                        FHfrac14 30

                                                                                        10 19 9frac14 0175

                                                                                        Hence from Figure 93

                                                                                        frac14 50

                                                                                        For F frac14 12

                                                                                        Ns frac14 30

                                                                                        12 19 9frac14 0146

                                                                                        frac14 27

                                                                                        93

                                                                                        Refer to Figure Q93

                                                                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                        1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                        74 m

                                                                                        214 1deg

                                                                                        213 1deg

                                                                                        39 m

                                                                                        WB

                                                                                        D

                                                                                        C

                                                                                        28 m

                                                                                        21 m

                                                                                        A

                                                                                        Q

                                                                                        Soil (1)Soil (2)

                                                                                        73deg

                                                                                        Figure Q91

                                                                                        Stability of slopes 75

                                                                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                        9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                        599 256 328 1372

                                                                                        Figure Q93

                                                                                        76 Stability of slopes

                                                                                        XW cos frac14 b

                                                                                        Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                        W sin frac14 bX

                                                                                        h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                        Arc length La frac14

                                                                                        180 57

                                                                                        1

                                                                                        2 326 frac14 327m

                                                                                        The factor of safety is given by

                                                                                        F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                        W sin

                                                                                        frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                        frac14 091

                                                                                        According to the limit state method

                                                                                        0d frac14 tan1tan 32

                                                                                        125

                                                                                        frac14 265

                                                                                        c0 frac14 8

                                                                                        160frac14 5 kN=m2

                                                                                        Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                        Design disturbing moment frac14 1075 kN=m

                                                                                        The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                        94

                                                                                        F frac14 1

                                                                                        W sin

                                                                                        Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                        1thorn ethtan tan0=FTHORN

                                                                                        c0 frac14 8 kN=m2

                                                                                        0 frac14 32

                                                                                        c0b frac14 8 2 frac14 16 kN=m

                                                                                        W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                        Try F frac14 100

                                                                                        tan0

                                                                                        Ffrac14 0625

                                                                                        Stability of slopes 77

                                                                                        Values of u are as obtained in Figure Q93

                                                                                        SliceNo

                                                                                        h(m)

                                                                                        W frac14 bh(kNm)

                                                                                        W sin(kNm)

                                                                                        ub(kNm)

                                                                                        c0bthorn (W ub) tan0(kNm)

                                                                                        sec

                                                                                        1thorn (tan tan0)FProduct(kNm)

                                                                                        1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                        23 33 30 1042 31

                                                                                        3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                        224 92 72 0931 67

                                                                                        6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                        12 58 112 90 0889 80

                                                                                        8 60 252 1812

                                                                                        80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                        10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                        2154 88 116 0853 99

                                                                                        14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                        1074 1091

                                                                                        F frac14 1091

                                                                                        1074frac14 102 (assumed value 100)

                                                                                        Thus

                                                                                        F frac14 101

                                                                                        95

                                                                                        F frac14 1

                                                                                        W sin

                                                                                        XfWeth1 ruTHORN tan0g sec

                                                                                        1thorn ethtan tan0THORN=F

                                                                                        0 frac14 33

                                                                                        ru frac14 020

                                                                                        W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                        eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                        Try F frac14 110

                                                                                        tan 0

                                                                                        Ffrac14 tan 33

                                                                                        110frac14 0590

                                                                                        78 Stability of slopes

                                                                                        Referring to Figure Q95

                                                                                        SliceNo

                                                                                        h(m)

                                                                                        W frac14 bh(kNm)

                                                                                        W sin(kNm)

                                                                                        W(1 ru) tan0(kNm)

                                                                                        sec

                                                                                        1thorn ( tan tan0)FProduct(kNm)

                                                                                        1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                        2120 234 0892 209

                                                                                        4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                        1185 1271

                                                                                        Figure Q95

                                                                                        Stability of slopes 79

                                                                                        F frac14 1271

                                                                                        1185frac14 107

                                                                                        The trial value was 110 therefore take F to be 108

                                                                                        96

                                                                                        (a) Water table at surface the factor of safety is given by Equation 912

                                                                                        F frac14 0

                                                                                        sat

                                                                                        tan0

                                                                                        tan

                                                                                        ptie 15 frac14 92

                                                                                        19

                                                                                        tan 36

                                                                                        tan

                                                                                        tan frac14 0234

                                                                                        frac14 13

                                                                                        Water table well below surface the factor of safety is given by Equation 911

                                                                                        F frac14 tan0

                                                                                        tan

                                                                                        frac14 tan 36

                                                                                        tan 13

                                                                                        frac14 31

                                                                                        (b) 0d frac14 tan1tan 36

                                                                                        125

                                                                                        frac14 30

                                                                                        Depth of potential failure surface frac14 z

                                                                                        Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                        frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                        frac14 504z kN

                                                                                        Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                        frac14 19 z sin 13 cos 13

                                                                                        frac14 416z kN

                                                                                        Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                        80 Stability of slopes

                                                                                        • Book Cover
                                                                                        • Title
                                                                                        • Contents
                                                                                        • Basic characteristics of soils
                                                                                        • Seepage
                                                                                        • Effective stress
                                                                                        • Shear strength
                                                                                        • Stresses and displacements
                                                                                        • Lateral earth pressure
                                                                                        • Consolidation theory
                                                                                        • Bearing capacity
                                                                                        • Stability of slopes

                                                                                          XM frac14 283d3 146d2 thorn 80d thorn 488 frac14 0

                                                                                          d3 thorn 516d2 283d 1724 frac14 0

                                                                                          d frac14 179m

                                                                                          Depth of penetration frac14 12eth179thorn 150THORN frac14 395mXF frac14 0 hence R frac14 715 kN ethsubstituting d frac14 179mTHORN

                                                                                          Over additional 20 embedded depth

                                                                                          pp pa frac14 eth385 17 45THORN eth026 17 15THORN thorn eth385 026THORNeth102 212THORNfrac14 3655 kN=m2

                                                                                          Net passive resistance frac14 3655 066 frac14 241 kN ethgtRTHORN

                                                                                          66

                                                                                          The design value of 0 frac14 36 ie the partial factor has been appliedThe active pressure coefficient is given by Equation 617 in which frac14 105 frac14 20 frac14 36 and frac14 25

                                                                                          Ka frac14 sin 69=sin 105

                                                                                          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 130p thorn

                                                                                          ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi( sin 61 sin 16)

                                                                                          pffiffiffiffiffiffiffiffiffiffiffiffiffiffisin 85p

                                                                                          26664

                                                                                          37775

                                                                                          2

                                                                                          frac14 050

                                                                                          The total active thrust (acting at 25 above the normal) is given by Equation 616

                                                                                          Pa frac14 1

                                                                                          2 050 19 7502 frac14 267 kN=m

                                                                                          Figure Q65

                                                                                          Lateral earth pressure 39

                                                                                          Horizontal component

                                                                                          Ph frac14 267 cos 40 frac14 205 kN=m

                                                                                          Vertical component

                                                                                          Pv frac14 267 sin 40 frac14 172 kN=m

                                                                                          Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                                          Force (kN) Arm (m) Moment (kN m)

                                                                                          (1)1

                                                                                          2 175 650 235 frac14 1337 258 345

                                                                                          (2) 050 650 235 frac14 764 175 134

                                                                                          (3)1

                                                                                          2 070 650 235 frac14 535 127 68

                                                                                          (4) 100 400 235 frac14 940 200 188

                                                                                          (5) 1

                                                                                          2 080 050 235 frac14 47 027 1

                                                                                          Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                                          Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                                          Lever arm of base resultant

                                                                                          M

                                                                                          Vfrac14 795

                                                                                          525frac14 151m

                                                                                          Eccentricity of base resultant

                                                                                          e frac14 200 151 frac14 049m

                                                                                          Figure Q66

                                                                                          40 Lateral earth pressure

                                                                                          Base pressures (Equation 627)

                                                                                          p frac14 525

                                                                                          41 6 049

                                                                                          4

                                                                                          frac14 228 kN=m2 and 35 kN=m2

                                                                                          The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                                          The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                                          The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                                          67

                                                                                          For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                                          Force (kN) Arm (m) Moment (kNm)

                                                                                          (1)1

                                                                                          2 027 17 52 frac14 574 183 1050

                                                                                          (2) 027 17 5 3 frac14 689 500 3445

                                                                                          (3)1

                                                                                          2 027 102 32 frac14 124 550 682

                                                                                          (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                                          (5)1

                                                                                          2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                                          (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                                          (7) 1

                                                                                          2 267

                                                                                          2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                                          (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                                          2 d frac14 163d d2thorn 650 82d2 1060d

                                                                                          Tie rod force per m frac14 T 0 0

                                                                                          XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                                          d3 thorn 77d2 269d 1438 frac14 0

                                                                                          d frac14 467m

                                                                                          Depth of penetration frac14 12d frac14 560m

                                                                                          Lateral earth pressure 41

                                                                                          Algebraic sum of forces for d frac14 467m isX

                                                                                          F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                                          T frac14 905 kN=m

                                                                                          Force in each tie rod frac14 25T frac14 226 kN

                                                                                          68

                                                                                          (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                                          0 frac14 21 98 frac14 112 kN=m3

                                                                                          The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                                          uC frac14 150

                                                                                          165 15 98 frac14 134 kN=m2

                                                                                          The average seepage pressure is

                                                                                          j frac14 15

                                                                                          165 98 frac14 09 kN=m3

                                                                                          Hence

                                                                                          0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                                          0 j frac14 112 09 frac14 103 kN=m3

                                                                                          Figure Q67

                                                                                          42 Lateral earth pressure

                                                                                          Consider moments about the anchor point A (per m)

                                                                                          Force (kN) Arm (m) Moment (kN m)

                                                                                          (1) 10 026 150 frac14 390 60 2340

                                                                                          (2)1

                                                                                          2 026 18 452 frac14 474 15 711

                                                                                          (3) 026 18 45 105 frac14 2211 825 18240

                                                                                          (4)1

                                                                                          2 026 121 1052 frac14 1734 100 17340

                                                                                          (5)1

                                                                                          2 134 15 frac14 101 40 404

                                                                                          (6) 134 30 frac14 402 60 2412

                                                                                          (7)1

                                                                                          2 134 60 frac14 402 95 3819

                                                                                          571 4527(8) Ppm

                                                                                          115 115PPm

                                                                                          XM frac14 0

                                                                                          Ppm frac144527

                                                                                          115frac14 394 kN=m

                                                                                          Available passive resistance

                                                                                          Pp frac14 1

                                                                                          2 385 103 62 frac14 714 kN=m

                                                                                          Factor of safety

                                                                                          Fp frac14 Pp

                                                                                          Ppm

                                                                                          frac14 714

                                                                                          394frac14 18

                                                                                          Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                                          Figure Q68

                                                                                          Lateral earth pressure 43

                                                                                          (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                                          Consider moments (per m) about the tie point A

                                                                                          Force (kN) Arm (m)

                                                                                          (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                                          (2)1

                                                                                          2 033 18 452 frac14 601 15

                                                                                          (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                                          (4)1

                                                                                          2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                                          (5)1

                                                                                          2 134 15 frac14 101 40

                                                                                          (6) 134 30 frac14 402 60

                                                                                          (7)1

                                                                                          2 134 d frac14 67d d3thorn 75

                                                                                          (8) 1

                                                                                          2 30 103 d2 frac141545d2 2d3thorn 75

                                                                                          Moment (kN m)

                                                                                          (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                                          XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                                          d3 thorn 827d2 466d 1518 frac14 0

                                                                                          By trial

                                                                                          d frac14 544m

                                                                                          The minimum depth of embedment required is 544m

                                                                                          69

                                                                                          For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                                          0 frac14 20 98 frac14 102 kN=m3

                                                                                          The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                                          44 Lateral earth pressure

                                                                                          uC frac14 147

                                                                                          173 26 98 frac14 216 kN=m2

                                                                                          and the average seepage pressure around the wall is

                                                                                          j frac14 26

                                                                                          173 98 frac14 15 kN=m3

                                                                                          Consider moments about the prop (A) (per m)

                                                                                          Force (kN) Arm (m) Moment (kN m)

                                                                                          (1)1

                                                                                          2 03 17 272 frac14 186 020 37

                                                                                          (2) 03 17 27 53 frac14 730 335 2445

                                                                                          (3)1

                                                                                          2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                                          (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                                          (5)1

                                                                                          2 216 26 frac14 281 243 684

                                                                                          (6) 216 27 frac14 583 465 2712

                                                                                          (7)1

                                                                                          2 216 60 frac14 648 800 5184

                                                                                          3055(8)

                                                                                          1

                                                                                          2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                                          Factor of safety

                                                                                          Fr frac14 6885

                                                                                          3055frac14 225

                                                                                          Figure Q69

                                                                                          Lateral earth pressure 45

                                                                                          610

                                                                                          For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                                          p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                                          Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                                          Using the recommendations of Twine and Roscoe

                                                                                          p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                                          Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                                          611

                                                                                          frac14 18 kN=m3 0 frac14 34

                                                                                          H frac14 350m nH frac14 335m mH frac14 185m

                                                                                          Consider a trial value of F frac14 20 Refer to Figure 635

                                                                                          0m frac14 tan1tan 34

                                                                                          20

                                                                                          frac14 186

                                                                                          Then

                                                                                          frac14 45 thorn 0m2frac14 543

                                                                                          W frac14 1

                                                                                          2 18 3502 cot 543 frac14 792 kN=m

                                                                                          Figure Q610

                                                                                          46 Lateral earth pressure

                                                                                          P frac14 1

                                                                                          2 s 3352 frac14 561s kN=m

                                                                                          U frac14 1

                                                                                          2 98 1852 cosec 543 frac14 206 kN=m

                                                                                          Equations 630 and 631 then become

                                                                                          561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                                          792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                                          ie

                                                                                          561s 0616N 405 frac14 0

                                                                                          792 0857N thorn 563 frac14 0

                                                                                          N frac14 848

                                                                                          0857frac14 989 kN=m

                                                                                          Then

                                                                                          561s 609 405 frac14 0

                                                                                          s frac14 649

                                                                                          561frac14 116 kN=m3

                                                                                          The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                                          F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                                          20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                                          s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                                          Figure Q611

                                                                                          Lateral earth pressure 47

                                                                                          612

                                                                                          For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                          45 thorn 0

                                                                                          2frac14 63

                                                                                          For the retained material between the surface and a depth of 36m

                                                                                          Pa frac14 1

                                                                                          2 030 18 362 frac14 350 kN=m

                                                                                          Weight of reinforced fill between the surface and a depth of 36m is

                                                                                          Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                          eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                          Eccentricity of Rv

                                                                                          e frac14 263 250 frac14 013m

                                                                                          The average vertical stress at a depth of 36m is

                                                                                          z frac14 Rv

                                                                                          L 2efrac14 324

                                                                                          474frac14 68 kN=m2

                                                                                          (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                          Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                          Tensile stress in the element frac14 138 103

                                                                                          65 3frac14 71N=mm2

                                                                                          Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                          Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                          Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                          The weight of ABC is

                                                                                          W frac14 1

                                                                                          2 18 52 265 frac14 124 kN=m

                                                                                          From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                          48 Lateral earth pressure

                                                                                          (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                          Tp frac14 032 68 120 065 frac14 170 kN

                                                                                          Tr frac14 213 420

                                                                                          418frac14 214 kN

                                                                                          Again the tensile failure and slipping limit states are satisfied for this element

                                                                                          Figure Q612

                                                                                          Lateral earth pressure 49

                                                                                          Chapter 7

                                                                                          Consolidation theory

                                                                                          71

                                                                                          Total change in thickness

                                                                                          H frac14 782 602 frac14 180mm

                                                                                          Average thickness frac14 1530thorn 180

                                                                                          2frac14 1620mm

                                                                                          Length of drainage path d frac14 1620

                                                                                          2frac14 810mm

                                                                                          Root time plot (Figure Q71a)

                                                                                          ffiffiffiffiffiffit90p frac14 33

                                                                                          t90 frac14 109min

                                                                                          cv frac14 0848d2

                                                                                          t90frac14 0848 8102

                                                                                          109 1440 365

                                                                                          106frac14 27m2=year

                                                                                          r0 frac14 782 764

                                                                                          782 602frac14 018

                                                                                          180frac14 0100

                                                                                          rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                          10 119

                                                                                          9 180frac14 0735

                                                                                          rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                          Log time plot (Figure Q71b)

                                                                                          t50 frac14 26min

                                                                                          cv frac14 0196d2

                                                                                          t50frac14 0196 8102

                                                                                          26 1440 365

                                                                                          106frac14 26m2=year

                                                                                          r0 frac14 782 763

                                                                                          782 602frac14 019

                                                                                          180frac14 0106

                                                                                          rp frac14 763 623

                                                                                          782 602frac14 140

                                                                                          180frac14 0778

                                                                                          rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                          Figure Q71(a)

                                                                                          Figure Q71(b)

                                                                                          Final void ratio

                                                                                          e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                          e

                                                                                          Hfrac14 1thorn e0

                                                                                          H0frac14 1thorn e1 thorne

                                                                                          H0

                                                                                          ie

                                                                                          e

                                                                                          180frac14 1631thorne

                                                                                          1710

                                                                                          e frac14 2936

                                                                                          1530frac14 0192

                                                                                          Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                          mv frac14 1

                                                                                          1thorn e0 e0 e101 00

                                                                                          frac14 1

                                                                                          1823 0192

                                                                                          0107frac14 098m2=MN

                                                                                          k frac14 cvmvw frac14 265 098 98

                                                                                          60 1440 365 103frac14 81 1010 m=s

                                                                                          72

                                                                                          Using Equation 77 (one-dimensional method)

                                                                                          sc frac14 e0 e11thorn e0 H

                                                                                          Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                          Figure Q72

                                                                                          52 Consolidation theory

                                                                                          Settlement

                                                                                          Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                          1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                          318

                                                                                          Notes 5 92y 460thorn 84

                                                                                          Heave

                                                                                          Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                          1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                          38

                                                                                          73

                                                                                          U frac14 f ethTvTHORN frac14 f cvt

                                                                                          d2

                                                                                          Hence if cv is constant

                                                                                          t1

                                                                                          t2frac14 d

                                                                                          21

                                                                                          d22

                                                                                          where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                          d1 frac14 95mm and d2 frac14 2500mm

                                                                                          for U frac14 050 t2 frac14 t1 d22

                                                                                          d21

                                                                                          frac14 20

                                                                                          60 24 365 25002

                                                                                          952frac14 263 years

                                                                                          for U lt 060 Tv frac14

                                                                                          4U2 (Equation 724(a))

                                                                                          t030 frac14 t050 0302

                                                                                          0502

                                                                                          frac14 263 036 frac14 095 years

                                                                                          Consolidation theory 53

                                                                                          74

                                                                                          The layer is open

                                                                                          d frac14 8

                                                                                          2frac14 4m

                                                                                          Tv frac14 cvtd2frac14 24 3

                                                                                          42frac14 0450

                                                                                          ui frac14 frac14 84 kN=m2

                                                                                          The excess pore water pressure is given by Equation 721

                                                                                          ue frac14Xmfrac141mfrac140

                                                                                          2ui

                                                                                          Msin

                                                                                          Mz

                                                                                          d

                                                                                          expethM2TvTHORN

                                                                                          In this case z frac14 d

                                                                                          sinMz

                                                                                          d

                                                                                          frac14 sinM

                                                                                          where

                                                                                          M frac14

                                                                                          23

                                                                                          25

                                                                                          2

                                                                                          M sin M M2Tv exp (M2Tv)

                                                                                          2thorn1 1110 0329

                                                                                          3

                                                                                          21 9993 457 105

                                                                                          ue frac14 2 84 2

                                                                                          1 0329 ethother terms negligibleTHORN

                                                                                          frac14 352 kN=m2

                                                                                          75

                                                                                          The layer is open

                                                                                          d frac14 6

                                                                                          2frac14 3m

                                                                                          Tv frac14 cvtd2frac14 10 3

                                                                                          32frac14 0333

                                                                                          The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                          54 Consolidation theory

                                                                                          For an open layer

                                                                                          Tv frac14 4n

                                                                                          m2

                                                                                          n frac14 0333 62

                                                                                          4frac14 300

                                                                                          The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                          ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                          i j

                                                                                          0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                          0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                          The initial and 3-year isochrones are plotted in Figure Q75

                                                                                          Area under initial isochrone frac14 180 units

                                                                                          Area under 3-year isochrone frac14 63 units

                                                                                          The average degree of consolidation is given by Equation 725Thus

                                                                                          U frac14 1 63

                                                                                          180frac14 065

                                                                                          Figure Q75

                                                                                          Consolidation theory 55

                                                                                          76

                                                                                          At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                          0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                          The final consolidation settlement (one-dimensional method) is

                                                                                          sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                          Corrected time t frac14 2 1

                                                                                          2

                                                                                          40

                                                                                          52

                                                                                          frac14 1615 years

                                                                                          Tv frac14 cvtd2frac14 44 1615

                                                                                          42frac14 0444

                                                                                          From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                          77

                                                                                          The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                          Figure Q77

                                                                                          56 Consolidation theory

                                                                                          Point m n Ir (kNm2) sc (mm)

                                                                                          13020frac14 15 20

                                                                                          20frac14 10 0194 (4) 113 124

                                                                                          260

                                                                                          20frac14 30

                                                                                          20

                                                                                          20frac14 10 0204 (2) 59 65

                                                                                          360

                                                                                          20frac14 30

                                                                                          40

                                                                                          20frac14 20 0238 (1) 35 38

                                                                                          430

                                                                                          20frac14 15

                                                                                          40

                                                                                          20frac14 20 0224 (2) 65 72

                                                                                          Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                          78

                                                                                          Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                          (a) Immediate settlement

                                                                                          H

                                                                                          Bfrac14 30

                                                                                          35frac14 086

                                                                                          D

                                                                                          Bfrac14 2

                                                                                          35frac14 006

                                                                                          Figure Q78

                                                                                          Consolidation theory 57

                                                                                          From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                          si frac14 130131qB

                                                                                          Eufrac14 10 032 105 35

                                                                                          40frac14 30mm

                                                                                          (b) Consolidation settlement

                                                                                          Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                          1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                          3150

                                                                                          Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                          Now

                                                                                          H

                                                                                          Bfrac14 30

                                                                                          35frac14 086 and A frac14 065

                                                                                          from Figure 712 13 frac14 079

                                                                                          sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                          Total settlement

                                                                                          s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                          79

                                                                                          Without sand drains

                                                                                          Uv frac14 025

                                                                                          Tv frac14 0049 ethfrom Figure 718THORN

                                                                                          t frac14 Tvd2

                                                                                          cvfrac14 0049 82

                                                                                          cvWith sand drains

                                                                                          R frac14 0564S frac14 0564 3 frac14 169m

                                                                                          n frac14 Rrfrac14 169

                                                                                          015frac14 113

                                                                                          Tr frac14 cht

                                                                                          4R2frac14 ch

                                                                                          4 1692 0049 82

                                                                                          cvethand ch frac14 cvTHORN

                                                                                          frac14 0275

                                                                                          Ur frac14 073 (from Figure 730)

                                                                                          58 Consolidation theory

                                                                                          Using Equation 740

                                                                                          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                          U frac14 080

                                                                                          710

                                                                                          Without sand drains

                                                                                          Uv frac14 090

                                                                                          Tv frac14 0848

                                                                                          t frac14 Tvd2

                                                                                          cvfrac14 0848 102

                                                                                          96frac14 88 years

                                                                                          With sand drains

                                                                                          R frac14 0564S frac14 0564 4 frac14 226m

                                                                                          n frac14 Rrfrac14 226

                                                                                          015frac14 15

                                                                                          Tr

                                                                                          Tvfrac14 chcv

                                                                                          d2

                                                                                          4R2ethsame tTHORN

                                                                                          Tr

                                                                                          Tvfrac14 140

                                                                                          96 102

                                                                                          4 2262frac14 714 eth1THORN

                                                                                          Using Equation 740

                                                                                          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                          An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                          Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                          040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                          Thus

                                                                                          Uv frac14 0295 and Ur frac14 086

                                                                                          t frac14 88 00683

                                                                                          0848frac14 07 years

                                                                                          Consolidation theory 59

                                                                                          Chapter 8

                                                                                          Bearing capacity

                                                                                          81

                                                                                          (a) The ultimate bearing capacity is given by Equation 83

                                                                                          qf frac14 cNc thorn DNq thorn 1

                                                                                          2BN

                                                                                          For u frac14 0

                                                                                          Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                          qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                          The net ultimate bearing capacity is

                                                                                          qnf frac14 qf D frac14 540 kN=m2

                                                                                          The net foundation pressure is

                                                                                          qn frac14 q D frac14 425

                                                                                          2 eth21 1THORN frac14 192 kN=m2

                                                                                          The factor of safety (Equation 86) is

                                                                                          F frac14 qnfqnfrac14 540

                                                                                          192frac14 28

                                                                                          (b) For 0 frac14 28

                                                                                          Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                          qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                          2 112 2 13

                                                                                          frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                          qnf frac14 574 112 frac14 563 kN=m2

                                                                                          F frac14 563

                                                                                          192frac14 29

                                                                                          (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                          82

                                                                                          For 0 frac14 38

                                                                                          Nq frac14 49 N frac14 67

                                                                                          qnf frac14 DethNq 1THORN thorn 1

                                                                                          2BN ethfrom Equation 83THORN

                                                                                          frac14 eth18 075 48THORN thorn 1

                                                                                          2 18 15 67

                                                                                          frac14 648thorn 905 frac14 1553 kN=m2

                                                                                          qn frac14 500

                                                                                          15 eth18 075THORN frac14 320 kN=m2

                                                                                          F frac14 qnfqnfrac14 1553

                                                                                          320frac14 48

                                                                                          0d frac14 tan1tan 38

                                                                                          125

                                                                                          frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                          Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                          2 18 15 25

                                                                                          frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                          Design load (action) Vd frac14 500 kN=m

                                                                                          The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                          83

                                                                                          D

                                                                                          Bfrac14 350

                                                                                          225frac14 155

                                                                                          From Figure 85 for a square foundation

                                                                                          Nc frac14 81

                                                                                          Bearing capacity 61

                                                                                          For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                          Nc frac14 084thorn 016B

                                                                                          L

                                                                                          81 frac14 745

                                                                                          Using Equation 810

                                                                                          qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                          For F frac14 3

                                                                                          qn frac14 1006

                                                                                          3frac14 335 kN=m2

                                                                                          q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                          Design load frac14 405 450 225 frac14 4100 kN

                                                                                          Design undrained strength cud frac14 135

                                                                                          14frac14 96 kN=m2

                                                                                          Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                          frac14 7241 kN

                                                                                          Design load Vd frac14 4100 kN

                                                                                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                          84

                                                                                          For 0 frac14 40

                                                                                          Nq frac14 64 N frac14 95

                                                                                          qnf frac14 DethNq 1THORN thorn 04BN

                                                                                          (a) Water table 5m below ground level

                                                                                          qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                          qn frac14 400 17 frac14 383 kN=m2

                                                                                          F frac14 2686

                                                                                          383frac14 70

                                                                                          (b) Water table 1m below ground level (ie at foundation level)

                                                                                          0 frac14 20 98 frac14 102 kN=m3

                                                                                          62 Bearing capacity

                                                                                          qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                          F frac14 2040

                                                                                          383frac14 53

                                                                                          (c) Water table at ground level with upward hydraulic gradient 02

                                                                                          eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                          qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                          F frac14 1296

                                                                                          392frac14 33

                                                                                          85

                                                                                          The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                          Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                          Design value of 0 frac14 tan1tan 39

                                                                                          125

                                                                                          frac14 33

                                                                                          For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                          Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                          86

                                                                                          (a) Undrained shear for u frac14 0

                                                                                          Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                          qnf frac14 12cuNc

                                                                                          frac14 12 100 514 frac14 617 kN=m2

                                                                                          qn frac14 qnfFfrac14 617

                                                                                          3frac14 206 kN=m2

                                                                                          q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                          Bearing capacity 63

                                                                                          Drained shear for 0 frac14 32

                                                                                          Nq frac14 23 N frac14 25

                                                                                          0 frac14 21 98 frac14 112 kN=m3

                                                                                          qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                          frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                          frac14 694 kN=m2

                                                                                          q frac14 694

                                                                                          3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                          Design load frac14 42 227 frac14 3632 kN

                                                                                          (b) Design undrained strength cud frac14 100

                                                                                          14frac14 71 kNm2

                                                                                          Design bearing resistance Rd frac14 12cudNe area

                                                                                          frac14 12 71 514 42

                                                                                          frac14 7007 kN

                                                                                          For drained shear 0d frac14 tan1tan 32

                                                                                          125

                                                                                          frac14 26

                                                                                          Nq frac14 12 N frac14 10

                                                                                          Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                          (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                          Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                          1 2 100 0175 0700qn 0182qn

                                                                                          2 6 033 0044 0176qn 0046qn

                                                                                          3 10 020 0017 0068qn 0018qn

                                                                                          0246qn

                                                                                          Diameter of equivalent circle B frac14 45m

                                                                                          H

                                                                                          Bfrac14 12

                                                                                          45frac14 27 and A frac14 042

                                                                                          13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                          64 Bearing capacity

                                                                                          For sc frac14 30mm

                                                                                          qn frac14 30

                                                                                          0147frac14 204 kN=m2

                                                                                          q frac14 204thorn 21 frac14 225 kN=m2

                                                                                          Design load frac14 42 225 frac14 3600 kN

                                                                                          The design load is 3600 kN settlement being the limiting criterion

                                                                                          87

                                                                                          D

                                                                                          Bfrac14 8

                                                                                          4frac14 20

                                                                                          From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                          F frac14 cuNc

                                                                                          Dfrac14 40 71

                                                                                          20 8frac14 18

                                                                                          88

                                                                                          Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                          Design value of 0 frac14 tan1tan 38

                                                                                          125

                                                                                          frac14 32

                                                                                          Figure Q86

                                                                                          Bearing capacity 65

                                                                                          For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                          Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                          The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                          Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                          For B frac14 250m qn frac14 3750

                                                                                          2502 17 frac14 583 kN=m2

                                                                                          From Figure 510 m frac14 n frac14 126

                                                                                          6frac14 021

                                                                                          Ir frac14 0019

                                                                                          Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                          Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                          The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                          89

                                                                                          Depth (m) N 0v (kNm2) CN N1

                                                                                          070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                          Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                          (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                          Cw frac14 05thorn 0530

                                                                                          47

                                                                                          frac14 082

                                                                                          66 Bearing capacity

                                                                                          Thus

                                                                                          qa frac14 150 082 frac14 120 kN=m2

                                                                                          (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                          Thus

                                                                                          qa frac14 90 15 frac14 135 kN=m2

                                                                                          (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                          Ic frac14 171

                                                                                          1014frac14 0068

                                                                                          From Equation 819(a) with s frac14 25mm

                                                                                          q frac14 25

                                                                                          3507 0068frac14 150 kN=m2

                                                                                          810

                                                                                          Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                          Izp frac14 05thorn 01130

                                                                                          16 27

                                                                                          05

                                                                                          frac14 067

                                                                                          Refer to Figure Q810

                                                                                          E frac14 25qc

                                                                                          Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                          Ez (mm3MN)

                                                                                          1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                          0203

                                                                                          C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                          130frac14 093

                                                                                          C2 frac14 1 ethsayTHORN

                                                                                          s frac14 C1C2qnX Iz

                                                                                          Ez frac14 093 1 130 0203 frac14 25mm

                                                                                          Bearing capacity 67

                                                                                          811

                                                                                          At pile base level

                                                                                          cu frac14 220 kN=m2

                                                                                          qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                          Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                          00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                          qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                          Then

                                                                                          Qf frac14 Abqb thorn Asqs

                                                                                          frac14

                                                                                          4 32 1980

                                                                                          thorn eth 105 139 86THORN

                                                                                          frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                          0 01 02 03 04 05 06 07

                                                                                          0 2 4 6 8 10 12 14

                                                                                          1

                                                                                          2

                                                                                          3

                                                                                          4

                                                                                          5

                                                                                          6

                                                                                          7

                                                                                          8

                                                                                          (1)

                                                                                          (2)

                                                                                          (3)

                                                                                          (4)

                                                                                          (5)

                                                                                          qc

                                                                                          qc

                                                                                          Iz

                                                                                          Iz

                                                                                          (MNm2)

                                                                                          z (m)

                                                                                          Figure Q810

                                                                                          68 Bearing capacity

                                                                                          Allowable load

                                                                                          ethaTHORN Qf

                                                                                          2frac14 17 937

                                                                                          2frac14 8968 kN

                                                                                          ethbTHORN Abqb

                                                                                          3thorn Asqs frac14 13 996

                                                                                          3thorn 3941 frac14 8606 kN

                                                                                          ie allowable load frac14 8600 kN

                                                                                          Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                          According to the limit state method

                                                                                          Characteristic undrained strength at base level cuk frac14 220

                                                                                          150kN=m2

                                                                                          Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                          150frac14 1320 kN=m2

                                                                                          Characteristic shaft resistance qsk frac14 00150

                                                                                          frac14 86

                                                                                          150frac14 57 kN=m2

                                                                                          Characteristic base and shaft resistances

                                                                                          Rbk frac14

                                                                                          4 32 1320 frac14 9330 kN

                                                                                          Rsk frac14 105 139 86

                                                                                          150frac14 2629 kN

                                                                                          For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                          Design bearing resistance Rcd frac14 9330

                                                                                          160thorn 2629

                                                                                          130

                                                                                          frac14 5831thorn 2022

                                                                                          frac14 7850 kN

                                                                                          Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                          812

                                                                                          ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                          qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                          For a single pile

                                                                                          Qf frac14 Abqb thorn Asqs

                                                                                          frac14

                                                                                          4 062 1305

                                                                                          thorn eth 06 15 42THORN

                                                                                          frac14 369thorn 1187 frac14 1556 kN

                                                                                          Bearing capacity 69

                                                                                          Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                          eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                          (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                          (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                          qbkfrac14 9cuk frac14 9 220

                                                                                          150frac14 1320 kN=m2

                                                                                          qskfrac14cuk frac14 040 105

                                                                                          150frac14 28 kN=m2

                                                                                          Rbkfrac14

                                                                                          4 0602 1320 frac14 373 kN

                                                                                          Rskfrac14 060 15 28 frac14 791 kN

                                                                                          Rcdfrac14 373

                                                                                          160thorn 791

                                                                                          130frac14 233thorn 608 frac14 841 kN

                                                                                          Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                          Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                          (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                          q frac14 21 000

                                                                                          1762frac14 68 kN=m2

                                                                                          Immediate settlement

                                                                                          H

                                                                                          Bfrac14 15

                                                                                          176frac14 085

                                                                                          D

                                                                                          Bfrac14 13

                                                                                          176frac14 074

                                                                                          L

                                                                                          Bfrac14 1

                                                                                          Hence from Figure 515

                                                                                          130 frac14 078 and 131 frac14 041

                                                                                          70 Bearing capacity

                                                                                          Thus using Equation 528

                                                                                          si frac14 078 041 68 176

                                                                                          65frac14 6mm

                                                                                          Consolidation settlement

                                                                                          Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                          1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                          434 (sod)

                                                                                          Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                          sc frac14 056 434 frac14 24mm

                                                                                          The total settlement is (6thorn 24) frac14 30mm

                                                                                          813

                                                                                          At base level N frac14 26 Then using Equation 830

                                                                                          qb frac14 40NDb

                                                                                          Bfrac14 40 26 2

                                                                                          025frac14 8320 kN=m2

                                                                                          ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                          Figure Q812

                                                                                          Bearing capacity 71

                                                                                          Over the length embedded in sand

                                                                                          N frac14 21 ie18thorn 24

                                                                                          2

                                                                                          Using Equation 831

                                                                                          qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                          For a single pile

                                                                                          Qf frac14 Abqb thorn Asqs

                                                                                          frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                          For the pile group assuming a group efficiency of 12

                                                                                          XQf frac14 12 9 604 frac14 6523 kN

                                                                                          Then the load factor is

                                                                                          F frac14 6523

                                                                                          2000thorn 1000frac14 21

                                                                                          (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                          Characteristic base resistance per unit area qbk frac14 8320

                                                                                          150frac14 5547 kNm2

                                                                                          Characteristic shaft resistance per unit area qsk frac14 42

                                                                                          150frac14 28 kNm2

                                                                                          Characteristic base and shaft resistances for a single pile

                                                                                          Rbk frac14 0252 5547 frac14 347 kN

                                                                                          Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                          For a driven pile the partial factors are b frac14 s frac14 130

                                                                                          Design bearing resistance Rcd frac14 347

                                                                                          130thorn 56

                                                                                          130frac14 310 kN

                                                                                          For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                          Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                          (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                          From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                          72 Bearing capacity

                                                                                          N frac14 24thorn 26thorn 34

                                                                                          3frac14 28

                                                                                          Ic frac14 171

                                                                                          2814frac14 0016 ethEquation 818THORN

                                                                                          s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                          814

                                                                                          Using Equation 841

                                                                                          Tf frac14 DLcu thorn

                                                                                          4ethD2 d2THORNcuNc

                                                                                          frac14 eth 02 5 06 110THORN thorn

                                                                                          4eth022 012THORN110 9

                                                                                          frac14 207thorn 23 frac14 230 kN

                                                                                          Figure Q813

                                                                                          Bearing capacity 73

                                                                                          Chapter 9

                                                                                          Stability of slopes

                                                                                          91

                                                                                          Referring to Figure Q91

                                                                                          W frac14 417 19 frac14 792 kN=m

                                                                                          Q frac14 20 28 frac14 56 kN=m

                                                                                          Arc lengthAB frac14

                                                                                          180 73 90 frac14 115m

                                                                                          Arc length BC frac14

                                                                                          180 28 90 frac14 44m

                                                                                          The factor of safety is given by

                                                                                          F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                          90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                          Depth of tension crack z0 frac14 2cu

                                                                                          frac14 2 20

                                                                                          19frac14 21m

                                                                                          Arc length BD frac14

                                                                                          180 13

                                                                                          1

                                                                                          2 90 frac14 21m

                                                                                          F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                          The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                          Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                          14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                          Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                          The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                          92

                                                                                          u frac14 0

                                                                                          Depth factor D frac14 11

                                                                                          9frac14 122

                                                                                          Using Equation 92 with F frac14 10

                                                                                          Ns frac14 cu

                                                                                          FHfrac14 30

                                                                                          10 19 9frac14 0175

                                                                                          Hence from Figure 93

                                                                                          frac14 50

                                                                                          For F frac14 12

                                                                                          Ns frac14 30

                                                                                          12 19 9frac14 0146

                                                                                          frac14 27

                                                                                          93

                                                                                          Refer to Figure Q93

                                                                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                          1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                          74 m

                                                                                          214 1deg

                                                                                          213 1deg

                                                                                          39 m

                                                                                          WB

                                                                                          D

                                                                                          C

                                                                                          28 m

                                                                                          21 m

                                                                                          A

                                                                                          Q

                                                                                          Soil (1)Soil (2)

                                                                                          73deg

                                                                                          Figure Q91

                                                                                          Stability of slopes 75

                                                                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                          9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                          599 256 328 1372

                                                                                          Figure Q93

                                                                                          76 Stability of slopes

                                                                                          XW cos frac14 b

                                                                                          Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                          W sin frac14 bX

                                                                                          h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                          Arc length La frac14

                                                                                          180 57

                                                                                          1

                                                                                          2 326 frac14 327m

                                                                                          The factor of safety is given by

                                                                                          F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                          W sin

                                                                                          frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                          frac14 091

                                                                                          According to the limit state method

                                                                                          0d frac14 tan1tan 32

                                                                                          125

                                                                                          frac14 265

                                                                                          c0 frac14 8

                                                                                          160frac14 5 kN=m2

                                                                                          Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                          Design disturbing moment frac14 1075 kN=m

                                                                                          The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                          94

                                                                                          F frac14 1

                                                                                          W sin

                                                                                          Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                          1thorn ethtan tan0=FTHORN

                                                                                          c0 frac14 8 kN=m2

                                                                                          0 frac14 32

                                                                                          c0b frac14 8 2 frac14 16 kN=m

                                                                                          W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                          Try F frac14 100

                                                                                          tan0

                                                                                          Ffrac14 0625

                                                                                          Stability of slopes 77

                                                                                          Values of u are as obtained in Figure Q93

                                                                                          SliceNo

                                                                                          h(m)

                                                                                          W frac14 bh(kNm)

                                                                                          W sin(kNm)

                                                                                          ub(kNm)

                                                                                          c0bthorn (W ub) tan0(kNm)

                                                                                          sec

                                                                                          1thorn (tan tan0)FProduct(kNm)

                                                                                          1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                          23 33 30 1042 31

                                                                                          3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                          224 92 72 0931 67

                                                                                          6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                          12 58 112 90 0889 80

                                                                                          8 60 252 1812

                                                                                          80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                          10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                          2154 88 116 0853 99

                                                                                          14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                          1074 1091

                                                                                          F frac14 1091

                                                                                          1074frac14 102 (assumed value 100)

                                                                                          Thus

                                                                                          F frac14 101

                                                                                          95

                                                                                          F frac14 1

                                                                                          W sin

                                                                                          XfWeth1 ruTHORN tan0g sec

                                                                                          1thorn ethtan tan0THORN=F

                                                                                          0 frac14 33

                                                                                          ru frac14 020

                                                                                          W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                          eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                          Try F frac14 110

                                                                                          tan 0

                                                                                          Ffrac14 tan 33

                                                                                          110frac14 0590

                                                                                          78 Stability of slopes

                                                                                          Referring to Figure Q95

                                                                                          SliceNo

                                                                                          h(m)

                                                                                          W frac14 bh(kNm)

                                                                                          W sin(kNm)

                                                                                          W(1 ru) tan0(kNm)

                                                                                          sec

                                                                                          1thorn ( tan tan0)FProduct(kNm)

                                                                                          1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                          2120 234 0892 209

                                                                                          4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                          1185 1271

                                                                                          Figure Q95

                                                                                          Stability of slopes 79

                                                                                          F frac14 1271

                                                                                          1185frac14 107

                                                                                          The trial value was 110 therefore take F to be 108

                                                                                          96

                                                                                          (a) Water table at surface the factor of safety is given by Equation 912

                                                                                          F frac14 0

                                                                                          sat

                                                                                          tan0

                                                                                          tan

                                                                                          ptie 15 frac14 92

                                                                                          19

                                                                                          tan 36

                                                                                          tan

                                                                                          tan frac14 0234

                                                                                          frac14 13

                                                                                          Water table well below surface the factor of safety is given by Equation 911

                                                                                          F frac14 tan0

                                                                                          tan

                                                                                          frac14 tan 36

                                                                                          tan 13

                                                                                          frac14 31

                                                                                          (b) 0d frac14 tan1tan 36

                                                                                          125

                                                                                          frac14 30

                                                                                          Depth of potential failure surface frac14 z

                                                                                          Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                          frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                          frac14 504z kN

                                                                                          Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                          frac14 19 z sin 13 cos 13

                                                                                          frac14 416z kN

                                                                                          Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                          80 Stability of slopes

                                                                                          • Book Cover
                                                                                          • Title
                                                                                          • Contents
                                                                                          • Basic characteristics of soils
                                                                                          • Seepage
                                                                                          • Effective stress
                                                                                          • Shear strength
                                                                                          • Stresses and displacements
                                                                                          • Lateral earth pressure
                                                                                          • Consolidation theory
                                                                                          • Bearing capacity
                                                                                          • Stability of slopes

                                                                                            Horizontal component

                                                                                            Ph frac14 267 cos 40 frac14 205 kN=m

                                                                                            Vertical component

                                                                                            Pv frac14 267 sin 40 frac14 172 kN=m

                                                                                            Consider moments about the toe of the wall (Figure Q66) (per m)

                                                                                            Force (kN) Arm (m) Moment (kN m)

                                                                                            (1)1

                                                                                            2 175 650 235 frac14 1337 258 345

                                                                                            (2) 050 650 235 frac14 764 175 134

                                                                                            (3)1

                                                                                            2 070 650 235 frac14 535 127 68

                                                                                            (4) 100 400 235 frac14 940 200 188

                                                                                            (5) 1

                                                                                            2 080 050 235 frac14 47 027 1

                                                                                            Pa sin 40 frac14 1720 333 573Vfrac14 525 MV frac141307

                                                                                            Pa cos 40 Hfrac14 205 250 MH frac14 512M frac14 795

                                                                                            Lever arm of base resultant

                                                                                            M

                                                                                            Vfrac14 795

                                                                                            525frac14 151m

                                                                                            Eccentricity of base resultant

                                                                                            e frac14 200 151 frac14 049m

                                                                                            Figure Q66

                                                                                            40 Lateral earth pressure

                                                                                            Base pressures (Equation 627)

                                                                                            p frac14 525

                                                                                            41 6 049

                                                                                            4

                                                                                            frac14 228 kN=m2 and 35 kN=m2

                                                                                            The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                                            The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                                            The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                                            67

                                                                                            For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                                            Force (kN) Arm (m) Moment (kNm)

                                                                                            (1)1

                                                                                            2 027 17 52 frac14 574 183 1050

                                                                                            (2) 027 17 5 3 frac14 689 500 3445

                                                                                            (3)1

                                                                                            2 027 102 32 frac14 124 550 682

                                                                                            (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                                            (5)1

                                                                                            2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                                            (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                                            (7) 1

                                                                                            2 267

                                                                                            2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                                            (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                                            2 d frac14 163d d2thorn 650 82d2 1060d

                                                                                            Tie rod force per m frac14 T 0 0

                                                                                            XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                                            d3 thorn 77d2 269d 1438 frac14 0

                                                                                            d frac14 467m

                                                                                            Depth of penetration frac14 12d frac14 560m

                                                                                            Lateral earth pressure 41

                                                                                            Algebraic sum of forces for d frac14 467m isX

                                                                                            F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                                            T frac14 905 kN=m

                                                                                            Force in each tie rod frac14 25T frac14 226 kN

                                                                                            68

                                                                                            (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                                            0 frac14 21 98 frac14 112 kN=m3

                                                                                            The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                                            uC frac14 150

                                                                                            165 15 98 frac14 134 kN=m2

                                                                                            The average seepage pressure is

                                                                                            j frac14 15

                                                                                            165 98 frac14 09 kN=m3

                                                                                            Hence

                                                                                            0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                                            0 j frac14 112 09 frac14 103 kN=m3

                                                                                            Figure Q67

                                                                                            42 Lateral earth pressure

                                                                                            Consider moments about the anchor point A (per m)

                                                                                            Force (kN) Arm (m) Moment (kN m)

                                                                                            (1) 10 026 150 frac14 390 60 2340

                                                                                            (2)1

                                                                                            2 026 18 452 frac14 474 15 711

                                                                                            (3) 026 18 45 105 frac14 2211 825 18240

                                                                                            (4)1

                                                                                            2 026 121 1052 frac14 1734 100 17340

                                                                                            (5)1

                                                                                            2 134 15 frac14 101 40 404

                                                                                            (6) 134 30 frac14 402 60 2412

                                                                                            (7)1

                                                                                            2 134 60 frac14 402 95 3819

                                                                                            571 4527(8) Ppm

                                                                                            115 115PPm

                                                                                            XM frac14 0

                                                                                            Ppm frac144527

                                                                                            115frac14 394 kN=m

                                                                                            Available passive resistance

                                                                                            Pp frac14 1

                                                                                            2 385 103 62 frac14 714 kN=m

                                                                                            Factor of safety

                                                                                            Fp frac14 Pp

                                                                                            Ppm

                                                                                            frac14 714

                                                                                            394frac14 18

                                                                                            Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                                            Figure Q68

                                                                                            Lateral earth pressure 43

                                                                                            (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                                            Consider moments (per m) about the tie point A

                                                                                            Force (kN) Arm (m)

                                                                                            (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                                            (2)1

                                                                                            2 033 18 452 frac14 601 15

                                                                                            (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                                            (4)1

                                                                                            2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                                            (5)1

                                                                                            2 134 15 frac14 101 40

                                                                                            (6) 134 30 frac14 402 60

                                                                                            (7)1

                                                                                            2 134 d frac14 67d d3thorn 75

                                                                                            (8) 1

                                                                                            2 30 103 d2 frac141545d2 2d3thorn 75

                                                                                            Moment (kN m)

                                                                                            (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                                            XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                                            d3 thorn 827d2 466d 1518 frac14 0

                                                                                            By trial

                                                                                            d frac14 544m

                                                                                            The minimum depth of embedment required is 544m

                                                                                            69

                                                                                            For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                                            0 frac14 20 98 frac14 102 kN=m3

                                                                                            The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                                            44 Lateral earth pressure

                                                                                            uC frac14 147

                                                                                            173 26 98 frac14 216 kN=m2

                                                                                            and the average seepage pressure around the wall is

                                                                                            j frac14 26

                                                                                            173 98 frac14 15 kN=m3

                                                                                            Consider moments about the prop (A) (per m)

                                                                                            Force (kN) Arm (m) Moment (kN m)

                                                                                            (1)1

                                                                                            2 03 17 272 frac14 186 020 37

                                                                                            (2) 03 17 27 53 frac14 730 335 2445

                                                                                            (3)1

                                                                                            2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                                            (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                                            (5)1

                                                                                            2 216 26 frac14 281 243 684

                                                                                            (6) 216 27 frac14 583 465 2712

                                                                                            (7)1

                                                                                            2 216 60 frac14 648 800 5184

                                                                                            3055(8)

                                                                                            1

                                                                                            2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                                            Factor of safety

                                                                                            Fr frac14 6885

                                                                                            3055frac14 225

                                                                                            Figure Q69

                                                                                            Lateral earth pressure 45

                                                                                            610

                                                                                            For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                                            p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                                            Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                                            Using the recommendations of Twine and Roscoe

                                                                                            p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                                            Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                                            611

                                                                                            frac14 18 kN=m3 0 frac14 34

                                                                                            H frac14 350m nH frac14 335m mH frac14 185m

                                                                                            Consider a trial value of F frac14 20 Refer to Figure 635

                                                                                            0m frac14 tan1tan 34

                                                                                            20

                                                                                            frac14 186

                                                                                            Then

                                                                                            frac14 45 thorn 0m2frac14 543

                                                                                            W frac14 1

                                                                                            2 18 3502 cot 543 frac14 792 kN=m

                                                                                            Figure Q610

                                                                                            46 Lateral earth pressure

                                                                                            P frac14 1

                                                                                            2 s 3352 frac14 561s kN=m

                                                                                            U frac14 1

                                                                                            2 98 1852 cosec 543 frac14 206 kN=m

                                                                                            Equations 630 and 631 then become

                                                                                            561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                                            792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                                            ie

                                                                                            561s 0616N 405 frac14 0

                                                                                            792 0857N thorn 563 frac14 0

                                                                                            N frac14 848

                                                                                            0857frac14 989 kN=m

                                                                                            Then

                                                                                            561s 609 405 frac14 0

                                                                                            s frac14 649

                                                                                            561frac14 116 kN=m3

                                                                                            The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                                            F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                                            20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                                            s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                                            Figure Q611

                                                                                            Lateral earth pressure 47

                                                                                            612

                                                                                            For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                            45 thorn 0

                                                                                            2frac14 63

                                                                                            For the retained material between the surface and a depth of 36m

                                                                                            Pa frac14 1

                                                                                            2 030 18 362 frac14 350 kN=m

                                                                                            Weight of reinforced fill between the surface and a depth of 36m is

                                                                                            Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                            eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                            Eccentricity of Rv

                                                                                            e frac14 263 250 frac14 013m

                                                                                            The average vertical stress at a depth of 36m is

                                                                                            z frac14 Rv

                                                                                            L 2efrac14 324

                                                                                            474frac14 68 kN=m2

                                                                                            (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                            Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                            Tensile stress in the element frac14 138 103

                                                                                            65 3frac14 71N=mm2

                                                                                            Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                            Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                            Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                            The weight of ABC is

                                                                                            W frac14 1

                                                                                            2 18 52 265 frac14 124 kN=m

                                                                                            From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                            48 Lateral earth pressure

                                                                                            (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                            Tp frac14 032 68 120 065 frac14 170 kN

                                                                                            Tr frac14 213 420

                                                                                            418frac14 214 kN

                                                                                            Again the tensile failure and slipping limit states are satisfied for this element

                                                                                            Figure Q612

                                                                                            Lateral earth pressure 49

                                                                                            Chapter 7

                                                                                            Consolidation theory

                                                                                            71

                                                                                            Total change in thickness

                                                                                            H frac14 782 602 frac14 180mm

                                                                                            Average thickness frac14 1530thorn 180

                                                                                            2frac14 1620mm

                                                                                            Length of drainage path d frac14 1620

                                                                                            2frac14 810mm

                                                                                            Root time plot (Figure Q71a)

                                                                                            ffiffiffiffiffiffit90p frac14 33

                                                                                            t90 frac14 109min

                                                                                            cv frac14 0848d2

                                                                                            t90frac14 0848 8102

                                                                                            109 1440 365

                                                                                            106frac14 27m2=year

                                                                                            r0 frac14 782 764

                                                                                            782 602frac14 018

                                                                                            180frac14 0100

                                                                                            rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                            10 119

                                                                                            9 180frac14 0735

                                                                                            rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                            Log time plot (Figure Q71b)

                                                                                            t50 frac14 26min

                                                                                            cv frac14 0196d2

                                                                                            t50frac14 0196 8102

                                                                                            26 1440 365

                                                                                            106frac14 26m2=year

                                                                                            r0 frac14 782 763

                                                                                            782 602frac14 019

                                                                                            180frac14 0106

                                                                                            rp frac14 763 623

                                                                                            782 602frac14 140

                                                                                            180frac14 0778

                                                                                            rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                            Figure Q71(a)

                                                                                            Figure Q71(b)

                                                                                            Final void ratio

                                                                                            e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                            e

                                                                                            Hfrac14 1thorn e0

                                                                                            H0frac14 1thorn e1 thorne

                                                                                            H0

                                                                                            ie

                                                                                            e

                                                                                            180frac14 1631thorne

                                                                                            1710

                                                                                            e frac14 2936

                                                                                            1530frac14 0192

                                                                                            Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                            mv frac14 1

                                                                                            1thorn e0 e0 e101 00

                                                                                            frac14 1

                                                                                            1823 0192

                                                                                            0107frac14 098m2=MN

                                                                                            k frac14 cvmvw frac14 265 098 98

                                                                                            60 1440 365 103frac14 81 1010 m=s

                                                                                            72

                                                                                            Using Equation 77 (one-dimensional method)

                                                                                            sc frac14 e0 e11thorn e0 H

                                                                                            Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                            Figure Q72

                                                                                            52 Consolidation theory

                                                                                            Settlement

                                                                                            Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                            1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                            318

                                                                                            Notes 5 92y 460thorn 84

                                                                                            Heave

                                                                                            Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                            1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                            38

                                                                                            73

                                                                                            U frac14 f ethTvTHORN frac14 f cvt

                                                                                            d2

                                                                                            Hence if cv is constant

                                                                                            t1

                                                                                            t2frac14 d

                                                                                            21

                                                                                            d22

                                                                                            where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                            d1 frac14 95mm and d2 frac14 2500mm

                                                                                            for U frac14 050 t2 frac14 t1 d22

                                                                                            d21

                                                                                            frac14 20

                                                                                            60 24 365 25002

                                                                                            952frac14 263 years

                                                                                            for U lt 060 Tv frac14

                                                                                            4U2 (Equation 724(a))

                                                                                            t030 frac14 t050 0302

                                                                                            0502

                                                                                            frac14 263 036 frac14 095 years

                                                                                            Consolidation theory 53

                                                                                            74

                                                                                            The layer is open

                                                                                            d frac14 8

                                                                                            2frac14 4m

                                                                                            Tv frac14 cvtd2frac14 24 3

                                                                                            42frac14 0450

                                                                                            ui frac14 frac14 84 kN=m2

                                                                                            The excess pore water pressure is given by Equation 721

                                                                                            ue frac14Xmfrac141mfrac140

                                                                                            2ui

                                                                                            Msin

                                                                                            Mz

                                                                                            d

                                                                                            expethM2TvTHORN

                                                                                            In this case z frac14 d

                                                                                            sinMz

                                                                                            d

                                                                                            frac14 sinM

                                                                                            where

                                                                                            M frac14

                                                                                            23

                                                                                            25

                                                                                            2

                                                                                            M sin M M2Tv exp (M2Tv)

                                                                                            2thorn1 1110 0329

                                                                                            3

                                                                                            21 9993 457 105

                                                                                            ue frac14 2 84 2

                                                                                            1 0329 ethother terms negligibleTHORN

                                                                                            frac14 352 kN=m2

                                                                                            75

                                                                                            The layer is open

                                                                                            d frac14 6

                                                                                            2frac14 3m

                                                                                            Tv frac14 cvtd2frac14 10 3

                                                                                            32frac14 0333

                                                                                            The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                            54 Consolidation theory

                                                                                            For an open layer

                                                                                            Tv frac14 4n

                                                                                            m2

                                                                                            n frac14 0333 62

                                                                                            4frac14 300

                                                                                            The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                            ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                            i j

                                                                                            0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                            0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                            The initial and 3-year isochrones are plotted in Figure Q75

                                                                                            Area under initial isochrone frac14 180 units

                                                                                            Area under 3-year isochrone frac14 63 units

                                                                                            The average degree of consolidation is given by Equation 725Thus

                                                                                            U frac14 1 63

                                                                                            180frac14 065

                                                                                            Figure Q75

                                                                                            Consolidation theory 55

                                                                                            76

                                                                                            At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                            0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                            The final consolidation settlement (one-dimensional method) is

                                                                                            sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                            Corrected time t frac14 2 1

                                                                                            2

                                                                                            40

                                                                                            52

                                                                                            frac14 1615 years

                                                                                            Tv frac14 cvtd2frac14 44 1615

                                                                                            42frac14 0444

                                                                                            From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                            77

                                                                                            The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                            Figure Q77

                                                                                            56 Consolidation theory

                                                                                            Point m n Ir (kNm2) sc (mm)

                                                                                            13020frac14 15 20

                                                                                            20frac14 10 0194 (4) 113 124

                                                                                            260

                                                                                            20frac14 30

                                                                                            20

                                                                                            20frac14 10 0204 (2) 59 65

                                                                                            360

                                                                                            20frac14 30

                                                                                            40

                                                                                            20frac14 20 0238 (1) 35 38

                                                                                            430

                                                                                            20frac14 15

                                                                                            40

                                                                                            20frac14 20 0224 (2) 65 72

                                                                                            Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                            78

                                                                                            Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                            (a) Immediate settlement

                                                                                            H

                                                                                            Bfrac14 30

                                                                                            35frac14 086

                                                                                            D

                                                                                            Bfrac14 2

                                                                                            35frac14 006

                                                                                            Figure Q78

                                                                                            Consolidation theory 57

                                                                                            From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                            si frac14 130131qB

                                                                                            Eufrac14 10 032 105 35

                                                                                            40frac14 30mm

                                                                                            (b) Consolidation settlement

                                                                                            Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                            1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                            3150

                                                                                            Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                            Now

                                                                                            H

                                                                                            Bfrac14 30

                                                                                            35frac14 086 and A frac14 065

                                                                                            from Figure 712 13 frac14 079

                                                                                            sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                            Total settlement

                                                                                            s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                            79

                                                                                            Without sand drains

                                                                                            Uv frac14 025

                                                                                            Tv frac14 0049 ethfrom Figure 718THORN

                                                                                            t frac14 Tvd2

                                                                                            cvfrac14 0049 82

                                                                                            cvWith sand drains

                                                                                            R frac14 0564S frac14 0564 3 frac14 169m

                                                                                            n frac14 Rrfrac14 169

                                                                                            015frac14 113

                                                                                            Tr frac14 cht

                                                                                            4R2frac14 ch

                                                                                            4 1692 0049 82

                                                                                            cvethand ch frac14 cvTHORN

                                                                                            frac14 0275

                                                                                            Ur frac14 073 (from Figure 730)

                                                                                            58 Consolidation theory

                                                                                            Using Equation 740

                                                                                            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                            U frac14 080

                                                                                            710

                                                                                            Without sand drains

                                                                                            Uv frac14 090

                                                                                            Tv frac14 0848

                                                                                            t frac14 Tvd2

                                                                                            cvfrac14 0848 102

                                                                                            96frac14 88 years

                                                                                            With sand drains

                                                                                            R frac14 0564S frac14 0564 4 frac14 226m

                                                                                            n frac14 Rrfrac14 226

                                                                                            015frac14 15

                                                                                            Tr

                                                                                            Tvfrac14 chcv

                                                                                            d2

                                                                                            4R2ethsame tTHORN

                                                                                            Tr

                                                                                            Tvfrac14 140

                                                                                            96 102

                                                                                            4 2262frac14 714 eth1THORN

                                                                                            Using Equation 740

                                                                                            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                            An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                            Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                            040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                            Thus

                                                                                            Uv frac14 0295 and Ur frac14 086

                                                                                            t frac14 88 00683

                                                                                            0848frac14 07 years

                                                                                            Consolidation theory 59

                                                                                            Chapter 8

                                                                                            Bearing capacity

                                                                                            81

                                                                                            (a) The ultimate bearing capacity is given by Equation 83

                                                                                            qf frac14 cNc thorn DNq thorn 1

                                                                                            2BN

                                                                                            For u frac14 0

                                                                                            Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                            qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                            The net ultimate bearing capacity is

                                                                                            qnf frac14 qf D frac14 540 kN=m2

                                                                                            The net foundation pressure is

                                                                                            qn frac14 q D frac14 425

                                                                                            2 eth21 1THORN frac14 192 kN=m2

                                                                                            The factor of safety (Equation 86) is

                                                                                            F frac14 qnfqnfrac14 540

                                                                                            192frac14 28

                                                                                            (b) For 0 frac14 28

                                                                                            Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                            qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                            2 112 2 13

                                                                                            frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                            qnf frac14 574 112 frac14 563 kN=m2

                                                                                            F frac14 563

                                                                                            192frac14 29

                                                                                            (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                            82

                                                                                            For 0 frac14 38

                                                                                            Nq frac14 49 N frac14 67

                                                                                            qnf frac14 DethNq 1THORN thorn 1

                                                                                            2BN ethfrom Equation 83THORN

                                                                                            frac14 eth18 075 48THORN thorn 1

                                                                                            2 18 15 67

                                                                                            frac14 648thorn 905 frac14 1553 kN=m2

                                                                                            qn frac14 500

                                                                                            15 eth18 075THORN frac14 320 kN=m2

                                                                                            F frac14 qnfqnfrac14 1553

                                                                                            320frac14 48

                                                                                            0d frac14 tan1tan 38

                                                                                            125

                                                                                            frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                            Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                            2 18 15 25

                                                                                            frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                            Design load (action) Vd frac14 500 kN=m

                                                                                            The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                            83

                                                                                            D

                                                                                            Bfrac14 350

                                                                                            225frac14 155

                                                                                            From Figure 85 for a square foundation

                                                                                            Nc frac14 81

                                                                                            Bearing capacity 61

                                                                                            For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                            Nc frac14 084thorn 016B

                                                                                            L

                                                                                            81 frac14 745

                                                                                            Using Equation 810

                                                                                            qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                            For F frac14 3

                                                                                            qn frac14 1006

                                                                                            3frac14 335 kN=m2

                                                                                            q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                            Design load frac14 405 450 225 frac14 4100 kN

                                                                                            Design undrained strength cud frac14 135

                                                                                            14frac14 96 kN=m2

                                                                                            Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                            frac14 7241 kN

                                                                                            Design load Vd frac14 4100 kN

                                                                                            Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                            84

                                                                                            For 0 frac14 40

                                                                                            Nq frac14 64 N frac14 95

                                                                                            qnf frac14 DethNq 1THORN thorn 04BN

                                                                                            (a) Water table 5m below ground level

                                                                                            qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                            qn frac14 400 17 frac14 383 kN=m2

                                                                                            F frac14 2686

                                                                                            383frac14 70

                                                                                            (b) Water table 1m below ground level (ie at foundation level)

                                                                                            0 frac14 20 98 frac14 102 kN=m3

                                                                                            62 Bearing capacity

                                                                                            qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                            F frac14 2040

                                                                                            383frac14 53

                                                                                            (c) Water table at ground level with upward hydraulic gradient 02

                                                                                            eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                            qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                            F frac14 1296

                                                                                            392frac14 33

                                                                                            85

                                                                                            The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                            Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                            Design value of 0 frac14 tan1tan 39

                                                                                            125

                                                                                            frac14 33

                                                                                            For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                            Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                            Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                            86

                                                                                            (a) Undrained shear for u frac14 0

                                                                                            Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                            qnf frac14 12cuNc

                                                                                            frac14 12 100 514 frac14 617 kN=m2

                                                                                            qn frac14 qnfFfrac14 617

                                                                                            3frac14 206 kN=m2

                                                                                            q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                            Bearing capacity 63

                                                                                            Drained shear for 0 frac14 32

                                                                                            Nq frac14 23 N frac14 25

                                                                                            0 frac14 21 98 frac14 112 kN=m3

                                                                                            qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                            frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                            frac14 694 kN=m2

                                                                                            q frac14 694

                                                                                            3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                            Design load frac14 42 227 frac14 3632 kN

                                                                                            (b) Design undrained strength cud frac14 100

                                                                                            14frac14 71 kNm2

                                                                                            Design bearing resistance Rd frac14 12cudNe area

                                                                                            frac14 12 71 514 42

                                                                                            frac14 7007 kN

                                                                                            For drained shear 0d frac14 tan1tan 32

                                                                                            125

                                                                                            frac14 26

                                                                                            Nq frac14 12 N frac14 10

                                                                                            Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                            (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                            Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                            1 2 100 0175 0700qn 0182qn

                                                                                            2 6 033 0044 0176qn 0046qn

                                                                                            3 10 020 0017 0068qn 0018qn

                                                                                            0246qn

                                                                                            Diameter of equivalent circle B frac14 45m

                                                                                            H

                                                                                            Bfrac14 12

                                                                                            45frac14 27 and A frac14 042

                                                                                            13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                            64 Bearing capacity

                                                                                            For sc frac14 30mm

                                                                                            qn frac14 30

                                                                                            0147frac14 204 kN=m2

                                                                                            q frac14 204thorn 21 frac14 225 kN=m2

                                                                                            Design load frac14 42 225 frac14 3600 kN

                                                                                            The design load is 3600 kN settlement being the limiting criterion

                                                                                            87

                                                                                            D

                                                                                            Bfrac14 8

                                                                                            4frac14 20

                                                                                            From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                            F frac14 cuNc

                                                                                            Dfrac14 40 71

                                                                                            20 8frac14 18

                                                                                            88

                                                                                            Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                            Design value of 0 frac14 tan1tan 38

                                                                                            125

                                                                                            frac14 32

                                                                                            Figure Q86

                                                                                            Bearing capacity 65

                                                                                            For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                            Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                            The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                            Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                            For B frac14 250m qn frac14 3750

                                                                                            2502 17 frac14 583 kN=m2

                                                                                            From Figure 510 m frac14 n frac14 126

                                                                                            6frac14 021

                                                                                            Ir frac14 0019

                                                                                            Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                            Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                            The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                            89

                                                                                            Depth (m) N 0v (kNm2) CN N1

                                                                                            070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                            Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                            (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                            Cw frac14 05thorn 0530

                                                                                            47

                                                                                            frac14 082

                                                                                            66 Bearing capacity

                                                                                            Thus

                                                                                            qa frac14 150 082 frac14 120 kN=m2

                                                                                            (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                            Thus

                                                                                            qa frac14 90 15 frac14 135 kN=m2

                                                                                            (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                            Ic frac14 171

                                                                                            1014frac14 0068

                                                                                            From Equation 819(a) with s frac14 25mm

                                                                                            q frac14 25

                                                                                            3507 0068frac14 150 kN=m2

                                                                                            810

                                                                                            Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                            Izp frac14 05thorn 01130

                                                                                            16 27

                                                                                            05

                                                                                            frac14 067

                                                                                            Refer to Figure Q810

                                                                                            E frac14 25qc

                                                                                            Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                            Ez (mm3MN)

                                                                                            1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                            0203

                                                                                            C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                            130frac14 093

                                                                                            C2 frac14 1 ethsayTHORN

                                                                                            s frac14 C1C2qnX Iz

                                                                                            Ez frac14 093 1 130 0203 frac14 25mm

                                                                                            Bearing capacity 67

                                                                                            811

                                                                                            At pile base level

                                                                                            cu frac14 220 kN=m2

                                                                                            qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                            Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                            00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                            qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                            Then

                                                                                            Qf frac14 Abqb thorn Asqs

                                                                                            frac14

                                                                                            4 32 1980

                                                                                            thorn eth 105 139 86THORN

                                                                                            frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                            0 01 02 03 04 05 06 07

                                                                                            0 2 4 6 8 10 12 14

                                                                                            1

                                                                                            2

                                                                                            3

                                                                                            4

                                                                                            5

                                                                                            6

                                                                                            7

                                                                                            8

                                                                                            (1)

                                                                                            (2)

                                                                                            (3)

                                                                                            (4)

                                                                                            (5)

                                                                                            qc

                                                                                            qc

                                                                                            Iz

                                                                                            Iz

                                                                                            (MNm2)

                                                                                            z (m)

                                                                                            Figure Q810

                                                                                            68 Bearing capacity

                                                                                            Allowable load

                                                                                            ethaTHORN Qf

                                                                                            2frac14 17 937

                                                                                            2frac14 8968 kN

                                                                                            ethbTHORN Abqb

                                                                                            3thorn Asqs frac14 13 996

                                                                                            3thorn 3941 frac14 8606 kN

                                                                                            ie allowable load frac14 8600 kN

                                                                                            Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                            According to the limit state method

                                                                                            Characteristic undrained strength at base level cuk frac14 220

                                                                                            150kN=m2

                                                                                            Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                            150frac14 1320 kN=m2

                                                                                            Characteristic shaft resistance qsk frac14 00150

                                                                                            frac14 86

                                                                                            150frac14 57 kN=m2

                                                                                            Characteristic base and shaft resistances

                                                                                            Rbk frac14

                                                                                            4 32 1320 frac14 9330 kN

                                                                                            Rsk frac14 105 139 86

                                                                                            150frac14 2629 kN

                                                                                            For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                            Design bearing resistance Rcd frac14 9330

                                                                                            160thorn 2629

                                                                                            130

                                                                                            frac14 5831thorn 2022

                                                                                            frac14 7850 kN

                                                                                            Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                            812

                                                                                            ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                            qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                            For a single pile

                                                                                            Qf frac14 Abqb thorn Asqs

                                                                                            frac14

                                                                                            4 062 1305

                                                                                            thorn eth 06 15 42THORN

                                                                                            frac14 369thorn 1187 frac14 1556 kN

                                                                                            Bearing capacity 69

                                                                                            Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                            eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                            (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                            (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                            qbkfrac14 9cuk frac14 9 220

                                                                                            150frac14 1320 kN=m2

                                                                                            qskfrac14cuk frac14 040 105

                                                                                            150frac14 28 kN=m2

                                                                                            Rbkfrac14

                                                                                            4 0602 1320 frac14 373 kN

                                                                                            Rskfrac14 060 15 28 frac14 791 kN

                                                                                            Rcdfrac14 373

                                                                                            160thorn 791

                                                                                            130frac14 233thorn 608 frac14 841 kN

                                                                                            Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                            Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                            (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                            q frac14 21 000

                                                                                            1762frac14 68 kN=m2

                                                                                            Immediate settlement

                                                                                            H

                                                                                            Bfrac14 15

                                                                                            176frac14 085

                                                                                            D

                                                                                            Bfrac14 13

                                                                                            176frac14 074

                                                                                            L

                                                                                            Bfrac14 1

                                                                                            Hence from Figure 515

                                                                                            130 frac14 078 and 131 frac14 041

                                                                                            70 Bearing capacity

                                                                                            Thus using Equation 528

                                                                                            si frac14 078 041 68 176

                                                                                            65frac14 6mm

                                                                                            Consolidation settlement

                                                                                            Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                            1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                            434 (sod)

                                                                                            Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                            sc frac14 056 434 frac14 24mm

                                                                                            The total settlement is (6thorn 24) frac14 30mm

                                                                                            813

                                                                                            At base level N frac14 26 Then using Equation 830

                                                                                            qb frac14 40NDb

                                                                                            Bfrac14 40 26 2

                                                                                            025frac14 8320 kN=m2

                                                                                            ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                            Figure Q812

                                                                                            Bearing capacity 71

                                                                                            Over the length embedded in sand

                                                                                            N frac14 21 ie18thorn 24

                                                                                            2

                                                                                            Using Equation 831

                                                                                            qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                            For a single pile

                                                                                            Qf frac14 Abqb thorn Asqs

                                                                                            frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                            For the pile group assuming a group efficiency of 12

                                                                                            XQf frac14 12 9 604 frac14 6523 kN

                                                                                            Then the load factor is

                                                                                            F frac14 6523

                                                                                            2000thorn 1000frac14 21

                                                                                            (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                            Characteristic base resistance per unit area qbk frac14 8320

                                                                                            150frac14 5547 kNm2

                                                                                            Characteristic shaft resistance per unit area qsk frac14 42

                                                                                            150frac14 28 kNm2

                                                                                            Characteristic base and shaft resistances for a single pile

                                                                                            Rbk frac14 0252 5547 frac14 347 kN

                                                                                            Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                            For a driven pile the partial factors are b frac14 s frac14 130

                                                                                            Design bearing resistance Rcd frac14 347

                                                                                            130thorn 56

                                                                                            130frac14 310 kN

                                                                                            For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                            Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                            (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                            From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                            72 Bearing capacity

                                                                                            N frac14 24thorn 26thorn 34

                                                                                            3frac14 28

                                                                                            Ic frac14 171

                                                                                            2814frac14 0016 ethEquation 818THORN

                                                                                            s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                            814

                                                                                            Using Equation 841

                                                                                            Tf frac14 DLcu thorn

                                                                                            4ethD2 d2THORNcuNc

                                                                                            frac14 eth 02 5 06 110THORN thorn

                                                                                            4eth022 012THORN110 9

                                                                                            frac14 207thorn 23 frac14 230 kN

                                                                                            Figure Q813

                                                                                            Bearing capacity 73

                                                                                            Chapter 9

                                                                                            Stability of slopes

                                                                                            91

                                                                                            Referring to Figure Q91

                                                                                            W frac14 417 19 frac14 792 kN=m

                                                                                            Q frac14 20 28 frac14 56 kN=m

                                                                                            Arc lengthAB frac14

                                                                                            180 73 90 frac14 115m

                                                                                            Arc length BC frac14

                                                                                            180 28 90 frac14 44m

                                                                                            The factor of safety is given by

                                                                                            F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                            90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                            Depth of tension crack z0 frac14 2cu

                                                                                            frac14 2 20

                                                                                            19frac14 21m

                                                                                            Arc length BD frac14

                                                                                            180 13

                                                                                            1

                                                                                            2 90 frac14 21m

                                                                                            F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                            The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                            Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                            14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                            Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                            The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                            92

                                                                                            u frac14 0

                                                                                            Depth factor D frac14 11

                                                                                            9frac14 122

                                                                                            Using Equation 92 with F frac14 10

                                                                                            Ns frac14 cu

                                                                                            FHfrac14 30

                                                                                            10 19 9frac14 0175

                                                                                            Hence from Figure 93

                                                                                            frac14 50

                                                                                            For F frac14 12

                                                                                            Ns frac14 30

                                                                                            12 19 9frac14 0146

                                                                                            frac14 27

                                                                                            93

                                                                                            Refer to Figure Q93

                                                                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                            1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                            74 m

                                                                                            214 1deg

                                                                                            213 1deg

                                                                                            39 m

                                                                                            WB

                                                                                            D

                                                                                            C

                                                                                            28 m

                                                                                            21 m

                                                                                            A

                                                                                            Q

                                                                                            Soil (1)Soil (2)

                                                                                            73deg

                                                                                            Figure Q91

                                                                                            Stability of slopes 75

                                                                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                            9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                            599 256 328 1372

                                                                                            Figure Q93

                                                                                            76 Stability of slopes

                                                                                            XW cos frac14 b

                                                                                            Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                            W sin frac14 bX

                                                                                            h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                            Arc length La frac14

                                                                                            180 57

                                                                                            1

                                                                                            2 326 frac14 327m

                                                                                            The factor of safety is given by

                                                                                            F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                            W sin

                                                                                            frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                            frac14 091

                                                                                            According to the limit state method

                                                                                            0d frac14 tan1tan 32

                                                                                            125

                                                                                            frac14 265

                                                                                            c0 frac14 8

                                                                                            160frac14 5 kN=m2

                                                                                            Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                            Design disturbing moment frac14 1075 kN=m

                                                                                            The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                            94

                                                                                            F frac14 1

                                                                                            W sin

                                                                                            Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                            1thorn ethtan tan0=FTHORN

                                                                                            c0 frac14 8 kN=m2

                                                                                            0 frac14 32

                                                                                            c0b frac14 8 2 frac14 16 kN=m

                                                                                            W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                            Try F frac14 100

                                                                                            tan0

                                                                                            Ffrac14 0625

                                                                                            Stability of slopes 77

                                                                                            Values of u are as obtained in Figure Q93

                                                                                            SliceNo

                                                                                            h(m)

                                                                                            W frac14 bh(kNm)

                                                                                            W sin(kNm)

                                                                                            ub(kNm)

                                                                                            c0bthorn (W ub) tan0(kNm)

                                                                                            sec

                                                                                            1thorn (tan tan0)FProduct(kNm)

                                                                                            1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                            23 33 30 1042 31

                                                                                            3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                            224 92 72 0931 67

                                                                                            6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                            12 58 112 90 0889 80

                                                                                            8 60 252 1812

                                                                                            80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                            10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                            2154 88 116 0853 99

                                                                                            14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                            1074 1091

                                                                                            F frac14 1091

                                                                                            1074frac14 102 (assumed value 100)

                                                                                            Thus

                                                                                            F frac14 101

                                                                                            95

                                                                                            F frac14 1

                                                                                            W sin

                                                                                            XfWeth1 ruTHORN tan0g sec

                                                                                            1thorn ethtan tan0THORN=F

                                                                                            0 frac14 33

                                                                                            ru frac14 020

                                                                                            W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                            eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                            Try F frac14 110

                                                                                            tan 0

                                                                                            Ffrac14 tan 33

                                                                                            110frac14 0590

                                                                                            78 Stability of slopes

                                                                                            Referring to Figure Q95

                                                                                            SliceNo

                                                                                            h(m)

                                                                                            W frac14 bh(kNm)

                                                                                            W sin(kNm)

                                                                                            W(1 ru) tan0(kNm)

                                                                                            sec

                                                                                            1thorn ( tan tan0)FProduct(kNm)

                                                                                            1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                            2120 234 0892 209

                                                                                            4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                            1185 1271

                                                                                            Figure Q95

                                                                                            Stability of slopes 79

                                                                                            F frac14 1271

                                                                                            1185frac14 107

                                                                                            The trial value was 110 therefore take F to be 108

                                                                                            96

                                                                                            (a) Water table at surface the factor of safety is given by Equation 912

                                                                                            F frac14 0

                                                                                            sat

                                                                                            tan0

                                                                                            tan

                                                                                            ptie 15 frac14 92

                                                                                            19

                                                                                            tan 36

                                                                                            tan

                                                                                            tan frac14 0234

                                                                                            frac14 13

                                                                                            Water table well below surface the factor of safety is given by Equation 911

                                                                                            F frac14 tan0

                                                                                            tan

                                                                                            frac14 tan 36

                                                                                            tan 13

                                                                                            frac14 31

                                                                                            (b) 0d frac14 tan1tan 36

                                                                                            125

                                                                                            frac14 30

                                                                                            Depth of potential failure surface frac14 z

                                                                                            Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                            frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                            frac14 504z kN

                                                                                            Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                            frac14 19 z sin 13 cos 13

                                                                                            frac14 416z kN

                                                                                            Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                            80 Stability of slopes

                                                                                            • Book Cover
                                                                                            • Title
                                                                                            • Contents
                                                                                            • Basic characteristics of soils
                                                                                            • Seepage
                                                                                            • Effective stress
                                                                                            • Shear strength
                                                                                            • Stresses and displacements
                                                                                            • Lateral earth pressure
                                                                                            • Consolidation theory
                                                                                            • Bearing capacity
                                                                                            • Stability of slopes

                                                                                              Base pressures (Equation 627)

                                                                                              p frac14 525

                                                                                              41 6 049

                                                                                              4

                                                                                              frac14 228 kN=m2 and 35 kN=m2

                                                                                              The overturning limit state is satisfied the restoring moment (1307 kNm) beinggreater than the overturning moment (512 kNm)

                                                                                              The bearing resistance limit state is satisfied the ultimate bearing capacity of thefoundation soil (250kNm2) being greater than the maximum base pressure (228 kNm2)

                                                                                              The sliding limit state is satisfied the restoring force (525 tan 25 frac14 245 kN) beinggreater than the disturbing force (205 kN)

                                                                                              67

                                                                                              For 0 frac14 35 Ka frac14 027for 0 frac14 27 Ka frac14 0375 and Kp frac14 267for soil 0 frac14 112 kNm3for backfill 0 frac14 102 kNm3The pressure distribution is shown in Figure Q67 Hydrostatic pressure is balancedConsider moments about the anchor point (A) per m

                                                                                              Force (kN) Arm (m) Moment (kNm)

                                                                                              (1)1

                                                                                              2 027 17 52 frac14 574 183 1050

                                                                                              (2) 027 17 5 3 frac14 689 500 3445

                                                                                              (3)1

                                                                                              2 027 102 32 frac14 124 550 682

                                                                                              (4) 0375 f(17 5)thorn (102 3)g d frac14 434d d2thorn 650 217d2 thorn 2821d

                                                                                              (5)1

                                                                                              2 0375 112 d2 frac14 21d2 2d3thorn 650 14d3 thorn 137d2

                                                                                              (6) 2 10 ffiffiffiffiffiffiffiffiffiffiffi0375p d frac14 122d d2thorn 650 61d2 793d

                                                                                              (7) 1

                                                                                              2 267

                                                                                              2 112 d2 frac14 75d2 2d3thorn 650 50d3 488d2

                                                                                              (8) 2 10ffiffiffiffiffiffiffiffiffi267p

                                                                                              2 d frac14 163d d2thorn 650 82d2 1060d

                                                                                              Tie rod force per m frac14 T 0 0

                                                                                              XM frac1436d3 277d2 thorn 968d thorn 5177 frac14 0

                                                                                              d3 thorn 77d2 269d 1438 frac14 0

                                                                                              d frac14 467m

                                                                                              Depth of penetration frac14 12d frac14 560m

                                                                                              Lateral earth pressure 41

                                                                                              Algebraic sum of forces for d frac14 467m isX

                                                                                              F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                                              T frac14 905 kN=m

                                                                                              Force in each tie rod frac14 25T frac14 226 kN

                                                                                              68

                                                                                              (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                                              0 frac14 21 98 frac14 112 kN=m3

                                                                                              The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                                              uC frac14 150

                                                                                              165 15 98 frac14 134 kN=m2

                                                                                              The average seepage pressure is

                                                                                              j frac14 15

                                                                                              165 98 frac14 09 kN=m3

                                                                                              Hence

                                                                                              0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                                              0 j frac14 112 09 frac14 103 kN=m3

                                                                                              Figure Q67

                                                                                              42 Lateral earth pressure

                                                                                              Consider moments about the anchor point A (per m)

                                                                                              Force (kN) Arm (m) Moment (kN m)

                                                                                              (1) 10 026 150 frac14 390 60 2340

                                                                                              (2)1

                                                                                              2 026 18 452 frac14 474 15 711

                                                                                              (3) 026 18 45 105 frac14 2211 825 18240

                                                                                              (4)1

                                                                                              2 026 121 1052 frac14 1734 100 17340

                                                                                              (5)1

                                                                                              2 134 15 frac14 101 40 404

                                                                                              (6) 134 30 frac14 402 60 2412

                                                                                              (7)1

                                                                                              2 134 60 frac14 402 95 3819

                                                                                              571 4527(8) Ppm

                                                                                              115 115PPm

                                                                                              XM frac14 0

                                                                                              Ppm frac144527

                                                                                              115frac14 394 kN=m

                                                                                              Available passive resistance

                                                                                              Pp frac14 1

                                                                                              2 385 103 62 frac14 714 kN=m

                                                                                              Factor of safety

                                                                                              Fp frac14 Pp

                                                                                              Ppm

                                                                                              frac14 714

                                                                                              394frac14 18

                                                                                              Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                                              Figure Q68

                                                                                              Lateral earth pressure 43

                                                                                              (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                                              Consider moments (per m) about the tie point A

                                                                                              Force (kN) Arm (m)

                                                                                              (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                                              (2)1

                                                                                              2 033 18 452 frac14 601 15

                                                                                              (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                                              (4)1

                                                                                              2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                                              (5)1

                                                                                              2 134 15 frac14 101 40

                                                                                              (6) 134 30 frac14 402 60

                                                                                              (7)1

                                                                                              2 134 d frac14 67d d3thorn 75

                                                                                              (8) 1

                                                                                              2 30 103 d2 frac141545d2 2d3thorn 75

                                                                                              Moment (kN m)

                                                                                              (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                                              XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                                              d3 thorn 827d2 466d 1518 frac14 0

                                                                                              By trial

                                                                                              d frac14 544m

                                                                                              The minimum depth of embedment required is 544m

                                                                                              69

                                                                                              For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                                              0 frac14 20 98 frac14 102 kN=m3

                                                                                              The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                                              44 Lateral earth pressure

                                                                                              uC frac14 147

                                                                                              173 26 98 frac14 216 kN=m2

                                                                                              and the average seepage pressure around the wall is

                                                                                              j frac14 26

                                                                                              173 98 frac14 15 kN=m3

                                                                                              Consider moments about the prop (A) (per m)

                                                                                              Force (kN) Arm (m) Moment (kN m)

                                                                                              (1)1

                                                                                              2 03 17 272 frac14 186 020 37

                                                                                              (2) 03 17 27 53 frac14 730 335 2445

                                                                                              (3)1

                                                                                              2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                                              (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                                              (5)1

                                                                                              2 216 26 frac14 281 243 684

                                                                                              (6) 216 27 frac14 583 465 2712

                                                                                              (7)1

                                                                                              2 216 60 frac14 648 800 5184

                                                                                              3055(8)

                                                                                              1

                                                                                              2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                                              Factor of safety

                                                                                              Fr frac14 6885

                                                                                              3055frac14 225

                                                                                              Figure Q69

                                                                                              Lateral earth pressure 45

                                                                                              610

                                                                                              For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                                              p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                                              Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                                              Using the recommendations of Twine and Roscoe

                                                                                              p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                                              Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                                              611

                                                                                              frac14 18 kN=m3 0 frac14 34

                                                                                              H frac14 350m nH frac14 335m mH frac14 185m

                                                                                              Consider a trial value of F frac14 20 Refer to Figure 635

                                                                                              0m frac14 tan1tan 34

                                                                                              20

                                                                                              frac14 186

                                                                                              Then

                                                                                              frac14 45 thorn 0m2frac14 543

                                                                                              W frac14 1

                                                                                              2 18 3502 cot 543 frac14 792 kN=m

                                                                                              Figure Q610

                                                                                              46 Lateral earth pressure

                                                                                              P frac14 1

                                                                                              2 s 3352 frac14 561s kN=m

                                                                                              U frac14 1

                                                                                              2 98 1852 cosec 543 frac14 206 kN=m

                                                                                              Equations 630 and 631 then become

                                                                                              561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                                              792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                                              ie

                                                                                              561s 0616N 405 frac14 0

                                                                                              792 0857N thorn 563 frac14 0

                                                                                              N frac14 848

                                                                                              0857frac14 989 kN=m

                                                                                              Then

                                                                                              561s 609 405 frac14 0

                                                                                              s frac14 649

                                                                                              561frac14 116 kN=m3

                                                                                              The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                                              F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                                              20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                                              s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                                              Figure Q611

                                                                                              Lateral earth pressure 47

                                                                                              612

                                                                                              For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                              45 thorn 0

                                                                                              2frac14 63

                                                                                              For the retained material between the surface and a depth of 36m

                                                                                              Pa frac14 1

                                                                                              2 030 18 362 frac14 350 kN=m

                                                                                              Weight of reinforced fill between the surface and a depth of 36m is

                                                                                              Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                              eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                              Eccentricity of Rv

                                                                                              e frac14 263 250 frac14 013m

                                                                                              The average vertical stress at a depth of 36m is

                                                                                              z frac14 Rv

                                                                                              L 2efrac14 324

                                                                                              474frac14 68 kN=m2

                                                                                              (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                              Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                              Tensile stress in the element frac14 138 103

                                                                                              65 3frac14 71N=mm2

                                                                                              Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                              Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                              Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                              The weight of ABC is

                                                                                              W frac14 1

                                                                                              2 18 52 265 frac14 124 kN=m

                                                                                              From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                              48 Lateral earth pressure

                                                                                              (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                              Tp frac14 032 68 120 065 frac14 170 kN

                                                                                              Tr frac14 213 420

                                                                                              418frac14 214 kN

                                                                                              Again the tensile failure and slipping limit states are satisfied for this element

                                                                                              Figure Q612

                                                                                              Lateral earth pressure 49

                                                                                              Chapter 7

                                                                                              Consolidation theory

                                                                                              71

                                                                                              Total change in thickness

                                                                                              H frac14 782 602 frac14 180mm

                                                                                              Average thickness frac14 1530thorn 180

                                                                                              2frac14 1620mm

                                                                                              Length of drainage path d frac14 1620

                                                                                              2frac14 810mm

                                                                                              Root time plot (Figure Q71a)

                                                                                              ffiffiffiffiffiffit90p frac14 33

                                                                                              t90 frac14 109min

                                                                                              cv frac14 0848d2

                                                                                              t90frac14 0848 8102

                                                                                              109 1440 365

                                                                                              106frac14 27m2=year

                                                                                              r0 frac14 782 764

                                                                                              782 602frac14 018

                                                                                              180frac14 0100

                                                                                              rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                              10 119

                                                                                              9 180frac14 0735

                                                                                              rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                              Log time plot (Figure Q71b)

                                                                                              t50 frac14 26min

                                                                                              cv frac14 0196d2

                                                                                              t50frac14 0196 8102

                                                                                              26 1440 365

                                                                                              106frac14 26m2=year

                                                                                              r0 frac14 782 763

                                                                                              782 602frac14 019

                                                                                              180frac14 0106

                                                                                              rp frac14 763 623

                                                                                              782 602frac14 140

                                                                                              180frac14 0778

                                                                                              rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                              Figure Q71(a)

                                                                                              Figure Q71(b)

                                                                                              Final void ratio

                                                                                              e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                              e

                                                                                              Hfrac14 1thorn e0

                                                                                              H0frac14 1thorn e1 thorne

                                                                                              H0

                                                                                              ie

                                                                                              e

                                                                                              180frac14 1631thorne

                                                                                              1710

                                                                                              e frac14 2936

                                                                                              1530frac14 0192

                                                                                              Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                              mv frac14 1

                                                                                              1thorn e0 e0 e101 00

                                                                                              frac14 1

                                                                                              1823 0192

                                                                                              0107frac14 098m2=MN

                                                                                              k frac14 cvmvw frac14 265 098 98

                                                                                              60 1440 365 103frac14 81 1010 m=s

                                                                                              72

                                                                                              Using Equation 77 (one-dimensional method)

                                                                                              sc frac14 e0 e11thorn e0 H

                                                                                              Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                              Figure Q72

                                                                                              52 Consolidation theory

                                                                                              Settlement

                                                                                              Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                              1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                              318

                                                                                              Notes 5 92y 460thorn 84

                                                                                              Heave

                                                                                              Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                              1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                              38

                                                                                              73

                                                                                              U frac14 f ethTvTHORN frac14 f cvt

                                                                                              d2

                                                                                              Hence if cv is constant

                                                                                              t1

                                                                                              t2frac14 d

                                                                                              21

                                                                                              d22

                                                                                              where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                              d1 frac14 95mm and d2 frac14 2500mm

                                                                                              for U frac14 050 t2 frac14 t1 d22

                                                                                              d21

                                                                                              frac14 20

                                                                                              60 24 365 25002

                                                                                              952frac14 263 years

                                                                                              for U lt 060 Tv frac14

                                                                                              4U2 (Equation 724(a))

                                                                                              t030 frac14 t050 0302

                                                                                              0502

                                                                                              frac14 263 036 frac14 095 years

                                                                                              Consolidation theory 53

                                                                                              74

                                                                                              The layer is open

                                                                                              d frac14 8

                                                                                              2frac14 4m

                                                                                              Tv frac14 cvtd2frac14 24 3

                                                                                              42frac14 0450

                                                                                              ui frac14 frac14 84 kN=m2

                                                                                              The excess pore water pressure is given by Equation 721

                                                                                              ue frac14Xmfrac141mfrac140

                                                                                              2ui

                                                                                              Msin

                                                                                              Mz

                                                                                              d

                                                                                              expethM2TvTHORN

                                                                                              In this case z frac14 d

                                                                                              sinMz

                                                                                              d

                                                                                              frac14 sinM

                                                                                              where

                                                                                              M frac14

                                                                                              23

                                                                                              25

                                                                                              2

                                                                                              M sin M M2Tv exp (M2Tv)

                                                                                              2thorn1 1110 0329

                                                                                              3

                                                                                              21 9993 457 105

                                                                                              ue frac14 2 84 2

                                                                                              1 0329 ethother terms negligibleTHORN

                                                                                              frac14 352 kN=m2

                                                                                              75

                                                                                              The layer is open

                                                                                              d frac14 6

                                                                                              2frac14 3m

                                                                                              Tv frac14 cvtd2frac14 10 3

                                                                                              32frac14 0333

                                                                                              The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                              54 Consolidation theory

                                                                                              For an open layer

                                                                                              Tv frac14 4n

                                                                                              m2

                                                                                              n frac14 0333 62

                                                                                              4frac14 300

                                                                                              The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                              ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                              i j

                                                                                              0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                              0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                              The initial and 3-year isochrones are plotted in Figure Q75

                                                                                              Area under initial isochrone frac14 180 units

                                                                                              Area under 3-year isochrone frac14 63 units

                                                                                              The average degree of consolidation is given by Equation 725Thus

                                                                                              U frac14 1 63

                                                                                              180frac14 065

                                                                                              Figure Q75

                                                                                              Consolidation theory 55

                                                                                              76

                                                                                              At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                              0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                              The final consolidation settlement (one-dimensional method) is

                                                                                              sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                              Corrected time t frac14 2 1

                                                                                              2

                                                                                              40

                                                                                              52

                                                                                              frac14 1615 years

                                                                                              Tv frac14 cvtd2frac14 44 1615

                                                                                              42frac14 0444

                                                                                              From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                              77

                                                                                              The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                              Figure Q77

                                                                                              56 Consolidation theory

                                                                                              Point m n Ir (kNm2) sc (mm)

                                                                                              13020frac14 15 20

                                                                                              20frac14 10 0194 (4) 113 124

                                                                                              260

                                                                                              20frac14 30

                                                                                              20

                                                                                              20frac14 10 0204 (2) 59 65

                                                                                              360

                                                                                              20frac14 30

                                                                                              40

                                                                                              20frac14 20 0238 (1) 35 38

                                                                                              430

                                                                                              20frac14 15

                                                                                              40

                                                                                              20frac14 20 0224 (2) 65 72

                                                                                              Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                              78

                                                                                              Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                              (a) Immediate settlement

                                                                                              H

                                                                                              Bfrac14 30

                                                                                              35frac14 086

                                                                                              D

                                                                                              Bfrac14 2

                                                                                              35frac14 006

                                                                                              Figure Q78

                                                                                              Consolidation theory 57

                                                                                              From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                              si frac14 130131qB

                                                                                              Eufrac14 10 032 105 35

                                                                                              40frac14 30mm

                                                                                              (b) Consolidation settlement

                                                                                              Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                              1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                              3150

                                                                                              Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                              Now

                                                                                              H

                                                                                              Bfrac14 30

                                                                                              35frac14 086 and A frac14 065

                                                                                              from Figure 712 13 frac14 079

                                                                                              sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                              Total settlement

                                                                                              s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                              79

                                                                                              Without sand drains

                                                                                              Uv frac14 025

                                                                                              Tv frac14 0049 ethfrom Figure 718THORN

                                                                                              t frac14 Tvd2

                                                                                              cvfrac14 0049 82

                                                                                              cvWith sand drains

                                                                                              R frac14 0564S frac14 0564 3 frac14 169m

                                                                                              n frac14 Rrfrac14 169

                                                                                              015frac14 113

                                                                                              Tr frac14 cht

                                                                                              4R2frac14 ch

                                                                                              4 1692 0049 82

                                                                                              cvethand ch frac14 cvTHORN

                                                                                              frac14 0275

                                                                                              Ur frac14 073 (from Figure 730)

                                                                                              58 Consolidation theory

                                                                                              Using Equation 740

                                                                                              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                              U frac14 080

                                                                                              710

                                                                                              Without sand drains

                                                                                              Uv frac14 090

                                                                                              Tv frac14 0848

                                                                                              t frac14 Tvd2

                                                                                              cvfrac14 0848 102

                                                                                              96frac14 88 years

                                                                                              With sand drains

                                                                                              R frac14 0564S frac14 0564 4 frac14 226m

                                                                                              n frac14 Rrfrac14 226

                                                                                              015frac14 15

                                                                                              Tr

                                                                                              Tvfrac14 chcv

                                                                                              d2

                                                                                              4R2ethsame tTHORN

                                                                                              Tr

                                                                                              Tvfrac14 140

                                                                                              96 102

                                                                                              4 2262frac14 714 eth1THORN

                                                                                              Using Equation 740

                                                                                              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                              An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                              Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                              040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                              Thus

                                                                                              Uv frac14 0295 and Ur frac14 086

                                                                                              t frac14 88 00683

                                                                                              0848frac14 07 years

                                                                                              Consolidation theory 59

                                                                                              Chapter 8

                                                                                              Bearing capacity

                                                                                              81

                                                                                              (a) The ultimate bearing capacity is given by Equation 83

                                                                                              qf frac14 cNc thorn DNq thorn 1

                                                                                              2BN

                                                                                              For u frac14 0

                                                                                              Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                              qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                              The net ultimate bearing capacity is

                                                                                              qnf frac14 qf D frac14 540 kN=m2

                                                                                              The net foundation pressure is

                                                                                              qn frac14 q D frac14 425

                                                                                              2 eth21 1THORN frac14 192 kN=m2

                                                                                              The factor of safety (Equation 86) is

                                                                                              F frac14 qnfqnfrac14 540

                                                                                              192frac14 28

                                                                                              (b) For 0 frac14 28

                                                                                              Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                              qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                              2 112 2 13

                                                                                              frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                              qnf frac14 574 112 frac14 563 kN=m2

                                                                                              F frac14 563

                                                                                              192frac14 29

                                                                                              (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                              82

                                                                                              For 0 frac14 38

                                                                                              Nq frac14 49 N frac14 67

                                                                                              qnf frac14 DethNq 1THORN thorn 1

                                                                                              2BN ethfrom Equation 83THORN

                                                                                              frac14 eth18 075 48THORN thorn 1

                                                                                              2 18 15 67

                                                                                              frac14 648thorn 905 frac14 1553 kN=m2

                                                                                              qn frac14 500

                                                                                              15 eth18 075THORN frac14 320 kN=m2

                                                                                              F frac14 qnfqnfrac14 1553

                                                                                              320frac14 48

                                                                                              0d frac14 tan1tan 38

                                                                                              125

                                                                                              frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                              Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                              2 18 15 25

                                                                                              frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                              Design load (action) Vd frac14 500 kN=m

                                                                                              The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                              83

                                                                                              D

                                                                                              Bfrac14 350

                                                                                              225frac14 155

                                                                                              From Figure 85 for a square foundation

                                                                                              Nc frac14 81

                                                                                              Bearing capacity 61

                                                                                              For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                              Nc frac14 084thorn 016B

                                                                                              L

                                                                                              81 frac14 745

                                                                                              Using Equation 810

                                                                                              qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                              For F frac14 3

                                                                                              qn frac14 1006

                                                                                              3frac14 335 kN=m2

                                                                                              q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                              Design load frac14 405 450 225 frac14 4100 kN

                                                                                              Design undrained strength cud frac14 135

                                                                                              14frac14 96 kN=m2

                                                                                              Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                              frac14 7241 kN

                                                                                              Design load Vd frac14 4100 kN

                                                                                              Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                              84

                                                                                              For 0 frac14 40

                                                                                              Nq frac14 64 N frac14 95

                                                                                              qnf frac14 DethNq 1THORN thorn 04BN

                                                                                              (a) Water table 5m below ground level

                                                                                              qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                              qn frac14 400 17 frac14 383 kN=m2

                                                                                              F frac14 2686

                                                                                              383frac14 70

                                                                                              (b) Water table 1m below ground level (ie at foundation level)

                                                                                              0 frac14 20 98 frac14 102 kN=m3

                                                                                              62 Bearing capacity

                                                                                              qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                              F frac14 2040

                                                                                              383frac14 53

                                                                                              (c) Water table at ground level with upward hydraulic gradient 02

                                                                                              eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                              qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                              F frac14 1296

                                                                                              392frac14 33

                                                                                              85

                                                                                              The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                              Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                              Design value of 0 frac14 tan1tan 39

                                                                                              125

                                                                                              frac14 33

                                                                                              For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                              Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                              Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                              86

                                                                                              (a) Undrained shear for u frac14 0

                                                                                              Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                              qnf frac14 12cuNc

                                                                                              frac14 12 100 514 frac14 617 kN=m2

                                                                                              qn frac14 qnfFfrac14 617

                                                                                              3frac14 206 kN=m2

                                                                                              q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                              Bearing capacity 63

                                                                                              Drained shear for 0 frac14 32

                                                                                              Nq frac14 23 N frac14 25

                                                                                              0 frac14 21 98 frac14 112 kN=m3

                                                                                              qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                              frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                              frac14 694 kN=m2

                                                                                              q frac14 694

                                                                                              3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                              Design load frac14 42 227 frac14 3632 kN

                                                                                              (b) Design undrained strength cud frac14 100

                                                                                              14frac14 71 kNm2

                                                                                              Design bearing resistance Rd frac14 12cudNe area

                                                                                              frac14 12 71 514 42

                                                                                              frac14 7007 kN

                                                                                              For drained shear 0d frac14 tan1tan 32

                                                                                              125

                                                                                              frac14 26

                                                                                              Nq frac14 12 N frac14 10

                                                                                              Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                              (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                              Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                              1 2 100 0175 0700qn 0182qn

                                                                                              2 6 033 0044 0176qn 0046qn

                                                                                              3 10 020 0017 0068qn 0018qn

                                                                                              0246qn

                                                                                              Diameter of equivalent circle B frac14 45m

                                                                                              H

                                                                                              Bfrac14 12

                                                                                              45frac14 27 and A frac14 042

                                                                                              13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                              64 Bearing capacity

                                                                                              For sc frac14 30mm

                                                                                              qn frac14 30

                                                                                              0147frac14 204 kN=m2

                                                                                              q frac14 204thorn 21 frac14 225 kN=m2

                                                                                              Design load frac14 42 225 frac14 3600 kN

                                                                                              The design load is 3600 kN settlement being the limiting criterion

                                                                                              87

                                                                                              D

                                                                                              Bfrac14 8

                                                                                              4frac14 20

                                                                                              From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                              F frac14 cuNc

                                                                                              Dfrac14 40 71

                                                                                              20 8frac14 18

                                                                                              88

                                                                                              Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                              Design value of 0 frac14 tan1tan 38

                                                                                              125

                                                                                              frac14 32

                                                                                              Figure Q86

                                                                                              Bearing capacity 65

                                                                                              For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                              Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                              The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                              Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                              For B frac14 250m qn frac14 3750

                                                                                              2502 17 frac14 583 kN=m2

                                                                                              From Figure 510 m frac14 n frac14 126

                                                                                              6frac14 021

                                                                                              Ir frac14 0019

                                                                                              Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                              Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                              The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                              89

                                                                                              Depth (m) N 0v (kNm2) CN N1

                                                                                              070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                              Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                              (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                              Cw frac14 05thorn 0530

                                                                                              47

                                                                                              frac14 082

                                                                                              66 Bearing capacity

                                                                                              Thus

                                                                                              qa frac14 150 082 frac14 120 kN=m2

                                                                                              (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                              Thus

                                                                                              qa frac14 90 15 frac14 135 kN=m2

                                                                                              (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                              Ic frac14 171

                                                                                              1014frac14 0068

                                                                                              From Equation 819(a) with s frac14 25mm

                                                                                              q frac14 25

                                                                                              3507 0068frac14 150 kN=m2

                                                                                              810

                                                                                              Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                              Izp frac14 05thorn 01130

                                                                                              16 27

                                                                                              05

                                                                                              frac14 067

                                                                                              Refer to Figure Q810

                                                                                              E frac14 25qc

                                                                                              Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                              Ez (mm3MN)

                                                                                              1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                              0203

                                                                                              C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                              130frac14 093

                                                                                              C2 frac14 1 ethsayTHORN

                                                                                              s frac14 C1C2qnX Iz

                                                                                              Ez frac14 093 1 130 0203 frac14 25mm

                                                                                              Bearing capacity 67

                                                                                              811

                                                                                              At pile base level

                                                                                              cu frac14 220 kN=m2

                                                                                              qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                              Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                              00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                              qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                              Then

                                                                                              Qf frac14 Abqb thorn Asqs

                                                                                              frac14

                                                                                              4 32 1980

                                                                                              thorn eth 105 139 86THORN

                                                                                              frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                              0 01 02 03 04 05 06 07

                                                                                              0 2 4 6 8 10 12 14

                                                                                              1

                                                                                              2

                                                                                              3

                                                                                              4

                                                                                              5

                                                                                              6

                                                                                              7

                                                                                              8

                                                                                              (1)

                                                                                              (2)

                                                                                              (3)

                                                                                              (4)

                                                                                              (5)

                                                                                              qc

                                                                                              qc

                                                                                              Iz

                                                                                              Iz

                                                                                              (MNm2)

                                                                                              z (m)

                                                                                              Figure Q810

                                                                                              68 Bearing capacity

                                                                                              Allowable load

                                                                                              ethaTHORN Qf

                                                                                              2frac14 17 937

                                                                                              2frac14 8968 kN

                                                                                              ethbTHORN Abqb

                                                                                              3thorn Asqs frac14 13 996

                                                                                              3thorn 3941 frac14 8606 kN

                                                                                              ie allowable load frac14 8600 kN

                                                                                              Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                              According to the limit state method

                                                                                              Characteristic undrained strength at base level cuk frac14 220

                                                                                              150kN=m2

                                                                                              Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                              150frac14 1320 kN=m2

                                                                                              Characteristic shaft resistance qsk frac14 00150

                                                                                              frac14 86

                                                                                              150frac14 57 kN=m2

                                                                                              Characteristic base and shaft resistances

                                                                                              Rbk frac14

                                                                                              4 32 1320 frac14 9330 kN

                                                                                              Rsk frac14 105 139 86

                                                                                              150frac14 2629 kN

                                                                                              For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                              Design bearing resistance Rcd frac14 9330

                                                                                              160thorn 2629

                                                                                              130

                                                                                              frac14 5831thorn 2022

                                                                                              frac14 7850 kN

                                                                                              Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                              812

                                                                                              ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                              qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                              For a single pile

                                                                                              Qf frac14 Abqb thorn Asqs

                                                                                              frac14

                                                                                              4 062 1305

                                                                                              thorn eth 06 15 42THORN

                                                                                              frac14 369thorn 1187 frac14 1556 kN

                                                                                              Bearing capacity 69

                                                                                              Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                              eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                              (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                              (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                              qbkfrac14 9cuk frac14 9 220

                                                                                              150frac14 1320 kN=m2

                                                                                              qskfrac14cuk frac14 040 105

                                                                                              150frac14 28 kN=m2

                                                                                              Rbkfrac14

                                                                                              4 0602 1320 frac14 373 kN

                                                                                              Rskfrac14 060 15 28 frac14 791 kN

                                                                                              Rcdfrac14 373

                                                                                              160thorn 791

                                                                                              130frac14 233thorn 608 frac14 841 kN

                                                                                              Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                              Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                              (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                              q frac14 21 000

                                                                                              1762frac14 68 kN=m2

                                                                                              Immediate settlement

                                                                                              H

                                                                                              Bfrac14 15

                                                                                              176frac14 085

                                                                                              D

                                                                                              Bfrac14 13

                                                                                              176frac14 074

                                                                                              L

                                                                                              Bfrac14 1

                                                                                              Hence from Figure 515

                                                                                              130 frac14 078 and 131 frac14 041

                                                                                              70 Bearing capacity

                                                                                              Thus using Equation 528

                                                                                              si frac14 078 041 68 176

                                                                                              65frac14 6mm

                                                                                              Consolidation settlement

                                                                                              Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                              1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                              434 (sod)

                                                                                              Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                              sc frac14 056 434 frac14 24mm

                                                                                              The total settlement is (6thorn 24) frac14 30mm

                                                                                              813

                                                                                              At base level N frac14 26 Then using Equation 830

                                                                                              qb frac14 40NDb

                                                                                              Bfrac14 40 26 2

                                                                                              025frac14 8320 kN=m2

                                                                                              ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                              Figure Q812

                                                                                              Bearing capacity 71

                                                                                              Over the length embedded in sand

                                                                                              N frac14 21 ie18thorn 24

                                                                                              2

                                                                                              Using Equation 831

                                                                                              qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                              For a single pile

                                                                                              Qf frac14 Abqb thorn Asqs

                                                                                              frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                              For the pile group assuming a group efficiency of 12

                                                                                              XQf frac14 12 9 604 frac14 6523 kN

                                                                                              Then the load factor is

                                                                                              F frac14 6523

                                                                                              2000thorn 1000frac14 21

                                                                                              (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                              Characteristic base resistance per unit area qbk frac14 8320

                                                                                              150frac14 5547 kNm2

                                                                                              Characteristic shaft resistance per unit area qsk frac14 42

                                                                                              150frac14 28 kNm2

                                                                                              Characteristic base and shaft resistances for a single pile

                                                                                              Rbk frac14 0252 5547 frac14 347 kN

                                                                                              Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                              For a driven pile the partial factors are b frac14 s frac14 130

                                                                                              Design bearing resistance Rcd frac14 347

                                                                                              130thorn 56

                                                                                              130frac14 310 kN

                                                                                              For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                              Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                              (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                              From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                              72 Bearing capacity

                                                                                              N frac14 24thorn 26thorn 34

                                                                                              3frac14 28

                                                                                              Ic frac14 171

                                                                                              2814frac14 0016 ethEquation 818THORN

                                                                                              s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                              814

                                                                                              Using Equation 841

                                                                                              Tf frac14 DLcu thorn

                                                                                              4ethD2 d2THORNcuNc

                                                                                              frac14 eth 02 5 06 110THORN thorn

                                                                                              4eth022 012THORN110 9

                                                                                              frac14 207thorn 23 frac14 230 kN

                                                                                              Figure Q813

                                                                                              Bearing capacity 73

                                                                                              Chapter 9

                                                                                              Stability of slopes

                                                                                              91

                                                                                              Referring to Figure Q91

                                                                                              W frac14 417 19 frac14 792 kN=m

                                                                                              Q frac14 20 28 frac14 56 kN=m

                                                                                              Arc lengthAB frac14

                                                                                              180 73 90 frac14 115m

                                                                                              Arc length BC frac14

                                                                                              180 28 90 frac14 44m

                                                                                              The factor of safety is given by

                                                                                              F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                              90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                              Depth of tension crack z0 frac14 2cu

                                                                                              frac14 2 20

                                                                                              19frac14 21m

                                                                                              Arc length BD frac14

                                                                                              180 13

                                                                                              1

                                                                                              2 90 frac14 21m

                                                                                              F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                              The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                              Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                              14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                              Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                              The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                              92

                                                                                              u frac14 0

                                                                                              Depth factor D frac14 11

                                                                                              9frac14 122

                                                                                              Using Equation 92 with F frac14 10

                                                                                              Ns frac14 cu

                                                                                              FHfrac14 30

                                                                                              10 19 9frac14 0175

                                                                                              Hence from Figure 93

                                                                                              frac14 50

                                                                                              For F frac14 12

                                                                                              Ns frac14 30

                                                                                              12 19 9frac14 0146

                                                                                              frac14 27

                                                                                              93

                                                                                              Refer to Figure Q93

                                                                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                              1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                              74 m

                                                                                              214 1deg

                                                                                              213 1deg

                                                                                              39 m

                                                                                              WB

                                                                                              D

                                                                                              C

                                                                                              28 m

                                                                                              21 m

                                                                                              A

                                                                                              Q

                                                                                              Soil (1)Soil (2)

                                                                                              73deg

                                                                                              Figure Q91

                                                                                              Stability of slopes 75

                                                                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                              9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                              599 256 328 1372

                                                                                              Figure Q93

                                                                                              76 Stability of slopes

                                                                                              XW cos frac14 b

                                                                                              Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                              W sin frac14 bX

                                                                                              h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                              Arc length La frac14

                                                                                              180 57

                                                                                              1

                                                                                              2 326 frac14 327m

                                                                                              The factor of safety is given by

                                                                                              F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                              W sin

                                                                                              frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                              frac14 091

                                                                                              According to the limit state method

                                                                                              0d frac14 tan1tan 32

                                                                                              125

                                                                                              frac14 265

                                                                                              c0 frac14 8

                                                                                              160frac14 5 kN=m2

                                                                                              Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                              Design disturbing moment frac14 1075 kN=m

                                                                                              The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                              94

                                                                                              F frac14 1

                                                                                              W sin

                                                                                              Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                              1thorn ethtan tan0=FTHORN

                                                                                              c0 frac14 8 kN=m2

                                                                                              0 frac14 32

                                                                                              c0b frac14 8 2 frac14 16 kN=m

                                                                                              W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                              Try F frac14 100

                                                                                              tan0

                                                                                              Ffrac14 0625

                                                                                              Stability of slopes 77

                                                                                              Values of u are as obtained in Figure Q93

                                                                                              SliceNo

                                                                                              h(m)

                                                                                              W frac14 bh(kNm)

                                                                                              W sin(kNm)

                                                                                              ub(kNm)

                                                                                              c0bthorn (W ub) tan0(kNm)

                                                                                              sec

                                                                                              1thorn (tan tan0)FProduct(kNm)

                                                                                              1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                              23 33 30 1042 31

                                                                                              3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                              224 92 72 0931 67

                                                                                              6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                              12 58 112 90 0889 80

                                                                                              8 60 252 1812

                                                                                              80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                              10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                              2154 88 116 0853 99

                                                                                              14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                              1074 1091

                                                                                              F frac14 1091

                                                                                              1074frac14 102 (assumed value 100)

                                                                                              Thus

                                                                                              F frac14 101

                                                                                              95

                                                                                              F frac14 1

                                                                                              W sin

                                                                                              XfWeth1 ruTHORN tan0g sec

                                                                                              1thorn ethtan tan0THORN=F

                                                                                              0 frac14 33

                                                                                              ru frac14 020

                                                                                              W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                              eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                              Try F frac14 110

                                                                                              tan 0

                                                                                              Ffrac14 tan 33

                                                                                              110frac14 0590

                                                                                              78 Stability of slopes

                                                                                              Referring to Figure Q95

                                                                                              SliceNo

                                                                                              h(m)

                                                                                              W frac14 bh(kNm)

                                                                                              W sin(kNm)

                                                                                              W(1 ru) tan0(kNm)

                                                                                              sec

                                                                                              1thorn ( tan tan0)FProduct(kNm)

                                                                                              1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                              2120 234 0892 209

                                                                                              4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                              1185 1271

                                                                                              Figure Q95

                                                                                              Stability of slopes 79

                                                                                              F frac14 1271

                                                                                              1185frac14 107

                                                                                              The trial value was 110 therefore take F to be 108

                                                                                              96

                                                                                              (a) Water table at surface the factor of safety is given by Equation 912

                                                                                              F frac14 0

                                                                                              sat

                                                                                              tan0

                                                                                              tan

                                                                                              ptie 15 frac14 92

                                                                                              19

                                                                                              tan 36

                                                                                              tan

                                                                                              tan frac14 0234

                                                                                              frac14 13

                                                                                              Water table well below surface the factor of safety is given by Equation 911

                                                                                              F frac14 tan0

                                                                                              tan

                                                                                              frac14 tan 36

                                                                                              tan 13

                                                                                              frac14 31

                                                                                              (b) 0d frac14 tan1tan 36

                                                                                              125

                                                                                              frac14 30

                                                                                              Depth of potential failure surface frac14 z

                                                                                              Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                              frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                              frac14 504z kN

                                                                                              Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                              frac14 19 z sin 13 cos 13

                                                                                              frac14 416z kN

                                                                                              Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                              80 Stability of slopes

                                                                                              • Book Cover
                                                                                              • Title
                                                                                              • Contents
                                                                                              • Basic characteristics of soils
                                                                                              • Seepage
                                                                                              • Effective stress
                                                                                              • Shear strength
                                                                                              • Stresses and displacements
                                                                                              • Lateral earth pressure
                                                                                              • Consolidation theory
                                                                                              • Bearing capacity
                                                                                              • Stability of slopes

                                                                                                Algebraic sum of forces for d frac14 467m isX

                                                                                                F frac14 574thorn 689thorn 124thorn 2027thorn 458 570 1635 761 T frac14 0

                                                                                                T frac14 905 kN=m

                                                                                                Force in each tie rod frac14 25T frac14 226 kN

                                                                                                68

                                                                                                (a) For 0 frac14 36 Ka frac14 026 and Kp frac14 385

                                                                                                0 frac14 21 98 frac14 112 kN=m3

                                                                                                The pressure distribution is shown in Figure Q68 In this case the net water pressure atC is given by

                                                                                                uC frac14 150

                                                                                                165 15 98 frac14 134 kN=m2

                                                                                                The average seepage pressure is

                                                                                                j frac14 15

                                                                                                165 98 frac14 09 kN=m3

                                                                                                Hence

                                                                                                0 thorn j frac14 112thorn 09 frac14 121 kN=m3

                                                                                                0 j frac14 112 09 frac14 103 kN=m3

                                                                                                Figure Q67

                                                                                                42 Lateral earth pressure

                                                                                                Consider moments about the anchor point A (per m)

                                                                                                Force (kN) Arm (m) Moment (kN m)

                                                                                                (1) 10 026 150 frac14 390 60 2340

                                                                                                (2)1

                                                                                                2 026 18 452 frac14 474 15 711

                                                                                                (3) 026 18 45 105 frac14 2211 825 18240

                                                                                                (4)1

                                                                                                2 026 121 1052 frac14 1734 100 17340

                                                                                                (5)1

                                                                                                2 134 15 frac14 101 40 404

                                                                                                (6) 134 30 frac14 402 60 2412

                                                                                                (7)1

                                                                                                2 134 60 frac14 402 95 3819

                                                                                                571 4527(8) Ppm

                                                                                                115 115PPm

                                                                                                XM frac14 0

                                                                                                Ppm frac144527

                                                                                                115frac14 394 kN=m

                                                                                                Available passive resistance

                                                                                                Pp frac14 1

                                                                                                2 385 103 62 frac14 714 kN=m

                                                                                                Factor of safety

                                                                                                Fp frac14 Pp

                                                                                                Ppm

                                                                                                frac14 714

                                                                                                394frac14 18

                                                                                                Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                                                Figure Q68

                                                                                                Lateral earth pressure 43

                                                                                                (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                                                Consider moments (per m) about the tie point A

                                                                                                Force (kN) Arm (m)

                                                                                                (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                                                (2)1

                                                                                                2 033 18 452 frac14 601 15

                                                                                                (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                                                (4)1

                                                                                                2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                                                (5)1

                                                                                                2 134 15 frac14 101 40

                                                                                                (6) 134 30 frac14 402 60

                                                                                                (7)1

                                                                                                2 134 d frac14 67d d3thorn 75

                                                                                                (8) 1

                                                                                                2 30 103 d2 frac141545d2 2d3thorn 75

                                                                                                Moment (kN m)

                                                                                                (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                                                XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                                                d3 thorn 827d2 466d 1518 frac14 0

                                                                                                By trial

                                                                                                d frac14 544m

                                                                                                The minimum depth of embedment required is 544m

                                                                                                69

                                                                                                For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                                                0 frac14 20 98 frac14 102 kN=m3

                                                                                                The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                                                44 Lateral earth pressure

                                                                                                uC frac14 147

                                                                                                173 26 98 frac14 216 kN=m2

                                                                                                and the average seepage pressure around the wall is

                                                                                                j frac14 26

                                                                                                173 98 frac14 15 kN=m3

                                                                                                Consider moments about the prop (A) (per m)

                                                                                                Force (kN) Arm (m) Moment (kN m)

                                                                                                (1)1

                                                                                                2 03 17 272 frac14 186 020 37

                                                                                                (2) 03 17 27 53 frac14 730 335 2445

                                                                                                (3)1

                                                                                                2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                                                (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                                                (5)1

                                                                                                2 216 26 frac14 281 243 684

                                                                                                (6) 216 27 frac14 583 465 2712

                                                                                                (7)1

                                                                                                2 216 60 frac14 648 800 5184

                                                                                                3055(8)

                                                                                                1

                                                                                                2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                                                Factor of safety

                                                                                                Fr frac14 6885

                                                                                                3055frac14 225

                                                                                                Figure Q69

                                                                                                Lateral earth pressure 45

                                                                                                610

                                                                                                For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                                                p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                                                Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                                                Using the recommendations of Twine and Roscoe

                                                                                                p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                                                Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                                                611

                                                                                                frac14 18 kN=m3 0 frac14 34

                                                                                                H frac14 350m nH frac14 335m mH frac14 185m

                                                                                                Consider a trial value of F frac14 20 Refer to Figure 635

                                                                                                0m frac14 tan1tan 34

                                                                                                20

                                                                                                frac14 186

                                                                                                Then

                                                                                                frac14 45 thorn 0m2frac14 543

                                                                                                W frac14 1

                                                                                                2 18 3502 cot 543 frac14 792 kN=m

                                                                                                Figure Q610

                                                                                                46 Lateral earth pressure

                                                                                                P frac14 1

                                                                                                2 s 3352 frac14 561s kN=m

                                                                                                U frac14 1

                                                                                                2 98 1852 cosec 543 frac14 206 kN=m

                                                                                                Equations 630 and 631 then become

                                                                                                561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                                                792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                                                ie

                                                                                                561s 0616N 405 frac14 0

                                                                                                792 0857N thorn 563 frac14 0

                                                                                                N frac14 848

                                                                                                0857frac14 989 kN=m

                                                                                                Then

                                                                                                561s 609 405 frac14 0

                                                                                                s frac14 649

                                                                                                561frac14 116 kN=m3

                                                                                                The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                                                F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                                                20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                                                s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                                                Figure Q611

                                                                                                Lateral earth pressure 47

                                                                                                612

                                                                                                For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                                45 thorn 0

                                                                                                2frac14 63

                                                                                                For the retained material between the surface and a depth of 36m

                                                                                                Pa frac14 1

                                                                                                2 030 18 362 frac14 350 kN=m

                                                                                                Weight of reinforced fill between the surface and a depth of 36m is

                                                                                                Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                                eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                                Eccentricity of Rv

                                                                                                e frac14 263 250 frac14 013m

                                                                                                The average vertical stress at a depth of 36m is

                                                                                                z frac14 Rv

                                                                                                L 2efrac14 324

                                                                                                474frac14 68 kN=m2

                                                                                                (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                                Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                                Tensile stress in the element frac14 138 103

                                                                                                65 3frac14 71N=mm2

                                                                                                Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                                Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                                Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                                The weight of ABC is

                                                                                                W frac14 1

                                                                                                2 18 52 265 frac14 124 kN=m

                                                                                                From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                                48 Lateral earth pressure

                                                                                                (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                                Tp frac14 032 68 120 065 frac14 170 kN

                                                                                                Tr frac14 213 420

                                                                                                418frac14 214 kN

                                                                                                Again the tensile failure and slipping limit states are satisfied for this element

                                                                                                Figure Q612

                                                                                                Lateral earth pressure 49

                                                                                                Chapter 7

                                                                                                Consolidation theory

                                                                                                71

                                                                                                Total change in thickness

                                                                                                H frac14 782 602 frac14 180mm

                                                                                                Average thickness frac14 1530thorn 180

                                                                                                2frac14 1620mm

                                                                                                Length of drainage path d frac14 1620

                                                                                                2frac14 810mm

                                                                                                Root time plot (Figure Q71a)

                                                                                                ffiffiffiffiffiffit90p frac14 33

                                                                                                t90 frac14 109min

                                                                                                cv frac14 0848d2

                                                                                                t90frac14 0848 8102

                                                                                                109 1440 365

                                                                                                106frac14 27m2=year

                                                                                                r0 frac14 782 764

                                                                                                782 602frac14 018

                                                                                                180frac14 0100

                                                                                                rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                                10 119

                                                                                                9 180frac14 0735

                                                                                                rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                                Log time plot (Figure Q71b)

                                                                                                t50 frac14 26min

                                                                                                cv frac14 0196d2

                                                                                                t50frac14 0196 8102

                                                                                                26 1440 365

                                                                                                106frac14 26m2=year

                                                                                                r0 frac14 782 763

                                                                                                782 602frac14 019

                                                                                                180frac14 0106

                                                                                                rp frac14 763 623

                                                                                                782 602frac14 140

                                                                                                180frac14 0778

                                                                                                rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                                Figure Q71(a)

                                                                                                Figure Q71(b)

                                                                                                Final void ratio

                                                                                                e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                                e

                                                                                                Hfrac14 1thorn e0

                                                                                                H0frac14 1thorn e1 thorne

                                                                                                H0

                                                                                                ie

                                                                                                e

                                                                                                180frac14 1631thorne

                                                                                                1710

                                                                                                e frac14 2936

                                                                                                1530frac14 0192

                                                                                                Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                                mv frac14 1

                                                                                                1thorn e0 e0 e101 00

                                                                                                frac14 1

                                                                                                1823 0192

                                                                                                0107frac14 098m2=MN

                                                                                                k frac14 cvmvw frac14 265 098 98

                                                                                                60 1440 365 103frac14 81 1010 m=s

                                                                                                72

                                                                                                Using Equation 77 (one-dimensional method)

                                                                                                sc frac14 e0 e11thorn e0 H

                                                                                                Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                                Figure Q72

                                                                                                52 Consolidation theory

                                                                                                Settlement

                                                                                                Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                                1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                                318

                                                                                                Notes 5 92y 460thorn 84

                                                                                                Heave

                                                                                                Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                                1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                                38

                                                                                                73

                                                                                                U frac14 f ethTvTHORN frac14 f cvt

                                                                                                d2

                                                                                                Hence if cv is constant

                                                                                                t1

                                                                                                t2frac14 d

                                                                                                21

                                                                                                d22

                                                                                                where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                                d1 frac14 95mm and d2 frac14 2500mm

                                                                                                for U frac14 050 t2 frac14 t1 d22

                                                                                                d21

                                                                                                frac14 20

                                                                                                60 24 365 25002

                                                                                                952frac14 263 years

                                                                                                for U lt 060 Tv frac14

                                                                                                4U2 (Equation 724(a))

                                                                                                t030 frac14 t050 0302

                                                                                                0502

                                                                                                frac14 263 036 frac14 095 years

                                                                                                Consolidation theory 53

                                                                                                74

                                                                                                The layer is open

                                                                                                d frac14 8

                                                                                                2frac14 4m

                                                                                                Tv frac14 cvtd2frac14 24 3

                                                                                                42frac14 0450

                                                                                                ui frac14 frac14 84 kN=m2

                                                                                                The excess pore water pressure is given by Equation 721

                                                                                                ue frac14Xmfrac141mfrac140

                                                                                                2ui

                                                                                                Msin

                                                                                                Mz

                                                                                                d

                                                                                                expethM2TvTHORN

                                                                                                In this case z frac14 d

                                                                                                sinMz

                                                                                                d

                                                                                                frac14 sinM

                                                                                                where

                                                                                                M frac14

                                                                                                23

                                                                                                25

                                                                                                2

                                                                                                M sin M M2Tv exp (M2Tv)

                                                                                                2thorn1 1110 0329

                                                                                                3

                                                                                                21 9993 457 105

                                                                                                ue frac14 2 84 2

                                                                                                1 0329 ethother terms negligibleTHORN

                                                                                                frac14 352 kN=m2

                                                                                                75

                                                                                                The layer is open

                                                                                                d frac14 6

                                                                                                2frac14 3m

                                                                                                Tv frac14 cvtd2frac14 10 3

                                                                                                32frac14 0333

                                                                                                The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                                54 Consolidation theory

                                                                                                For an open layer

                                                                                                Tv frac14 4n

                                                                                                m2

                                                                                                n frac14 0333 62

                                                                                                4frac14 300

                                                                                                The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                                ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                                i j

                                                                                                0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                                0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                                The initial and 3-year isochrones are plotted in Figure Q75

                                                                                                Area under initial isochrone frac14 180 units

                                                                                                Area under 3-year isochrone frac14 63 units

                                                                                                The average degree of consolidation is given by Equation 725Thus

                                                                                                U frac14 1 63

                                                                                                180frac14 065

                                                                                                Figure Q75

                                                                                                Consolidation theory 55

                                                                                                76

                                                                                                At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                The final consolidation settlement (one-dimensional method) is

                                                                                                sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                Corrected time t frac14 2 1

                                                                                                2

                                                                                                40

                                                                                                52

                                                                                                frac14 1615 years

                                                                                                Tv frac14 cvtd2frac14 44 1615

                                                                                                42frac14 0444

                                                                                                From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                77

                                                                                                The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                Figure Q77

                                                                                                56 Consolidation theory

                                                                                                Point m n Ir (kNm2) sc (mm)

                                                                                                13020frac14 15 20

                                                                                                20frac14 10 0194 (4) 113 124

                                                                                                260

                                                                                                20frac14 30

                                                                                                20

                                                                                                20frac14 10 0204 (2) 59 65

                                                                                                360

                                                                                                20frac14 30

                                                                                                40

                                                                                                20frac14 20 0238 (1) 35 38

                                                                                                430

                                                                                                20frac14 15

                                                                                                40

                                                                                                20frac14 20 0224 (2) 65 72

                                                                                                Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                78

                                                                                                Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                (a) Immediate settlement

                                                                                                H

                                                                                                Bfrac14 30

                                                                                                35frac14 086

                                                                                                D

                                                                                                Bfrac14 2

                                                                                                35frac14 006

                                                                                                Figure Q78

                                                                                                Consolidation theory 57

                                                                                                From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                si frac14 130131qB

                                                                                                Eufrac14 10 032 105 35

                                                                                                40frac14 30mm

                                                                                                (b) Consolidation settlement

                                                                                                Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                3150

                                                                                                Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                Now

                                                                                                H

                                                                                                Bfrac14 30

                                                                                                35frac14 086 and A frac14 065

                                                                                                from Figure 712 13 frac14 079

                                                                                                sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                Total settlement

                                                                                                s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                79

                                                                                                Without sand drains

                                                                                                Uv frac14 025

                                                                                                Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                t frac14 Tvd2

                                                                                                cvfrac14 0049 82

                                                                                                cvWith sand drains

                                                                                                R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                n frac14 Rrfrac14 169

                                                                                                015frac14 113

                                                                                                Tr frac14 cht

                                                                                                4R2frac14 ch

                                                                                                4 1692 0049 82

                                                                                                cvethand ch frac14 cvTHORN

                                                                                                frac14 0275

                                                                                                Ur frac14 073 (from Figure 730)

                                                                                                58 Consolidation theory

                                                                                                Using Equation 740

                                                                                                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                U frac14 080

                                                                                                710

                                                                                                Without sand drains

                                                                                                Uv frac14 090

                                                                                                Tv frac14 0848

                                                                                                t frac14 Tvd2

                                                                                                cvfrac14 0848 102

                                                                                                96frac14 88 years

                                                                                                With sand drains

                                                                                                R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                n frac14 Rrfrac14 226

                                                                                                015frac14 15

                                                                                                Tr

                                                                                                Tvfrac14 chcv

                                                                                                d2

                                                                                                4R2ethsame tTHORN

                                                                                                Tr

                                                                                                Tvfrac14 140

                                                                                                96 102

                                                                                                4 2262frac14 714 eth1THORN

                                                                                                Using Equation 740

                                                                                                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                Thus

                                                                                                Uv frac14 0295 and Ur frac14 086

                                                                                                t frac14 88 00683

                                                                                                0848frac14 07 years

                                                                                                Consolidation theory 59

                                                                                                Chapter 8

                                                                                                Bearing capacity

                                                                                                81

                                                                                                (a) The ultimate bearing capacity is given by Equation 83

                                                                                                qf frac14 cNc thorn DNq thorn 1

                                                                                                2BN

                                                                                                For u frac14 0

                                                                                                Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                The net ultimate bearing capacity is

                                                                                                qnf frac14 qf D frac14 540 kN=m2

                                                                                                The net foundation pressure is

                                                                                                qn frac14 q D frac14 425

                                                                                                2 eth21 1THORN frac14 192 kN=m2

                                                                                                The factor of safety (Equation 86) is

                                                                                                F frac14 qnfqnfrac14 540

                                                                                                192frac14 28

                                                                                                (b) For 0 frac14 28

                                                                                                Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                2 112 2 13

                                                                                                frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                qnf frac14 574 112 frac14 563 kN=m2

                                                                                                F frac14 563

                                                                                                192frac14 29

                                                                                                (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                82

                                                                                                For 0 frac14 38

                                                                                                Nq frac14 49 N frac14 67

                                                                                                qnf frac14 DethNq 1THORN thorn 1

                                                                                                2BN ethfrom Equation 83THORN

                                                                                                frac14 eth18 075 48THORN thorn 1

                                                                                                2 18 15 67

                                                                                                frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                qn frac14 500

                                                                                                15 eth18 075THORN frac14 320 kN=m2

                                                                                                F frac14 qnfqnfrac14 1553

                                                                                                320frac14 48

                                                                                                0d frac14 tan1tan 38

                                                                                                125

                                                                                                frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                2 18 15 25

                                                                                                frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                Design load (action) Vd frac14 500 kN=m

                                                                                                The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                83

                                                                                                D

                                                                                                Bfrac14 350

                                                                                                225frac14 155

                                                                                                From Figure 85 for a square foundation

                                                                                                Nc frac14 81

                                                                                                Bearing capacity 61

                                                                                                For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                Nc frac14 084thorn 016B

                                                                                                L

                                                                                                81 frac14 745

                                                                                                Using Equation 810

                                                                                                qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                For F frac14 3

                                                                                                qn frac14 1006

                                                                                                3frac14 335 kN=m2

                                                                                                q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                Design load frac14 405 450 225 frac14 4100 kN

                                                                                                Design undrained strength cud frac14 135

                                                                                                14frac14 96 kN=m2

                                                                                                Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                frac14 7241 kN

                                                                                                Design load Vd frac14 4100 kN

                                                                                                Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                84

                                                                                                For 0 frac14 40

                                                                                                Nq frac14 64 N frac14 95

                                                                                                qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                (a) Water table 5m below ground level

                                                                                                qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                qn frac14 400 17 frac14 383 kN=m2

                                                                                                F frac14 2686

                                                                                                383frac14 70

                                                                                                (b) Water table 1m below ground level (ie at foundation level)

                                                                                                0 frac14 20 98 frac14 102 kN=m3

                                                                                                62 Bearing capacity

                                                                                                qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                F frac14 2040

                                                                                                383frac14 53

                                                                                                (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                F frac14 1296

                                                                                                392frac14 33

                                                                                                85

                                                                                                The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                Design value of 0 frac14 tan1tan 39

                                                                                                125

                                                                                                frac14 33

                                                                                                For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                86

                                                                                                (a) Undrained shear for u frac14 0

                                                                                                Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                qnf frac14 12cuNc

                                                                                                frac14 12 100 514 frac14 617 kN=m2

                                                                                                qn frac14 qnfFfrac14 617

                                                                                                3frac14 206 kN=m2

                                                                                                q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                Bearing capacity 63

                                                                                                Drained shear for 0 frac14 32

                                                                                                Nq frac14 23 N frac14 25

                                                                                                0 frac14 21 98 frac14 112 kN=m3

                                                                                                qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                frac14 694 kN=m2

                                                                                                q frac14 694

                                                                                                3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                Design load frac14 42 227 frac14 3632 kN

                                                                                                (b) Design undrained strength cud frac14 100

                                                                                                14frac14 71 kNm2

                                                                                                Design bearing resistance Rd frac14 12cudNe area

                                                                                                frac14 12 71 514 42

                                                                                                frac14 7007 kN

                                                                                                For drained shear 0d frac14 tan1tan 32

                                                                                                125

                                                                                                frac14 26

                                                                                                Nq frac14 12 N frac14 10

                                                                                                Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                1 2 100 0175 0700qn 0182qn

                                                                                                2 6 033 0044 0176qn 0046qn

                                                                                                3 10 020 0017 0068qn 0018qn

                                                                                                0246qn

                                                                                                Diameter of equivalent circle B frac14 45m

                                                                                                H

                                                                                                Bfrac14 12

                                                                                                45frac14 27 and A frac14 042

                                                                                                13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                64 Bearing capacity

                                                                                                For sc frac14 30mm

                                                                                                qn frac14 30

                                                                                                0147frac14 204 kN=m2

                                                                                                q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                Design load frac14 42 225 frac14 3600 kN

                                                                                                The design load is 3600 kN settlement being the limiting criterion

                                                                                                87

                                                                                                D

                                                                                                Bfrac14 8

                                                                                                4frac14 20

                                                                                                From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                F frac14 cuNc

                                                                                                Dfrac14 40 71

                                                                                                20 8frac14 18

                                                                                                88

                                                                                                Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                Design value of 0 frac14 tan1tan 38

                                                                                                125

                                                                                                frac14 32

                                                                                                Figure Q86

                                                                                                Bearing capacity 65

                                                                                                For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                For B frac14 250m qn frac14 3750

                                                                                                2502 17 frac14 583 kN=m2

                                                                                                From Figure 510 m frac14 n frac14 126

                                                                                                6frac14 021

                                                                                                Ir frac14 0019

                                                                                                Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                89

                                                                                                Depth (m) N 0v (kNm2) CN N1

                                                                                                070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                Cw frac14 05thorn 0530

                                                                                                47

                                                                                                frac14 082

                                                                                                66 Bearing capacity

                                                                                                Thus

                                                                                                qa frac14 150 082 frac14 120 kN=m2

                                                                                                (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                Thus

                                                                                                qa frac14 90 15 frac14 135 kN=m2

                                                                                                (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                Ic frac14 171

                                                                                                1014frac14 0068

                                                                                                From Equation 819(a) with s frac14 25mm

                                                                                                q frac14 25

                                                                                                3507 0068frac14 150 kN=m2

                                                                                                810

                                                                                                Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                Izp frac14 05thorn 01130

                                                                                                16 27

                                                                                                05

                                                                                                frac14 067

                                                                                                Refer to Figure Q810

                                                                                                E frac14 25qc

                                                                                                Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                Ez (mm3MN)

                                                                                                1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                0203

                                                                                                C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                130frac14 093

                                                                                                C2 frac14 1 ethsayTHORN

                                                                                                s frac14 C1C2qnX Iz

                                                                                                Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                Bearing capacity 67

                                                                                                811

                                                                                                At pile base level

                                                                                                cu frac14 220 kN=m2

                                                                                                qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                Then

                                                                                                Qf frac14 Abqb thorn Asqs

                                                                                                frac14

                                                                                                4 32 1980

                                                                                                thorn eth 105 139 86THORN

                                                                                                frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                0 01 02 03 04 05 06 07

                                                                                                0 2 4 6 8 10 12 14

                                                                                                1

                                                                                                2

                                                                                                3

                                                                                                4

                                                                                                5

                                                                                                6

                                                                                                7

                                                                                                8

                                                                                                (1)

                                                                                                (2)

                                                                                                (3)

                                                                                                (4)

                                                                                                (5)

                                                                                                qc

                                                                                                qc

                                                                                                Iz

                                                                                                Iz

                                                                                                (MNm2)

                                                                                                z (m)

                                                                                                Figure Q810

                                                                                                68 Bearing capacity

                                                                                                Allowable load

                                                                                                ethaTHORN Qf

                                                                                                2frac14 17 937

                                                                                                2frac14 8968 kN

                                                                                                ethbTHORN Abqb

                                                                                                3thorn Asqs frac14 13 996

                                                                                                3thorn 3941 frac14 8606 kN

                                                                                                ie allowable load frac14 8600 kN

                                                                                                Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                According to the limit state method

                                                                                                Characteristic undrained strength at base level cuk frac14 220

                                                                                                150kN=m2

                                                                                                Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                150frac14 1320 kN=m2

                                                                                                Characteristic shaft resistance qsk frac14 00150

                                                                                                frac14 86

                                                                                                150frac14 57 kN=m2

                                                                                                Characteristic base and shaft resistances

                                                                                                Rbk frac14

                                                                                                4 32 1320 frac14 9330 kN

                                                                                                Rsk frac14 105 139 86

                                                                                                150frac14 2629 kN

                                                                                                For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                Design bearing resistance Rcd frac14 9330

                                                                                                160thorn 2629

                                                                                                130

                                                                                                frac14 5831thorn 2022

                                                                                                frac14 7850 kN

                                                                                                Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                812

                                                                                                ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                For a single pile

                                                                                                Qf frac14 Abqb thorn Asqs

                                                                                                frac14

                                                                                                4 062 1305

                                                                                                thorn eth 06 15 42THORN

                                                                                                frac14 369thorn 1187 frac14 1556 kN

                                                                                                Bearing capacity 69

                                                                                                Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                qbkfrac14 9cuk frac14 9 220

                                                                                                150frac14 1320 kN=m2

                                                                                                qskfrac14cuk frac14 040 105

                                                                                                150frac14 28 kN=m2

                                                                                                Rbkfrac14

                                                                                                4 0602 1320 frac14 373 kN

                                                                                                Rskfrac14 060 15 28 frac14 791 kN

                                                                                                Rcdfrac14 373

                                                                                                160thorn 791

                                                                                                130frac14 233thorn 608 frac14 841 kN

                                                                                                Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                q frac14 21 000

                                                                                                1762frac14 68 kN=m2

                                                                                                Immediate settlement

                                                                                                H

                                                                                                Bfrac14 15

                                                                                                176frac14 085

                                                                                                D

                                                                                                Bfrac14 13

                                                                                                176frac14 074

                                                                                                L

                                                                                                Bfrac14 1

                                                                                                Hence from Figure 515

                                                                                                130 frac14 078 and 131 frac14 041

                                                                                                70 Bearing capacity

                                                                                                Thus using Equation 528

                                                                                                si frac14 078 041 68 176

                                                                                                65frac14 6mm

                                                                                                Consolidation settlement

                                                                                                Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                434 (sod)

                                                                                                Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                sc frac14 056 434 frac14 24mm

                                                                                                The total settlement is (6thorn 24) frac14 30mm

                                                                                                813

                                                                                                At base level N frac14 26 Then using Equation 830

                                                                                                qb frac14 40NDb

                                                                                                Bfrac14 40 26 2

                                                                                                025frac14 8320 kN=m2

                                                                                                ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                Figure Q812

                                                                                                Bearing capacity 71

                                                                                                Over the length embedded in sand

                                                                                                N frac14 21 ie18thorn 24

                                                                                                2

                                                                                                Using Equation 831

                                                                                                qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                For a single pile

                                                                                                Qf frac14 Abqb thorn Asqs

                                                                                                frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                For the pile group assuming a group efficiency of 12

                                                                                                XQf frac14 12 9 604 frac14 6523 kN

                                                                                                Then the load factor is

                                                                                                F frac14 6523

                                                                                                2000thorn 1000frac14 21

                                                                                                (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                Characteristic base resistance per unit area qbk frac14 8320

                                                                                                150frac14 5547 kNm2

                                                                                                Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                150frac14 28 kNm2

                                                                                                Characteristic base and shaft resistances for a single pile

                                                                                                Rbk frac14 0252 5547 frac14 347 kN

                                                                                                Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                Design bearing resistance Rcd frac14 347

                                                                                                130thorn 56

                                                                                                130frac14 310 kN

                                                                                                For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                72 Bearing capacity

                                                                                                N frac14 24thorn 26thorn 34

                                                                                                3frac14 28

                                                                                                Ic frac14 171

                                                                                                2814frac14 0016 ethEquation 818THORN

                                                                                                s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                814

                                                                                                Using Equation 841

                                                                                                Tf frac14 DLcu thorn

                                                                                                4ethD2 d2THORNcuNc

                                                                                                frac14 eth 02 5 06 110THORN thorn

                                                                                                4eth022 012THORN110 9

                                                                                                frac14 207thorn 23 frac14 230 kN

                                                                                                Figure Q813

                                                                                                Bearing capacity 73

                                                                                                Chapter 9

                                                                                                Stability of slopes

                                                                                                91

                                                                                                Referring to Figure Q91

                                                                                                W frac14 417 19 frac14 792 kN=m

                                                                                                Q frac14 20 28 frac14 56 kN=m

                                                                                                Arc lengthAB frac14

                                                                                                180 73 90 frac14 115m

                                                                                                Arc length BC frac14

                                                                                                180 28 90 frac14 44m

                                                                                                The factor of safety is given by

                                                                                                F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                Depth of tension crack z0 frac14 2cu

                                                                                                frac14 2 20

                                                                                                19frac14 21m

                                                                                                Arc length BD frac14

                                                                                                180 13

                                                                                                1

                                                                                                2 90 frac14 21m

                                                                                                F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                92

                                                                                                u frac14 0

                                                                                                Depth factor D frac14 11

                                                                                                9frac14 122

                                                                                                Using Equation 92 with F frac14 10

                                                                                                Ns frac14 cu

                                                                                                FHfrac14 30

                                                                                                10 19 9frac14 0175

                                                                                                Hence from Figure 93

                                                                                                frac14 50

                                                                                                For F frac14 12

                                                                                                Ns frac14 30

                                                                                                12 19 9frac14 0146

                                                                                                frac14 27

                                                                                                93

                                                                                                Refer to Figure Q93

                                                                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                74 m

                                                                                                214 1deg

                                                                                                213 1deg

                                                                                                39 m

                                                                                                WB

                                                                                                D

                                                                                                C

                                                                                                28 m

                                                                                                21 m

                                                                                                A

                                                                                                Q

                                                                                                Soil (1)Soil (2)

                                                                                                73deg

                                                                                                Figure Q91

                                                                                                Stability of slopes 75

                                                                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                599 256 328 1372

                                                                                                Figure Q93

                                                                                                76 Stability of slopes

                                                                                                XW cos frac14 b

                                                                                                Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                W sin frac14 bX

                                                                                                h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                Arc length La frac14

                                                                                                180 57

                                                                                                1

                                                                                                2 326 frac14 327m

                                                                                                The factor of safety is given by

                                                                                                F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                W sin

                                                                                                frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                frac14 091

                                                                                                According to the limit state method

                                                                                                0d frac14 tan1tan 32

                                                                                                125

                                                                                                frac14 265

                                                                                                c0 frac14 8

                                                                                                160frac14 5 kN=m2

                                                                                                Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                Design disturbing moment frac14 1075 kN=m

                                                                                                The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                94

                                                                                                F frac14 1

                                                                                                W sin

                                                                                                Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                1thorn ethtan tan0=FTHORN

                                                                                                c0 frac14 8 kN=m2

                                                                                                0 frac14 32

                                                                                                c0b frac14 8 2 frac14 16 kN=m

                                                                                                W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                Try F frac14 100

                                                                                                tan0

                                                                                                Ffrac14 0625

                                                                                                Stability of slopes 77

                                                                                                Values of u are as obtained in Figure Q93

                                                                                                SliceNo

                                                                                                h(m)

                                                                                                W frac14 bh(kNm)

                                                                                                W sin(kNm)

                                                                                                ub(kNm)

                                                                                                c0bthorn (W ub) tan0(kNm)

                                                                                                sec

                                                                                                1thorn (tan tan0)FProduct(kNm)

                                                                                                1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                23 33 30 1042 31

                                                                                                3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                224 92 72 0931 67

                                                                                                6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                12 58 112 90 0889 80

                                                                                                8 60 252 1812

                                                                                                80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                2154 88 116 0853 99

                                                                                                14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                1074 1091

                                                                                                F frac14 1091

                                                                                                1074frac14 102 (assumed value 100)

                                                                                                Thus

                                                                                                F frac14 101

                                                                                                95

                                                                                                F frac14 1

                                                                                                W sin

                                                                                                XfWeth1 ruTHORN tan0g sec

                                                                                                1thorn ethtan tan0THORN=F

                                                                                                0 frac14 33

                                                                                                ru frac14 020

                                                                                                W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                Try F frac14 110

                                                                                                tan 0

                                                                                                Ffrac14 tan 33

                                                                                                110frac14 0590

                                                                                                78 Stability of slopes

                                                                                                Referring to Figure Q95

                                                                                                SliceNo

                                                                                                h(m)

                                                                                                W frac14 bh(kNm)

                                                                                                W sin(kNm)

                                                                                                W(1 ru) tan0(kNm)

                                                                                                sec

                                                                                                1thorn ( tan tan0)FProduct(kNm)

                                                                                                1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                2120 234 0892 209

                                                                                                4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                1185 1271

                                                                                                Figure Q95

                                                                                                Stability of slopes 79

                                                                                                F frac14 1271

                                                                                                1185frac14 107

                                                                                                The trial value was 110 therefore take F to be 108

                                                                                                96

                                                                                                (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                F frac14 0

                                                                                                sat

                                                                                                tan0

                                                                                                tan

                                                                                                ptie 15 frac14 92

                                                                                                19

                                                                                                tan 36

                                                                                                tan

                                                                                                tan frac14 0234

                                                                                                frac14 13

                                                                                                Water table well below surface the factor of safety is given by Equation 911

                                                                                                F frac14 tan0

                                                                                                tan

                                                                                                frac14 tan 36

                                                                                                tan 13

                                                                                                frac14 31

                                                                                                (b) 0d frac14 tan1tan 36

                                                                                                125

                                                                                                frac14 30

                                                                                                Depth of potential failure surface frac14 z

                                                                                                Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                frac14 504z kN

                                                                                                Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                frac14 19 z sin 13 cos 13

                                                                                                frac14 416z kN

                                                                                                Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                80 Stability of slopes

                                                                                                • Book Cover
                                                                                                • Title
                                                                                                • Contents
                                                                                                • Basic characteristics of soils
                                                                                                • Seepage
                                                                                                • Effective stress
                                                                                                • Shear strength
                                                                                                • Stresses and displacements
                                                                                                • Lateral earth pressure
                                                                                                • Consolidation theory
                                                                                                • Bearing capacity
                                                                                                • Stability of slopes

                                                                                                  Consider moments about the anchor point A (per m)

                                                                                                  Force (kN) Arm (m) Moment (kN m)

                                                                                                  (1) 10 026 150 frac14 390 60 2340

                                                                                                  (2)1

                                                                                                  2 026 18 452 frac14 474 15 711

                                                                                                  (3) 026 18 45 105 frac14 2211 825 18240

                                                                                                  (4)1

                                                                                                  2 026 121 1052 frac14 1734 100 17340

                                                                                                  (5)1

                                                                                                  2 134 15 frac14 101 40 404

                                                                                                  (6) 134 30 frac14 402 60 2412

                                                                                                  (7)1

                                                                                                  2 134 60 frac14 402 95 3819

                                                                                                  571 4527(8) Ppm

                                                                                                  115 115PPm

                                                                                                  XM frac14 0

                                                                                                  Ppm frac144527

                                                                                                  115frac14 394 kN=m

                                                                                                  Available passive resistance

                                                                                                  Pp frac14 1

                                                                                                  2 385 103 62 frac14 714 kN=m

                                                                                                  Factor of safety

                                                                                                  Fp frac14 Pp

                                                                                                  Ppm

                                                                                                  frac14 714

                                                                                                  394frac14 18

                                                                                                  Force in each tie frac14 2T frac14 2(571 394) frac14 354 kN

                                                                                                  Figure Q68

                                                                                                  Lateral earth pressure 43

                                                                                                  (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                                                  Consider moments (per m) about the tie point A

                                                                                                  Force (kN) Arm (m)

                                                                                                  (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                                                  (2)1

                                                                                                  2 033 18 452 frac14 601 15

                                                                                                  (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                                                  (4)1

                                                                                                  2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                                                  (5)1

                                                                                                  2 134 15 frac14 101 40

                                                                                                  (6) 134 30 frac14 402 60

                                                                                                  (7)1

                                                                                                  2 134 d frac14 67d d3thorn 75

                                                                                                  (8) 1

                                                                                                  2 30 103 d2 frac141545d2 2d3thorn 75

                                                                                                  Moment (kN m)

                                                                                                  (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                                                  XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                                                  d3 thorn 827d2 466d 1518 frac14 0

                                                                                                  By trial

                                                                                                  d frac14 544m

                                                                                                  The minimum depth of embedment required is 544m

                                                                                                  69

                                                                                                  For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                                                  0 frac14 20 98 frac14 102 kN=m3

                                                                                                  The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                                                  44 Lateral earth pressure

                                                                                                  uC frac14 147

                                                                                                  173 26 98 frac14 216 kN=m2

                                                                                                  and the average seepage pressure around the wall is

                                                                                                  j frac14 26

                                                                                                  173 98 frac14 15 kN=m3

                                                                                                  Consider moments about the prop (A) (per m)

                                                                                                  Force (kN) Arm (m) Moment (kN m)

                                                                                                  (1)1

                                                                                                  2 03 17 272 frac14 186 020 37

                                                                                                  (2) 03 17 27 53 frac14 730 335 2445

                                                                                                  (3)1

                                                                                                  2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                                                  (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                                                  (5)1

                                                                                                  2 216 26 frac14 281 243 684

                                                                                                  (6) 216 27 frac14 583 465 2712

                                                                                                  (7)1

                                                                                                  2 216 60 frac14 648 800 5184

                                                                                                  3055(8)

                                                                                                  1

                                                                                                  2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                                                  Factor of safety

                                                                                                  Fr frac14 6885

                                                                                                  3055frac14 225

                                                                                                  Figure Q69

                                                                                                  Lateral earth pressure 45

                                                                                                  610

                                                                                                  For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                                                  p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                                                  Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                                                  Using the recommendations of Twine and Roscoe

                                                                                                  p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                                                  Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                                                  611

                                                                                                  frac14 18 kN=m3 0 frac14 34

                                                                                                  H frac14 350m nH frac14 335m mH frac14 185m

                                                                                                  Consider a trial value of F frac14 20 Refer to Figure 635

                                                                                                  0m frac14 tan1tan 34

                                                                                                  20

                                                                                                  frac14 186

                                                                                                  Then

                                                                                                  frac14 45 thorn 0m2frac14 543

                                                                                                  W frac14 1

                                                                                                  2 18 3502 cot 543 frac14 792 kN=m

                                                                                                  Figure Q610

                                                                                                  46 Lateral earth pressure

                                                                                                  P frac14 1

                                                                                                  2 s 3352 frac14 561s kN=m

                                                                                                  U frac14 1

                                                                                                  2 98 1852 cosec 543 frac14 206 kN=m

                                                                                                  Equations 630 and 631 then become

                                                                                                  561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                                                  792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                                                  ie

                                                                                                  561s 0616N 405 frac14 0

                                                                                                  792 0857N thorn 563 frac14 0

                                                                                                  N frac14 848

                                                                                                  0857frac14 989 kN=m

                                                                                                  Then

                                                                                                  561s 609 405 frac14 0

                                                                                                  s frac14 649

                                                                                                  561frac14 116 kN=m3

                                                                                                  The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                                                  F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                                                  20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                                                  s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                                                  Figure Q611

                                                                                                  Lateral earth pressure 47

                                                                                                  612

                                                                                                  For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                                  45 thorn 0

                                                                                                  2frac14 63

                                                                                                  For the retained material between the surface and a depth of 36m

                                                                                                  Pa frac14 1

                                                                                                  2 030 18 362 frac14 350 kN=m

                                                                                                  Weight of reinforced fill between the surface and a depth of 36m is

                                                                                                  Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                                  eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                                  Eccentricity of Rv

                                                                                                  e frac14 263 250 frac14 013m

                                                                                                  The average vertical stress at a depth of 36m is

                                                                                                  z frac14 Rv

                                                                                                  L 2efrac14 324

                                                                                                  474frac14 68 kN=m2

                                                                                                  (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                                  Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                                  Tensile stress in the element frac14 138 103

                                                                                                  65 3frac14 71N=mm2

                                                                                                  Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                                  Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                                  Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                                  The weight of ABC is

                                                                                                  W frac14 1

                                                                                                  2 18 52 265 frac14 124 kN=m

                                                                                                  From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                                  48 Lateral earth pressure

                                                                                                  (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                                  Tp frac14 032 68 120 065 frac14 170 kN

                                                                                                  Tr frac14 213 420

                                                                                                  418frac14 214 kN

                                                                                                  Again the tensile failure and slipping limit states are satisfied for this element

                                                                                                  Figure Q612

                                                                                                  Lateral earth pressure 49

                                                                                                  Chapter 7

                                                                                                  Consolidation theory

                                                                                                  71

                                                                                                  Total change in thickness

                                                                                                  H frac14 782 602 frac14 180mm

                                                                                                  Average thickness frac14 1530thorn 180

                                                                                                  2frac14 1620mm

                                                                                                  Length of drainage path d frac14 1620

                                                                                                  2frac14 810mm

                                                                                                  Root time plot (Figure Q71a)

                                                                                                  ffiffiffiffiffiffit90p frac14 33

                                                                                                  t90 frac14 109min

                                                                                                  cv frac14 0848d2

                                                                                                  t90frac14 0848 8102

                                                                                                  109 1440 365

                                                                                                  106frac14 27m2=year

                                                                                                  r0 frac14 782 764

                                                                                                  782 602frac14 018

                                                                                                  180frac14 0100

                                                                                                  rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                                  10 119

                                                                                                  9 180frac14 0735

                                                                                                  rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                                  Log time plot (Figure Q71b)

                                                                                                  t50 frac14 26min

                                                                                                  cv frac14 0196d2

                                                                                                  t50frac14 0196 8102

                                                                                                  26 1440 365

                                                                                                  106frac14 26m2=year

                                                                                                  r0 frac14 782 763

                                                                                                  782 602frac14 019

                                                                                                  180frac14 0106

                                                                                                  rp frac14 763 623

                                                                                                  782 602frac14 140

                                                                                                  180frac14 0778

                                                                                                  rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                                  Figure Q71(a)

                                                                                                  Figure Q71(b)

                                                                                                  Final void ratio

                                                                                                  e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                                  e

                                                                                                  Hfrac14 1thorn e0

                                                                                                  H0frac14 1thorn e1 thorne

                                                                                                  H0

                                                                                                  ie

                                                                                                  e

                                                                                                  180frac14 1631thorne

                                                                                                  1710

                                                                                                  e frac14 2936

                                                                                                  1530frac14 0192

                                                                                                  Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                                  mv frac14 1

                                                                                                  1thorn e0 e0 e101 00

                                                                                                  frac14 1

                                                                                                  1823 0192

                                                                                                  0107frac14 098m2=MN

                                                                                                  k frac14 cvmvw frac14 265 098 98

                                                                                                  60 1440 365 103frac14 81 1010 m=s

                                                                                                  72

                                                                                                  Using Equation 77 (one-dimensional method)

                                                                                                  sc frac14 e0 e11thorn e0 H

                                                                                                  Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                                  Figure Q72

                                                                                                  52 Consolidation theory

                                                                                                  Settlement

                                                                                                  Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                                  1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                                  318

                                                                                                  Notes 5 92y 460thorn 84

                                                                                                  Heave

                                                                                                  Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                                  1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                                  38

                                                                                                  73

                                                                                                  U frac14 f ethTvTHORN frac14 f cvt

                                                                                                  d2

                                                                                                  Hence if cv is constant

                                                                                                  t1

                                                                                                  t2frac14 d

                                                                                                  21

                                                                                                  d22

                                                                                                  where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                                  d1 frac14 95mm and d2 frac14 2500mm

                                                                                                  for U frac14 050 t2 frac14 t1 d22

                                                                                                  d21

                                                                                                  frac14 20

                                                                                                  60 24 365 25002

                                                                                                  952frac14 263 years

                                                                                                  for U lt 060 Tv frac14

                                                                                                  4U2 (Equation 724(a))

                                                                                                  t030 frac14 t050 0302

                                                                                                  0502

                                                                                                  frac14 263 036 frac14 095 years

                                                                                                  Consolidation theory 53

                                                                                                  74

                                                                                                  The layer is open

                                                                                                  d frac14 8

                                                                                                  2frac14 4m

                                                                                                  Tv frac14 cvtd2frac14 24 3

                                                                                                  42frac14 0450

                                                                                                  ui frac14 frac14 84 kN=m2

                                                                                                  The excess pore water pressure is given by Equation 721

                                                                                                  ue frac14Xmfrac141mfrac140

                                                                                                  2ui

                                                                                                  Msin

                                                                                                  Mz

                                                                                                  d

                                                                                                  expethM2TvTHORN

                                                                                                  In this case z frac14 d

                                                                                                  sinMz

                                                                                                  d

                                                                                                  frac14 sinM

                                                                                                  where

                                                                                                  M frac14

                                                                                                  23

                                                                                                  25

                                                                                                  2

                                                                                                  M sin M M2Tv exp (M2Tv)

                                                                                                  2thorn1 1110 0329

                                                                                                  3

                                                                                                  21 9993 457 105

                                                                                                  ue frac14 2 84 2

                                                                                                  1 0329 ethother terms negligibleTHORN

                                                                                                  frac14 352 kN=m2

                                                                                                  75

                                                                                                  The layer is open

                                                                                                  d frac14 6

                                                                                                  2frac14 3m

                                                                                                  Tv frac14 cvtd2frac14 10 3

                                                                                                  32frac14 0333

                                                                                                  The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                                  54 Consolidation theory

                                                                                                  For an open layer

                                                                                                  Tv frac14 4n

                                                                                                  m2

                                                                                                  n frac14 0333 62

                                                                                                  4frac14 300

                                                                                                  The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                                  ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                                  i j

                                                                                                  0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                                  0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                                  The initial and 3-year isochrones are plotted in Figure Q75

                                                                                                  Area under initial isochrone frac14 180 units

                                                                                                  Area under 3-year isochrone frac14 63 units

                                                                                                  The average degree of consolidation is given by Equation 725Thus

                                                                                                  U frac14 1 63

                                                                                                  180frac14 065

                                                                                                  Figure Q75

                                                                                                  Consolidation theory 55

                                                                                                  76

                                                                                                  At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                  0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                  The final consolidation settlement (one-dimensional method) is

                                                                                                  sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                  Corrected time t frac14 2 1

                                                                                                  2

                                                                                                  40

                                                                                                  52

                                                                                                  frac14 1615 years

                                                                                                  Tv frac14 cvtd2frac14 44 1615

                                                                                                  42frac14 0444

                                                                                                  From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                  77

                                                                                                  The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                  Figure Q77

                                                                                                  56 Consolidation theory

                                                                                                  Point m n Ir (kNm2) sc (mm)

                                                                                                  13020frac14 15 20

                                                                                                  20frac14 10 0194 (4) 113 124

                                                                                                  260

                                                                                                  20frac14 30

                                                                                                  20

                                                                                                  20frac14 10 0204 (2) 59 65

                                                                                                  360

                                                                                                  20frac14 30

                                                                                                  40

                                                                                                  20frac14 20 0238 (1) 35 38

                                                                                                  430

                                                                                                  20frac14 15

                                                                                                  40

                                                                                                  20frac14 20 0224 (2) 65 72

                                                                                                  Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                  78

                                                                                                  Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                  (a) Immediate settlement

                                                                                                  H

                                                                                                  Bfrac14 30

                                                                                                  35frac14 086

                                                                                                  D

                                                                                                  Bfrac14 2

                                                                                                  35frac14 006

                                                                                                  Figure Q78

                                                                                                  Consolidation theory 57

                                                                                                  From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                  si frac14 130131qB

                                                                                                  Eufrac14 10 032 105 35

                                                                                                  40frac14 30mm

                                                                                                  (b) Consolidation settlement

                                                                                                  Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                  1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                  3150

                                                                                                  Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                  Now

                                                                                                  H

                                                                                                  Bfrac14 30

                                                                                                  35frac14 086 and A frac14 065

                                                                                                  from Figure 712 13 frac14 079

                                                                                                  sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                  Total settlement

                                                                                                  s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                  79

                                                                                                  Without sand drains

                                                                                                  Uv frac14 025

                                                                                                  Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                  t frac14 Tvd2

                                                                                                  cvfrac14 0049 82

                                                                                                  cvWith sand drains

                                                                                                  R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                  n frac14 Rrfrac14 169

                                                                                                  015frac14 113

                                                                                                  Tr frac14 cht

                                                                                                  4R2frac14 ch

                                                                                                  4 1692 0049 82

                                                                                                  cvethand ch frac14 cvTHORN

                                                                                                  frac14 0275

                                                                                                  Ur frac14 073 (from Figure 730)

                                                                                                  58 Consolidation theory

                                                                                                  Using Equation 740

                                                                                                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                  U frac14 080

                                                                                                  710

                                                                                                  Without sand drains

                                                                                                  Uv frac14 090

                                                                                                  Tv frac14 0848

                                                                                                  t frac14 Tvd2

                                                                                                  cvfrac14 0848 102

                                                                                                  96frac14 88 years

                                                                                                  With sand drains

                                                                                                  R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                  n frac14 Rrfrac14 226

                                                                                                  015frac14 15

                                                                                                  Tr

                                                                                                  Tvfrac14 chcv

                                                                                                  d2

                                                                                                  4R2ethsame tTHORN

                                                                                                  Tr

                                                                                                  Tvfrac14 140

                                                                                                  96 102

                                                                                                  4 2262frac14 714 eth1THORN

                                                                                                  Using Equation 740

                                                                                                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                  An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                  Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                  040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                  Thus

                                                                                                  Uv frac14 0295 and Ur frac14 086

                                                                                                  t frac14 88 00683

                                                                                                  0848frac14 07 years

                                                                                                  Consolidation theory 59

                                                                                                  Chapter 8

                                                                                                  Bearing capacity

                                                                                                  81

                                                                                                  (a) The ultimate bearing capacity is given by Equation 83

                                                                                                  qf frac14 cNc thorn DNq thorn 1

                                                                                                  2BN

                                                                                                  For u frac14 0

                                                                                                  Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                  qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                  The net ultimate bearing capacity is

                                                                                                  qnf frac14 qf D frac14 540 kN=m2

                                                                                                  The net foundation pressure is

                                                                                                  qn frac14 q D frac14 425

                                                                                                  2 eth21 1THORN frac14 192 kN=m2

                                                                                                  The factor of safety (Equation 86) is

                                                                                                  F frac14 qnfqnfrac14 540

                                                                                                  192frac14 28

                                                                                                  (b) For 0 frac14 28

                                                                                                  Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                  qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                  2 112 2 13

                                                                                                  frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                  qnf frac14 574 112 frac14 563 kN=m2

                                                                                                  F frac14 563

                                                                                                  192frac14 29

                                                                                                  (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                  82

                                                                                                  For 0 frac14 38

                                                                                                  Nq frac14 49 N frac14 67

                                                                                                  qnf frac14 DethNq 1THORN thorn 1

                                                                                                  2BN ethfrom Equation 83THORN

                                                                                                  frac14 eth18 075 48THORN thorn 1

                                                                                                  2 18 15 67

                                                                                                  frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                  qn frac14 500

                                                                                                  15 eth18 075THORN frac14 320 kN=m2

                                                                                                  F frac14 qnfqnfrac14 1553

                                                                                                  320frac14 48

                                                                                                  0d frac14 tan1tan 38

                                                                                                  125

                                                                                                  frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                  Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                  2 18 15 25

                                                                                                  frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                  Design load (action) Vd frac14 500 kN=m

                                                                                                  The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                  83

                                                                                                  D

                                                                                                  Bfrac14 350

                                                                                                  225frac14 155

                                                                                                  From Figure 85 for a square foundation

                                                                                                  Nc frac14 81

                                                                                                  Bearing capacity 61

                                                                                                  For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                  Nc frac14 084thorn 016B

                                                                                                  L

                                                                                                  81 frac14 745

                                                                                                  Using Equation 810

                                                                                                  qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                  For F frac14 3

                                                                                                  qn frac14 1006

                                                                                                  3frac14 335 kN=m2

                                                                                                  q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                  Design load frac14 405 450 225 frac14 4100 kN

                                                                                                  Design undrained strength cud frac14 135

                                                                                                  14frac14 96 kN=m2

                                                                                                  Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                  frac14 7241 kN

                                                                                                  Design load Vd frac14 4100 kN

                                                                                                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                  84

                                                                                                  For 0 frac14 40

                                                                                                  Nq frac14 64 N frac14 95

                                                                                                  qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                  (a) Water table 5m below ground level

                                                                                                  qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                  qn frac14 400 17 frac14 383 kN=m2

                                                                                                  F frac14 2686

                                                                                                  383frac14 70

                                                                                                  (b) Water table 1m below ground level (ie at foundation level)

                                                                                                  0 frac14 20 98 frac14 102 kN=m3

                                                                                                  62 Bearing capacity

                                                                                                  qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                  F frac14 2040

                                                                                                  383frac14 53

                                                                                                  (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                  eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                  qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                  F frac14 1296

                                                                                                  392frac14 33

                                                                                                  85

                                                                                                  The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                  Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                  Design value of 0 frac14 tan1tan 39

                                                                                                  125

                                                                                                  frac14 33

                                                                                                  For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                  Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                  86

                                                                                                  (a) Undrained shear for u frac14 0

                                                                                                  Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                  qnf frac14 12cuNc

                                                                                                  frac14 12 100 514 frac14 617 kN=m2

                                                                                                  qn frac14 qnfFfrac14 617

                                                                                                  3frac14 206 kN=m2

                                                                                                  q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                  Bearing capacity 63

                                                                                                  Drained shear for 0 frac14 32

                                                                                                  Nq frac14 23 N frac14 25

                                                                                                  0 frac14 21 98 frac14 112 kN=m3

                                                                                                  qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                  frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                  frac14 694 kN=m2

                                                                                                  q frac14 694

                                                                                                  3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                  Design load frac14 42 227 frac14 3632 kN

                                                                                                  (b) Design undrained strength cud frac14 100

                                                                                                  14frac14 71 kNm2

                                                                                                  Design bearing resistance Rd frac14 12cudNe area

                                                                                                  frac14 12 71 514 42

                                                                                                  frac14 7007 kN

                                                                                                  For drained shear 0d frac14 tan1tan 32

                                                                                                  125

                                                                                                  frac14 26

                                                                                                  Nq frac14 12 N frac14 10

                                                                                                  Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                  (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                  Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                  1 2 100 0175 0700qn 0182qn

                                                                                                  2 6 033 0044 0176qn 0046qn

                                                                                                  3 10 020 0017 0068qn 0018qn

                                                                                                  0246qn

                                                                                                  Diameter of equivalent circle B frac14 45m

                                                                                                  H

                                                                                                  Bfrac14 12

                                                                                                  45frac14 27 and A frac14 042

                                                                                                  13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                  64 Bearing capacity

                                                                                                  For sc frac14 30mm

                                                                                                  qn frac14 30

                                                                                                  0147frac14 204 kN=m2

                                                                                                  q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                  Design load frac14 42 225 frac14 3600 kN

                                                                                                  The design load is 3600 kN settlement being the limiting criterion

                                                                                                  87

                                                                                                  D

                                                                                                  Bfrac14 8

                                                                                                  4frac14 20

                                                                                                  From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                  F frac14 cuNc

                                                                                                  Dfrac14 40 71

                                                                                                  20 8frac14 18

                                                                                                  88

                                                                                                  Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                  Design value of 0 frac14 tan1tan 38

                                                                                                  125

                                                                                                  frac14 32

                                                                                                  Figure Q86

                                                                                                  Bearing capacity 65

                                                                                                  For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                  Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                  The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                  Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                  For B frac14 250m qn frac14 3750

                                                                                                  2502 17 frac14 583 kN=m2

                                                                                                  From Figure 510 m frac14 n frac14 126

                                                                                                  6frac14 021

                                                                                                  Ir frac14 0019

                                                                                                  Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                  Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                  The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                  89

                                                                                                  Depth (m) N 0v (kNm2) CN N1

                                                                                                  070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                  Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                  (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                  Cw frac14 05thorn 0530

                                                                                                  47

                                                                                                  frac14 082

                                                                                                  66 Bearing capacity

                                                                                                  Thus

                                                                                                  qa frac14 150 082 frac14 120 kN=m2

                                                                                                  (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                  Thus

                                                                                                  qa frac14 90 15 frac14 135 kN=m2

                                                                                                  (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                  Ic frac14 171

                                                                                                  1014frac14 0068

                                                                                                  From Equation 819(a) with s frac14 25mm

                                                                                                  q frac14 25

                                                                                                  3507 0068frac14 150 kN=m2

                                                                                                  810

                                                                                                  Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                  Izp frac14 05thorn 01130

                                                                                                  16 27

                                                                                                  05

                                                                                                  frac14 067

                                                                                                  Refer to Figure Q810

                                                                                                  E frac14 25qc

                                                                                                  Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                  Ez (mm3MN)

                                                                                                  1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                  0203

                                                                                                  C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                  130frac14 093

                                                                                                  C2 frac14 1 ethsayTHORN

                                                                                                  s frac14 C1C2qnX Iz

                                                                                                  Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                  Bearing capacity 67

                                                                                                  811

                                                                                                  At pile base level

                                                                                                  cu frac14 220 kN=m2

                                                                                                  qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                  Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                  00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                  qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                  Then

                                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                                  frac14

                                                                                                  4 32 1980

                                                                                                  thorn eth 105 139 86THORN

                                                                                                  frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                  0 01 02 03 04 05 06 07

                                                                                                  0 2 4 6 8 10 12 14

                                                                                                  1

                                                                                                  2

                                                                                                  3

                                                                                                  4

                                                                                                  5

                                                                                                  6

                                                                                                  7

                                                                                                  8

                                                                                                  (1)

                                                                                                  (2)

                                                                                                  (3)

                                                                                                  (4)

                                                                                                  (5)

                                                                                                  qc

                                                                                                  qc

                                                                                                  Iz

                                                                                                  Iz

                                                                                                  (MNm2)

                                                                                                  z (m)

                                                                                                  Figure Q810

                                                                                                  68 Bearing capacity

                                                                                                  Allowable load

                                                                                                  ethaTHORN Qf

                                                                                                  2frac14 17 937

                                                                                                  2frac14 8968 kN

                                                                                                  ethbTHORN Abqb

                                                                                                  3thorn Asqs frac14 13 996

                                                                                                  3thorn 3941 frac14 8606 kN

                                                                                                  ie allowable load frac14 8600 kN

                                                                                                  Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                  According to the limit state method

                                                                                                  Characteristic undrained strength at base level cuk frac14 220

                                                                                                  150kN=m2

                                                                                                  Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                  150frac14 1320 kN=m2

                                                                                                  Characteristic shaft resistance qsk frac14 00150

                                                                                                  frac14 86

                                                                                                  150frac14 57 kN=m2

                                                                                                  Characteristic base and shaft resistances

                                                                                                  Rbk frac14

                                                                                                  4 32 1320 frac14 9330 kN

                                                                                                  Rsk frac14 105 139 86

                                                                                                  150frac14 2629 kN

                                                                                                  For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                  Design bearing resistance Rcd frac14 9330

                                                                                                  160thorn 2629

                                                                                                  130

                                                                                                  frac14 5831thorn 2022

                                                                                                  frac14 7850 kN

                                                                                                  Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                  812

                                                                                                  ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                  qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                  For a single pile

                                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                                  frac14

                                                                                                  4 062 1305

                                                                                                  thorn eth 06 15 42THORN

                                                                                                  frac14 369thorn 1187 frac14 1556 kN

                                                                                                  Bearing capacity 69

                                                                                                  Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                  eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                  (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                  (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                  qbkfrac14 9cuk frac14 9 220

                                                                                                  150frac14 1320 kN=m2

                                                                                                  qskfrac14cuk frac14 040 105

                                                                                                  150frac14 28 kN=m2

                                                                                                  Rbkfrac14

                                                                                                  4 0602 1320 frac14 373 kN

                                                                                                  Rskfrac14 060 15 28 frac14 791 kN

                                                                                                  Rcdfrac14 373

                                                                                                  160thorn 791

                                                                                                  130frac14 233thorn 608 frac14 841 kN

                                                                                                  Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                  Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                  (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                  q frac14 21 000

                                                                                                  1762frac14 68 kN=m2

                                                                                                  Immediate settlement

                                                                                                  H

                                                                                                  Bfrac14 15

                                                                                                  176frac14 085

                                                                                                  D

                                                                                                  Bfrac14 13

                                                                                                  176frac14 074

                                                                                                  L

                                                                                                  Bfrac14 1

                                                                                                  Hence from Figure 515

                                                                                                  130 frac14 078 and 131 frac14 041

                                                                                                  70 Bearing capacity

                                                                                                  Thus using Equation 528

                                                                                                  si frac14 078 041 68 176

                                                                                                  65frac14 6mm

                                                                                                  Consolidation settlement

                                                                                                  Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                  1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                  434 (sod)

                                                                                                  Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                  sc frac14 056 434 frac14 24mm

                                                                                                  The total settlement is (6thorn 24) frac14 30mm

                                                                                                  813

                                                                                                  At base level N frac14 26 Then using Equation 830

                                                                                                  qb frac14 40NDb

                                                                                                  Bfrac14 40 26 2

                                                                                                  025frac14 8320 kN=m2

                                                                                                  ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                  Figure Q812

                                                                                                  Bearing capacity 71

                                                                                                  Over the length embedded in sand

                                                                                                  N frac14 21 ie18thorn 24

                                                                                                  2

                                                                                                  Using Equation 831

                                                                                                  qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                  For a single pile

                                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                                  frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                  For the pile group assuming a group efficiency of 12

                                                                                                  XQf frac14 12 9 604 frac14 6523 kN

                                                                                                  Then the load factor is

                                                                                                  F frac14 6523

                                                                                                  2000thorn 1000frac14 21

                                                                                                  (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                  Characteristic base resistance per unit area qbk frac14 8320

                                                                                                  150frac14 5547 kNm2

                                                                                                  Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                  150frac14 28 kNm2

                                                                                                  Characteristic base and shaft resistances for a single pile

                                                                                                  Rbk frac14 0252 5547 frac14 347 kN

                                                                                                  Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                  For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                  Design bearing resistance Rcd frac14 347

                                                                                                  130thorn 56

                                                                                                  130frac14 310 kN

                                                                                                  For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                  Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                  (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                  From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                  72 Bearing capacity

                                                                                                  N frac14 24thorn 26thorn 34

                                                                                                  3frac14 28

                                                                                                  Ic frac14 171

                                                                                                  2814frac14 0016 ethEquation 818THORN

                                                                                                  s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                  814

                                                                                                  Using Equation 841

                                                                                                  Tf frac14 DLcu thorn

                                                                                                  4ethD2 d2THORNcuNc

                                                                                                  frac14 eth 02 5 06 110THORN thorn

                                                                                                  4eth022 012THORN110 9

                                                                                                  frac14 207thorn 23 frac14 230 kN

                                                                                                  Figure Q813

                                                                                                  Bearing capacity 73

                                                                                                  Chapter 9

                                                                                                  Stability of slopes

                                                                                                  91

                                                                                                  Referring to Figure Q91

                                                                                                  W frac14 417 19 frac14 792 kN=m

                                                                                                  Q frac14 20 28 frac14 56 kN=m

                                                                                                  Arc lengthAB frac14

                                                                                                  180 73 90 frac14 115m

                                                                                                  Arc length BC frac14

                                                                                                  180 28 90 frac14 44m

                                                                                                  The factor of safety is given by

                                                                                                  F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                  90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                  Depth of tension crack z0 frac14 2cu

                                                                                                  frac14 2 20

                                                                                                  19frac14 21m

                                                                                                  Arc length BD frac14

                                                                                                  180 13

                                                                                                  1

                                                                                                  2 90 frac14 21m

                                                                                                  F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                  The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                  Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                  14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                  Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                  The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                  92

                                                                                                  u frac14 0

                                                                                                  Depth factor D frac14 11

                                                                                                  9frac14 122

                                                                                                  Using Equation 92 with F frac14 10

                                                                                                  Ns frac14 cu

                                                                                                  FHfrac14 30

                                                                                                  10 19 9frac14 0175

                                                                                                  Hence from Figure 93

                                                                                                  frac14 50

                                                                                                  For F frac14 12

                                                                                                  Ns frac14 30

                                                                                                  12 19 9frac14 0146

                                                                                                  frac14 27

                                                                                                  93

                                                                                                  Refer to Figure Q93

                                                                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                  1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                  74 m

                                                                                                  214 1deg

                                                                                                  213 1deg

                                                                                                  39 m

                                                                                                  WB

                                                                                                  D

                                                                                                  C

                                                                                                  28 m

                                                                                                  21 m

                                                                                                  A

                                                                                                  Q

                                                                                                  Soil (1)Soil (2)

                                                                                                  73deg

                                                                                                  Figure Q91

                                                                                                  Stability of slopes 75

                                                                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                  9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                  599 256 328 1372

                                                                                                  Figure Q93

                                                                                                  76 Stability of slopes

                                                                                                  XW cos frac14 b

                                                                                                  Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                  W sin frac14 bX

                                                                                                  h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                  Arc length La frac14

                                                                                                  180 57

                                                                                                  1

                                                                                                  2 326 frac14 327m

                                                                                                  The factor of safety is given by

                                                                                                  F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                  W sin

                                                                                                  frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                  frac14 091

                                                                                                  According to the limit state method

                                                                                                  0d frac14 tan1tan 32

                                                                                                  125

                                                                                                  frac14 265

                                                                                                  c0 frac14 8

                                                                                                  160frac14 5 kN=m2

                                                                                                  Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                  Design disturbing moment frac14 1075 kN=m

                                                                                                  The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                  94

                                                                                                  F frac14 1

                                                                                                  W sin

                                                                                                  Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                  1thorn ethtan tan0=FTHORN

                                                                                                  c0 frac14 8 kN=m2

                                                                                                  0 frac14 32

                                                                                                  c0b frac14 8 2 frac14 16 kN=m

                                                                                                  W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                  Try F frac14 100

                                                                                                  tan0

                                                                                                  Ffrac14 0625

                                                                                                  Stability of slopes 77

                                                                                                  Values of u are as obtained in Figure Q93

                                                                                                  SliceNo

                                                                                                  h(m)

                                                                                                  W frac14 bh(kNm)

                                                                                                  W sin(kNm)

                                                                                                  ub(kNm)

                                                                                                  c0bthorn (W ub) tan0(kNm)

                                                                                                  sec

                                                                                                  1thorn (tan tan0)FProduct(kNm)

                                                                                                  1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                  23 33 30 1042 31

                                                                                                  3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                  224 92 72 0931 67

                                                                                                  6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                  12 58 112 90 0889 80

                                                                                                  8 60 252 1812

                                                                                                  80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                  10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                  2154 88 116 0853 99

                                                                                                  14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                  1074 1091

                                                                                                  F frac14 1091

                                                                                                  1074frac14 102 (assumed value 100)

                                                                                                  Thus

                                                                                                  F frac14 101

                                                                                                  95

                                                                                                  F frac14 1

                                                                                                  W sin

                                                                                                  XfWeth1 ruTHORN tan0g sec

                                                                                                  1thorn ethtan tan0THORN=F

                                                                                                  0 frac14 33

                                                                                                  ru frac14 020

                                                                                                  W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                  eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                  Try F frac14 110

                                                                                                  tan 0

                                                                                                  Ffrac14 tan 33

                                                                                                  110frac14 0590

                                                                                                  78 Stability of slopes

                                                                                                  Referring to Figure Q95

                                                                                                  SliceNo

                                                                                                  h(m)

                                                                                                  W frac14 bh(kNm)

                                                                                                  W sin(kNm)

                                                                                                  W(1 ru) tan0(kNm)

                                                                                                  sec

                                                                                                  1thorn ( tan tan0)FProduct(kNm)

                                                                                                  1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                  2120 234 0892 209

                                                                                                  4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                  1185 1271

                                                                                                  Figure Q95

                                                                                                  Stability of slopes 79

                                                                                                  F frac14 1271

                                                                                                  1185frac14 107

                                                                                                  The trial value was 110 therefore take F to be 108

                                                                                                  96

                                                                                                  (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                  F frac14 0

                                                                                                  sat

                                                                                                  tan0

                                                                                                  tan

                                                                                                  ptie 15 frac14 92

                                                                                                  19

                                                                                                  tan 36

                                                                                                  tan

                                                                                                  tan frac14 0234

                                                                                                  frac14 13

                                                                                                  Water table well below surface the factor of safety is given by Equation 911

                                                                                                  F frac14 tan0

                                                                                                  tan

                                                                                                  frac14 tan 36

                                                                                                  tan 13

                                                                                                  frac14 31

                                                                                                  (b) 0d frac14 tan1tan 36

                                                                                                  125

                                                                                                  frac14 30

                                                                                                  Depth of potential failure surface frac14 z

                                                                                                  Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                  frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                  frac14 504z kN

                                                                                                  Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                  frac14 19 z sin 13 cos 13

                                                                                                  frac14 416z kN

                                                                                                  Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                  80 Stability of slopes

                                                                                                  • Book Cover
                                                                                                  • Title
                                                                                                  • Contents
                                                                                                  • Basic characteristics of soils
                                                                                                  • Seepage
                                                                                                  • Effective stress
                                                                                                  • Shear strength
                                                                                                  • Stresses and displacements
                                                                                                  • Lateral earth pressure
                                                                                                  • Consolidation theory
                                                                                                  • Bearing capacity
                                                                                                  • Stability of slopes

                                                                                                    (b) 0 frac14 tan1 ( tan 36125) frac14 30 therefore Ka frac14 033 and Kp frac14 30The surcharge is a variable action therefore a partial factor of 130 applies toforce (1)In this calculation the depth d in Figure Q68 is unknown

                                                                                                    Consider moments (per m) about the tie point A

                                                                                                    Force (kN) Arm (m)

                                                                                                    (1) 033 10 (dthorn 90) 130 frac14 43dthorn 386 d2thorn 30

                                                                                                    (2)1

                                                                                                    2 033 18 452 frac14 601 15

                                                                                                    (3) 033 18 45 (dthorn 45) frac14 267dthorn 1203 d2thorn 525

                                                                                                    (4)1

                                                                                                    2 033 121 (dthorn 45)2 frac14 200d2 thorn 180dthorn 404 2d3thorn 60

                                                                                                    (5)1

                                                                                                    2 134 15 frac14 101 40

                                                                                                    (6) 134 30 frac14 402 60

                                                                                                    (7)1

                                                                                                    2 134 d frac14 67d d3thorn 75

                                                                                                    (8) 1

                                                                                                    2 30 103 d2 frac141545d2 2d3thorn 75

                                                                                                    Moment (kN m)

                                                                                                    (1) 215d2 thorn 322dthorn 1158(2) 902(3) 1335d2 thorn 2003dthorn 6316(4) 133d3 thorn 240d2 thorn 1349dthorn 2424(5) 404(6) 2412(7) 220d2 thorn 502d(8) 103d3 1159d2

                                                                                                    XM frac14 897d3 742d2 thorn 4176d thorn 13616 frac14 0

                                                                                                    d3 thorn 827d2 466d 1518 frac14 0

                                                                                                    By trial

                                                                                                    d frac14 544m

                                                                                                    The minimum depth of embedment required is 544m

                                                                                                    69

                                                                                                    For 0 frac14 30 and frac14 15 Ka frac14 030 and Kp frac14 48

                                                                                                    0 frac14 20 98 frac14 102 kN=m3

                                                                                                    The pressure distribution is shown in Figure Q69 Assuming uniform loss in totalhead along the wall the net water pressure at C is

                                                                                                    44 Lateral earth pressure

                                                                                                    uC frac14 147

                                                                                                    173 26 98 frac14 216 kN=m2

                                                                                                    and the average seepage pressure around the wall is

                                                                                                    j frac14 26

                                                                                                    173 98 frac14 15 kN=m3

                                                                                                    Consider moments about the prop (A) (per m)

                                                                                                    Force (kN) Arm (m) Moment (kN m)

                                                                                                    (1)1

                                                                                                    2 03 17 272 frac14 186 020 37

                                                                                                    (2) 03 17 27 53 frac14 730 335 2445

                                                                                                    (3)1

                                                                                                    2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                                                    (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                                                    (5)1

                                                                                                    2 216 26 frac14 281 243 684

                                                                                                    (6) 216 27 frac14 583 465 2712

                                                                                                    (7)1

                                                                                                    2 216 60 frac14 648 800 5184

                                                                                                    3055(8)

                                                                                                    1

                                                                                                    2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                                                    Factor of safety

                                                                                                    Fr frac14 6885

                                                                                                    3055frac14 225

                                                                                                    Figure Q69

                                                                                                    Lateral earth pressure 45

                                                                                                    610

                                                                                                    For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                                                    p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                                                    Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                                                    Using the recommendations of Twine and Roscoe

                                                                                                    p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                                                    Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                                                    611

                                                                                                    frac14 18 kN=m3 0 frac14 34

                                                                                                    H frac14 350m nH frac14 335m mH frac14 185m

                                                                                                    Consider a trial value of F frac14 20 Refer to Figure 635

                                                                                                    0m frac14 tan1tan 34

                                                                                                    20

                                                                                                    frac14 186

                                                                                                    Then

                                                                                                    frac14 45 thorn 0m2frac14 543

                                                                                                    W frac14 1

                                                                                                    2 18 3502 cot 543 frac14 792 kN=m

                                                                                                    Figure Q610

                                                                                                    46 Lateral earth pressure

                                                                                                    P frac14 1

                                                                                                    2 s 3352 frac14 561s kN=m

                                                                                                    U frac14 1

                                                                                                    2 98 1852 cosec 543 frac14 206 kN=m

                                                                                                    Equations 630 and 631 then become

                                                                                                    561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                                                    792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                                                    ie

                                                                                                    561s 0616N 405 frac14 0

                                                                                                    792 0857N thorn 563 frac14 0

                                                                                                    N frac14 848

                                                                                                    0857frac14 989 kN=m

                                                                                                    Then

                                                                                                    561s 609 405 frac14 0

                                                                                                    s frac14 649

                                                                                                    561frac14 116 kN=m3

                                                                                                    The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                                                    F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                                                    20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                                                    s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                                                    Figure Q611

                                                                                                    Lateral earth pressure 47

                                                                                                    612

                                                                                                    For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                                    45 thorn 0

                                                                                                    2frac14 63

                                                                                                    For the retained material between the surface and a depth of 36m

                                                                                                    Pa frac14 1

                                                                                                    2 030 18 362 frac14 350 kN=m

                                                                                                    Weight of reinforced fill between the surface and a depth of 36m is

                                                                                                    Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                                    eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                                    Eccentricity of Rv

                                                                                                    e frac14 263 250 frac14 013m

                                                                                                    The average vertical stress at a depth of 36m is

                                                                                                    z frac14 Rv

                                                                                                    L 2efrac14 324

                                                                                                    474frac14 68 kN=m2

                                                                                                    (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                                    Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                                    Tensile stress in the element frac14 138 103

                                                                                                    65 3frac14 71N=mm2

                                                                                                    Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                                    Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                                    Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                                    The weight of ABC is

                                                                                                    W frac14 1

                                                                                                    2 18 52 265 frac14 124 kN=m

                                                                                                    From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                                    48 Lateral earth pressure

                                                                                                    (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                                    Tp frac14 032 68 120 065 frac14 170 kN

                                                                                                    Tr frac14 213 420

                                                                                                    418frac14 214 kN

                                                                                                    Again the tensile failure and slipping limit states are satisfied for this element

                                                                                                    Figure Q612

                                                                                                    Lateral earth pressure 49

                                                                                                    Chapter 7

                                                                                                    Consolidation theory

                                                                                                    71

                                                                                                    Total change in thickness

                                                                                                    H frac14 782 602 frac14 180mm

                                                                                                    Average thickness frac14 1530thorn 180

                                                                                                    2frac14 1620mm

                                                                                                    Length of drainage path d frac14 1620

                                                                                                    2frac14 810mm

                                                                                                    Root time plot (Figure Q71a)

                                                                                                    ffiffiffiffiffiffit90p frac14 33

                                                                                                    t90 frac14 109min

                                                                                                    cv frac14 0848d2

                                                                                                    t90frac14 0848 8102

                                                                                                    109 1440 365

                                                                                                    106frac14 27m2=year

                                                                                                    r0 frac14 782 764

                                                                                                    782 602frac14 018

                                                                                                    180frac14 0100

                                                                                                    rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                                    10 119

                                                                                                    9 180frac14 0735

                                                                                                    rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                                    Log time plot (Figure Q71b)

                                                                                                    t50 frac14 26min

                                                                                                    cv frac14 0196d2

                                                                                                    t50frac14 0196 8102

                                                                                                    26 1440 365

                                                                                                    106frac14 26m2=year

                                                                                                    r0 frac14 782 763

                                                                                                    782 602frac14 019

                                                                                                    180frac14 0106

                                                                                                    rp frac14 763 623

                                                                                                    782 602frac14 140

                                                                                                    180frac14 0778

                                                                                                    rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                                    Figure Q71(a)

                                                                                                    Figure Q71(b)

                                                                                                    Final void ratio

                                                                                                    e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                                    e

                                                                                                    Hfrac14 1thorn e0

                                                                                                    H0frac14 1thorn e1 thorne

                                                                                                    H0

                                                                                                    ie

                                                                                                    e

                                                                                                    180frac14 1631thorne

                                                                                                    1710

                                                                                                    e frac14 2936

                                                                                                    1530frac14 0192

                                                                                                    Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                                    mv frac14 1

                                                                                                    1thorn e0 e0 e101 00

                                                                                                    frac14 1

                                                                                                    1823 0192

                                                                                                    0107frac14 098m2=MN

                                                                                                    k frac14 cvmvw frac14 265 098 98

                                                                                                    60 1440 365 103frac14 81 1010 m=s

                                                                                                    72

                                                                                                    Using Equation 77 (one-dimensional method)

                                                                                                    sc frac14 e0 e11thorn e0 H

                                                                                                    Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                                    Figure Q72

                                                                                                    52 Consolidation theory

                                                                                                    Settlement

                                                                                                    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                                    1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                                    318

                                                                                                    Notes 5 92y 460thorn 84

                                                                                                    Heave

                                                                                                    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                                    1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                                    38

                                                                                                    73

                                                                                                    U frac14 f ethTvTHORN frac14 f cvt

                                                                                                    d2

                                                                                                    Hence if cv is constant

                                                                                                    t1

                                                                                                    t2frac14 d

                                                                                                    21

                                                                                                    d22

                                                                                                    where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                                    d1 frac14 95mm and d2 frac14 2500mm

                                                                                                    for U frac14 050 t2 frac14 t1 d22

                                                                                                    d21

                                                                                                    frac14 20

                                                                                                    60 24 365 25002

                                                                                                    952frac14 263 years

                                                                                                    for U lt 060 Tv frac14

                                                                                                    4U2 (Equation 724(a))

                                                                                                    t030 frac14 t050 0302

                                                                                                    0502

                                                                                                    frac14 263 036 frac14 095 years

                                                                                                    Consolidation theory 53

                                                                                                    74

                                                                                                    The layer is open

                                                                                                    d frac14 8

                                                                                                    2frac14 4m

                                                                                                    Tv frac14 cvtd2frac14 24 3

                                                                                                    42frac14 0450

                                                                                                    ui frac14 frac14 84 kN=m2

                                                                                                    The excess pore water pressure is given by Equation 721

                                                                                                    ue frac14Xmfrac141mfrac140

                                                                                                    2ui

                                                                                                    Msin

                                                                                                    Mz

                                                                                                    d

                                                                                                    expethM2TvTHORN

                                                                                                    In this case z frac14 d

                                                                                                    sinMz

                                                                                                    d

                                                                                                    frac14 sinM

                                                                                                    where

                                                                                                    M frac14

                                                                                                    23

                                                                                                    25

                                                                                                    2

                                                                                                    M sin M M2Tv exp (M2Tv)

                                                                                                    2thorn1 1110 0329

                                                                                                    3

                                                                                                    21 9993 457 105

                                                                                                    ue frac14 2 84 2

                                                                                                    1 0329 ethother terms negligibleTHORN

                                                                                                    frac14 352 kN=m2

                                                                                                    75

                                                                                                    The layer is open

                                                                                                    d frac14 6

                                                                                                    2frac14 3m

                                                                                                    Tv frac14 cvtd2frac14 10 3

                                                                                                    32frac14 0333

                                                                                                    The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                                    54 Consolidation theory

                                                                                                    For an open layer

                                                                                                    Tv frac14 4n

                                                                                                    m2

                                                                                                    n frac14 0333 62

                                                                                                    4frac14 300

                                                                                                    The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                                    ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                                    i j

                                                                                                    0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                                    0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                                    The initial and 3-year isochrones are plotted in Figure Q75

                                                                                                    Area under initial isochrone frac14 180 units

                                                                                                    Area under 3-year isochrone frac14 63 units

                                                                                                    The average degree of consolidation is given by Equation 725Thus

                                                                                                    U frac14 1 63

                                                                                                    180frac14 065

                                                                                                    Figure Q75

                                                                                                    Consolidation theory 55

                                                                                                    76

                                                                                                    At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                    0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                    The final consolidation settlement (one-dimensional method) is

                                                                                                    sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                    Corrected time t frac14 2 1

                                                                                                    2

                                                                                                    40

                                                                                                    52

                                                                                                    frac14 1615 years

                                                                                                    Tv frac14 cvtd2frac14 44 1615

                                                                                                    42frac14 0444

                                                                                                    From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                    77

                                                                                                    The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                    Figure Q77

                                                                                                    56 Consolidation theory

                                                                                                    Point m n Ir (kNm2) sc (mm)

                                                                                                    13020frac14 15 20

                                                                                                    20frac14 10 0194 (4) 113 124

                                                                                                    260

                                                                                                    20frac14 30

                                                                                                    20

                                                                                                    20frac14 10 0204 (2) 59 65

                                                                                                    360

                                                                                                    20frac14 30

                                                                                                    40

                                                                                                    20frac14 20 0238 (1) 35 38

                                                                                                    430

                                                                                                    20frac14 15

                                                                                                    40

                                                                                                    20frac14 20 0224 (2) 65 72

                                                                                                    Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                    78

                                                                                                    Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                    (a) Immediate settlement

                                                                                                    H

                                                                                                    Bfrac14 30

                                                                                                    35frac14 086

                                                                                                    D

                                                                                                    Bfrac14 2

                                                                                                    35frac14 006

                                                                                                    Figure Q78

                                                                                                    Consolidation theory 57

                                                                                                    From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                    si frac14 130131qB

                                                                                                    Eufrac14 10 032 105 35

                                                                                                    40frac14 30mm

                                                                                                    (b) Consolidation settlement

                                                                                                    Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                    1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                    3150

                                                                                                    Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                    Now

                                                                                                    H

                                                                                                    Bfrac14 30

                                                                                                    35frac14 086 and A frac14 065

                                                                                                    from Figure 712 13 frac14 079

                                                                                                    sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                    Total settlement

                                                                                                    s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                    79

                                                                                                    Without sand drains

                                                                                                    Uv frac14 025

                                                                                                    Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                    t frac14 Tvd2

                                                                                                    cvfrac14 0049 82

                                                                                                    cvWith sand drains

                                                                                                    R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                    n frac14 Rrfrac14 169

                                                                                                    015frac14 113

                                                                                                    Tr frac14 cht

                                                                                                    4R2frac14 ch

                                                                                                    4 1692 0049 82

                                                                                                    cvethand ch frac14 cvTHORN

                                                                                                    frac14 0275

                                                                                                    Ur frac14 073 (from Figure 730)

                                                                                                    58 Consolidation theory

                                                                                                    Using Equation 740

                                                                                                    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                    U frac14 080

                                                                                                    710

                                                                                                    Without sand drains

                                                                                                    Uv frac14 090

                                                                                                    Tv frac14 0848

                                                                                                    t frac14 Tvd2

                                                                                                    cvfrac14 0848 102

                                                                                                    96frac14 88 years

                                                                                                    With sand drains

                                                                                                    R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                    n frac14 Rrfrac14 226

                                                                                                    015frac14 15

                                                                                                    Tr

                                                                                                    Tvfrac14 chcv

                                                                                                    d2

                                                                                                    4R2ethsame tTHORN

                                                                                                    Tr

                                                                                                    Tvfrac14 140

                                                                                                    96 102

                                                                                                    4 2262frac14 714 eth1THORN

                                                                                                    Using Equation 740

                                                                                                    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                    An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                    Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                    040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                    Thus

                                                                                                    Uv frac14 0295 and Ur frac14 086

                                                                                                    t frac14 88 00683

                                                                                                    0848frac14 07 years

                                                                                                    Consolidation theory 59

                                                                                                    Chapter 8

                                                                                                    Bearing capacity

                                                                                                    81

                                                                                                    (a) The ultimate bearing capacity is given by Equation 83

                                                                                                    qf frac14 cNc thorn DNq thorn 1

                                                                                                    2BN

                                                                                                    For u frac14 0

                                                                                                    Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                    qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                    The net ultimate bearing capacity is

                                                                                                    qnf frac14 qf D frac14 540 kN=m2

                                                                                                    The net foundation pressure is

                                                                                                    qn frac14 q D frac14 425

                                                                                                    2 eth21 1THORN frac14 192 kN=m2

                                                                                                    The factor of safety (Equation 86) is

                                                                                                    F frac14 qnfqnfrac14 540

                                                                                                    192frac14 28

                                                                                                    (b) For 0 frac14 28

                                                                                                    Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                    qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                    2 112 2 13

                                                                                                    frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                    qnf frac14 574 112 frac14 563 kN=m2

                                                                                                    F frac14 563

                                                                                                    192frac14 29

                                                                                                    (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                    82

                                                                                                    For 0 frac14 38

                                                                                                    Nq frac14 49 N frac14 67

                                                                                                    qnf frac14 DethNq 1THORN thorn 1

                                                                                                    2BN ethfrom Equation 83THORN

                                                                                                    frac14 eth18 075 48THORN thorn 1

                                                                                                    2 18 15 67

                                                                                                    frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                    qn frac14 500

                                                                                                    15 eth18 075THORN frac14 320 kN=m2

                                                                                                    F frac14 qnfqnfrac14 1553

                                                                                                    320frac14 48

                                                                                                    0d frac14 tan1tan 38

                                                                                                    125

                                                                                                    frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                    Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                    2 18 15 25

                                                                                                    frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                    Design load (action) Vd frac14 500 kN=m

                                                                                                    The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                    83

                                                                                                    D

                                                                                                    Bfrac14 350

                                                                                                    225frac14 155

                                                                                                    From Figure 85 for a square foundation

                                                                                                    Nc frac14 81

                                                                                                    Bearing capacity 61

                                                                                                    For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                    Nc frac14 084thorn 016B

                                                                                                    L

                                                                                                    81 frac14 745

                                                                                                    Using Equation 810

                                                                                                    qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                    For F frac14 3

                                                                                                    qn frac14 1006

                                                                                                    3frac14 335 kN=m2

                                                                                                    q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                    Design load frac14 405 450 225 frac14 4100 kN

                                                                                                    Design undrained strength cud frac14 135

                                                                                                    14frac14 96 kN=m2

                                                                                                    Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                    frac14 7241 kN

                                                                                                    Design load Vd frac14 4100 kN

                                                                                                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                    84

                                                                                                    For 0 frac14 40

                                                                                                    Nq frac14 64 N frac14 95

                                                                                                    qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                    (a) Water table 5m below ground level

                                                                                                    qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                    qn frac14 400 17 frac14 383 kN=m2

                                                                                                    F frac14 2686

                                                                                                    383frac14 70

                                                                                                    (b) Water table 1m below ground level (ie at foundation level)

                                                                                                    0 frac14 20 98 frac14 102 kN=m3

                                                                                                    62 Bearing capacity

                                                                                                    qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                    F frac14 2040

                                                                                                    383frac14 53

                                                                                                    (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                    eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                    qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                    F frac14 1296

                                                                                                    392frac14 33

                                                                                                    85

                                                                                                    The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                    Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                    Design value of 0 frac14 tan1tan 39

                                                                                                    125

                                                                                                    frac14 33

                                                                                                    For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                    Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                    86

                                                                                                    (a) Undrained shear for u frac14 0

                                                                                                    Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                    qnf frac14 12cuNc

                                                                                                    frac14 12 100 514 frac14 617 kN=m2

                                                                                                    qn frac14 qnfFfrac14 617

                                                                                                    3frac14 206 kN=m2

                                                                                                    q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                    Bearing capacity 63

                                                                                                    Drained shear for 0 frac14 32

                                                                                                    Nq frac14 23 N frac14 25

                                                                                                    0 frac14 21 98 frac14 112 kN=m3

                                                                                                    qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                    frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                    frac14 694 kN=m2

                                                                                                    q frac14 694

                                                                                                    3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                    Design load frac14 42 227 frac14 3632 kN

                                                                                                    (b) Design undrained strength cud frac14 100

                                                                                                    14frac14 71 kNm2

                                                                                                    Design bearing resistance Rd frac14 12cudNe area

                                                                                                    frac14 12 71 514 42

                                                                                                    frac14 7007 kN

                                                                                                    For drained shear 0d frac14 tan1tan 32

                                                                                                    125

                                                                                                    frac14 26

                                                                                                    Nq frac14 12 N frac14 10

                                                                                                    Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                    (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                    Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                    1 2 100 0175 0700qn 0182qn

                                                                                                    2 6 033 0044 0176qn 0046qn

                                                                                                    3 10 020 0017 0068qn 0018qn

                                                                                                    0246qn

                                                                                                    Diameter of equivalent circle B frac14 45m

                                                                                                    H

                                                                                                    Bfrac14 12

                                                                                                    45frac14 27 and A frac14 042

                                                                                                    13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                    64 Bearing capacity

                                                                                                    For sc frac14 30mm

                                                                                                    qn frac14 30

                                                                                                    0147frac14 204 kN=m2

                                                                                                    q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                    Design load frac14 42 225 frac14 3600 kN

                                                                                                    The design load is 3600 kN settlement being the limiting criterion

                                                                                                    87

                                                                                                    D

                                                                                                    Bfrac14 8

                                                                                                    4frac14 20

                                                                                                    From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                    F frac14 cuNc

                                                                                                    Dfrac14 40 71

                                                                                                    20 8frac14 18

                                                                                                    88

                                                                                                    Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                    Design value of 0 frac14 tan1tan 38

                                                                                                    125

                                                                                                    frac14 32

                                                                                                    Figure Q86

                                                                                                    Bearing capacity 65

                                                                                                    For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                    Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                    The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                    Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                    For B frac14 250m qn frac14 3750

                                                                                                    2502 17 frac14 583 kN=m2

                                                                                                    From Figure 510 m frac14 n frac14 126

                                                                                                    6frac14 021

                                                                                                    Ir frac14 0019

                                                                                                    Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                    Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                    The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                    89

                                                                                                    Depth (m) N 0v (kNm2) CN N1

                                                                                                    070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                    Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                    (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                    Cw frac14 05thorn 0530

                                                                                                    47

                                                                                                    frac14 082

                                                                                                    66 Bearing capacity

                                                                                                    Thus

                                                                                                    qa frac14 150 082 frac14 120 kN=m2

                                                                                                    (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                    Thus

                                                                                                    qa frac14 90 15 frac14 135 kN=m2

                                                                                                    (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                    Ic frac14 171

                                                                                                    1014frac14 0068

                                                                                                    From Equation 819(a) with s frac14 25mm

                                                                                                    q frac14 25

                                                                                                    3507 0068frac14 150 kN=m2

                                                                                                    810

                                                                                                    Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                    Izp frac14 05thorn 01130

                                                                                                    16 27

                                                                                                    05

                                                                                                    frac14 067

                                                                                                    Refer to Figure Q810

                                                                                                    E frac14 25qc

                                                                                                    Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                    Ez (mm3MN)

                                                                                                    1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                    0203

                                                                                                    C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                    130frac14 093

                                                                                                    C2 frac14 1 ethsayTHORN

                                                                                                    s frac14 C1C2qnX Iz

                                                                                                    Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                    Bearing capacity 67

                                                                                                    811

                                                                                                    At pile base level

                                                                                                    cu frac14 220 kN=m2

                                                                                                    qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                    Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                    00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                    qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                    Then

                                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                                    frac14

                                                                                                    4 32 1980

                                                                                                    thorn eth 105 139 86THORN

                                                                                                    frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                    0 01 02 03 04 05 06 07

                                                                                                    0 2 4 6 8 10 12 14

                                                                                                    1

                                                                                                    2

                                                                                                    3

                                                                                                    4

                                                                                                    5

                                                                                                    6

                                                                                                    7

                                                                                                    8

                                                                                                    (1)

                                                                                                    (2)

                                                                                                    (3)

                                                                                                    (4)

                                                                                                    (5)

                                                                                                    qc

                                                                                                    qc

                                                                                                    Iz

                                                                                                    Iz

                                                                                                    (MNm2)

                                                                                                    z (m)

                                                                                                    Figure Q810

                                                                                                    68 Bearing capacity

                                                                                                    Allowable load

                                                                                                    ethaTHORN Qf

                                                                                                    2frac14 17 937

                                                                                                    2frac14 8968 kN

                                                                                                    ethbTHORN Abqb

                                                                                                    3thorn Asqs frac14 13 996

                                                                                                    3thorn 3941 frac14 8606 kN

                                                                                                    ie allowable load frac14 8600 kN

                                                                                                    Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                    According to the limit state method

                                                                                                    Characteristic undrained strength at base level cuk frac14 220

                                                                                                    150kN=m2

                                                                                                    Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                    150frac14 1320 kN=m2

                                                                                                    Characteristic shaft resistance qsk frac14 00150

                                                                                                    frac14 86

                                                                                                    150frac14 57 kN=m2

                                                                                                    Characteristic base and shaft resistances

                                                                                                    Rbk frac14

                                                                                                    4 32 1320 frac14 9330 kN

                                                                                                    Rsk frac14 105 139 86

                                                                                                    150frac14 2629 kN

                                                                                                    For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                    Design bearing resistance Rcd frac14 9330

                                                                                                    160thorn 2629

                                                                                                    130

                                                                                                    frac14 5831thorn 2022

                                                                                                    frac14 7850 kN

                                                                                                    Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                    812

                                                                                                    ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                    qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                    For a single pile

                                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                                    frac14

                                                                                                    4 062 1305

                                                                                                    thorn eth 06 15 42THORN

                                                                                                    frac14 369thorn 1187 frac14 1556 kN

                                                                                                    Bearing capacity 69

                                                                                                    Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                    eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                    (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                    (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                    qbkfrac14 9cuk frac14 9 220

                                                                                                    150frac14 1320 kN=m2

                                                                                                    qskfrac14cuk frac14 040 105

                                                                                                    150frac14 28 kN=m2

                                                                                                    Rbkfrac14

                                                                                                    4 0602 1320 frac14 373 kN

                                                                                                    Rskfrac14 060 15 28 frac14 791 kN

                                                                                                    Rcdfrac14 373

                                                                                                    160thorn 791

                                                                                                    130frac14 233thorn 608 frac14 841 kN

                                                                                                    Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                    Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                    (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                    q frac14 21 000

                                                                                                    1762frac14 68 kN=m2

                                                                                                    Immediate settlement

                                                                                                    H

                                                                                                    Bfrac14 15

                                                                                                    176frac14 085

                                                                                                    D

                                                                                                    Bfrac14 13

                                                                                                    176frac14 074

                                                                                                    L

                                                                                                    Bfrac14 1

                                                                                                    Hence from Figure 515

                                                                                                    130 frac14 078 and 131 frac14 041

                                                                                                    70 Bearing capacity

                                                                                                    Thus using Equation 528

                                                                                                    si frac14 078 041 68 176

                                                                                                    65frac14 6mm

                                                                                                    Consolidation settlement

                                                                                                    Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                    1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                    434 (sod)

                                                                                                    Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                    sc frac14 056 434 frac14 24mm

                                                                                                    The total settlement is (6thorn 24) frac14 30mm

                                                                                                    813

                                                                                                    At base level N frac14 26 Then using Equation 830

                                                                                                    qb frac14 40NDb

                                                                                                    Bfrac14 40 26 2

                                                                                                    025frac14 8320 kN=m2

                                                                                                    ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                    Figure Q812

                                                                                                    Bearing capacity 71

                                                                                                    Over the length embedded in sand

                                                                                                    N frac14 21 ie18thorn 24

                                                                                                    2

                                                                                                    Using Equation 831

                                                                                                    qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                    For a single pile

                                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                                    frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                    For the pile group assuming a group efficiency of 12

                                                                                                    XQf frac14 12 9 604 frac14 6523 kN

                                                                                                    Then the load factor is

                                                                                                    F frac14 6523

                                                                                                    2000thorn 1000frac14 21

                                                                                                    (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                    Characteristic base resistance per unit area qbk frac14 8320

                                                                                                    150frac14 5547 kNm2

                                                                                                    Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                    150frac14 28 kNm2

                                                                                                    Characteristic base and shaft resistances for a single pile

                                                                                                    Rbk frac14 0252 5547 frac14 347 kN

                                                                                                    Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                    For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                    Design bearing resistance Rcd frac14 347

                                                                                                    130thorn 56

                                                                                                    130frac14 310 kN

                                                                                                    For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                    Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                    (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                    From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                    72 Bearing capacity

                                                                                                    N frac14 24thorn 26thorn 34

                                                                                                    3frac14 28

                                                                                                    Ic frac14 171

                                                                                                    2814frac14 0016 ethEquation 818THORN

                                                                                                    s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                    814

                                                                                                    Using Equation 841

                                                                                                    Tf frac14 DLcu thorn

                                                                                                    4ethD2 d2THORNcuNc

                                                                                                    frac14 eth 02 5 06 110THORN thorn

                                                                                                    4eth022 012THORN110 9

                                                                                                    frac14 207thorn 23 frac14 230 kN

                                                                                                    Figure Q813

                                                                                                    Bearing capacity 73

                                                                                                    Chapter 9

                                                                                                    Stability of slopes

                                                                                                    91

                                                                                                    Referring to Figure Q91

                                                                                                    W frac14 417 19 frac14 792 kN=m

                                                                                                    Q frac14 20 28 frac14 56 kN=m

                                                                                                    Arc lengthAB frac14

                                                                                                    180 73 90 frac14 115m

                                                                                                    Arc length BC frac14

                                                                                                    180 28 90 frac14 44m

                                                                                                    The factor of safety is given by

                                                                                                    F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                    90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                    Depth of tension crack z0 frac14 2cu

                                                                                                    frac14 2 20

                                                                                                    19frac14 21m

                                                                                                    Arc length BD frac14

                                                                                                    180 13

                                                                                                    1

                                                                                                    2 90 frac14 21m

                                                                                                    F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                    The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                    Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                    14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                    Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                    The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                    92

                                                                                                    u frac14 0

                                                                                                    Depth factor D frac14 11

                                                                                                    9frac14 122

                                                                                                    Using Equation 92 with F frac14 10

                                                                                                    Ns frac14 cu

                                                                                                    FHfrac14 30

                                                                                                    10 19 9frac14 0175

                                                                                                    Hence from Figure 93

                                                                                                    frac14 50

                                                                                                    For F frac14 12

                                                                                                    Ns frac14 30

                                                                                                    12 19 9frac14 0146

                                                                                                    frac14 27

                                                                                                    93

                                                                                                    Refer to Figure Q93

                                                                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                    1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                    74 m

                                                                                                    214 1deg

                                                                                                    213 1deg

                                                                                                    39 m

                                                                                                    WB

                                                                                                    D

                                                                                                    C

                                                                                                    28 m

                                                                                                    21 m

                                                                                                    A

                                                                                                    Q

                                                                                                    Soil (1)Soil (2)

                                                                                                    73deg

                                                                                                    Figure Q91

                                                                                                    Stability of slopes 75

                                                                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                    9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                    599 256 328 1372

                                                                                                    Figure Q93

                                                                                                    76 Stability of slopes

                                                                                                    XW cos frac14 b

                                                                                                    Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                    W sin frac14 bX

                                                                                                    h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                    Arc length La frac14

                                                                                                    180 57

                                                                                                    1

                                                                                                    2 326 frac14 327m

                                                                                                    The factor of safety is given by

                                                                                                    F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                    W sin

                                                                                                    frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                    frac14 091

                                                                                                    According to the limit state method

                                                                                                    0d frac14 tan1tan 32

                                                                                                    125

                                                                                                    frac14 265

                                                                                                    c0 frac14 8

                                                                                                    160frac14 5 kN=m2

                                                                                                    Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                    Design disturbing moment frac14 1075 kN=m

                                                                                                    The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                    94

                                                                                                    F frac14 1

                                                                                                    W sin

                                                                                                    Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                    1thorn ethtan tan0=FTHORN

                                                                                                    c0 frac14 8 kN=m2

                                                                                                    0 frac14 32

                                                                                                    c0b frac14 8 2 frac14 16 kN=m

                                                                                                    W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                    Try F frac14 100

                                                                                                    tan0

                                                                                                    Ffrac14 0625

                                                                                                    Stability of slopes 77

                                                                                                    Values of u are as obtained in Figure Q93

                                                                                                    SliceNo

                                                                                                    h(m)

                                                                                                    W frac14 bh(kNm)

                                                                                                    W sin(kNm)

                                                                                                    ub(kNm)

                                                                                                    c0bthorn (W ub) tan0(kNm)

                                                                                                    sec

                                                                                                    1thorn (tan tan0)FProduct(kNm)

                                                                                                    1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                    23 33 30 1042 31

                                                                                                    3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                    224 92 72 0931 67

                                                                                                    6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                    12 58 112 90 0889 80

                                                                                                    8 60 252 1812

                                                                                                    80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                    10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                    2154 88 116 0853 99

                                                                                                    14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                    1074 1091

                                                                                                    F frac14 1091

                                                                                                    1074frac14 102 (assumed value 100)

                                                                                                    Thus

                                                                                                    F frac14 101

                                                                                                    95

                                                                                                    F frac14 1

                                                                                                    W sin

                                                                                                    XfWeth1 ruTHORN tan0g sec

                                                                                                    1thorn ethtan tan0THORN=F

                                                                                                    0 frac14 33

                                                                                                    ru frac14 020

                                                                                                    W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                    eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                    Try F frac14 110

                                                                                                    tan 0

                                                                                                    Ffrac14 tan 33

                                                                                                    110frac14 0590

                                                                                                    78 Stability of slopes

                                                                                                    Referring to Figure Q95

                                                                                                    SliceNo

                                                                                                    h(m)

                                                                                                    W frac14 bh(kNm)

                                                                                                    W sin(kNm)

                                                                                                    W(1 ru) tan0(kNm)

                                                                                                    sec

                                                                                                    1thorn ( tan tan0)FProduct(kNm)

                                                                                                    1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                    2120 234 0892 209

                                                                                                    4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                    1185 1271

                                                                                                    Figure Q95

                                                                                                    Stability of slopes 79

                                                                                                    F frac14 1271

                                                                                                    1185frac14 107

                                                                                                    The trial value was 110 therefore take F to be 108

                                                                                                    96

                                                                                                    (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                    F frac14 0

                                                                                                    sat

                                                                                                    tan0

                                                                                                    tan

                                                                                                    ptie 15 frac14 92

                                                                                                    19

                                                                                                    tan 36

                                                                                                    tan

                                                                                                    tan frac14 0234

                                                                                                    frac14 13

                                                                                                    Water table well below surface the factor of safety is given by Equation 911

                                                                                                    F frac14 tan0

                                                                                                    tan

                                                                                                    frac14 tan 36

                                                                                                    tan 13

                                                                                                    frac14 31

                                                                                                    (b) 0d frac14 tan1tan 36

                                                                                                    125

                                                                                                    frac14 30

                                                                                                    Depth of potential failure surface frac14 z

                                                                                                    Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                    frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                    frac14 504z kN

                                                                                                    Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                    frac14 19 z sin 13 cos 13

                                                                                                    frac14 416z kN

                                                                                                    Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                    80 Stability of slopes

                                                                                                    • Book Cover
                                                                                                    • Title
                                                                                                    • Contents
                                                                                                    • Basic characteristics of soils
                                                                                                    • Seepage
                                                                                                    • Effective stress
                                                                                                    • Shear strength
                                                                                                    • Stresses and displacements
                                                                                                    • Lateral earth pressure
                                                                                                    • Consolidation theory
                                                                                                    • Bearing capacity
                                                                                                    • Stability of slopes

                                                                                                      uC frac14 147

                                                                                                      173 26 98 frac14 216 kN=m2

                                                                                                      and the average seepage pressure around the wall is

                                                                                                      j frac14 26

                                                                                                      173 98 frac14 15 kN=m3

                                                                                                      Consider moments about the prop (A) (per m)

                                                                                                      Force (kN) Arm (m) Moment (kN m)

                                                                                                      (1)1

                                                                                                      2 03 17 272 frac14 186 020 37

                                                                                                      (2) 03 17 27 53 frac14 730 335 2445

                                                                                                      (3)1

                                                                                                      2 03 (102thorn 15) 532 frac14 493 423 2085

                                                                                                      (4) 03 f(17 27)thorn (117 53)g 60 frac141942 900 17478

                                                                                                      (5)1

                                                                                                      2 216 26 frac14 281 243 684

                                                                                                      (6) 216 27 frac14 583 465 2712

                                                                                                      (7)1

                                                                                                      2 216 60 frac14 648 800 5184

                                                                                                      3055(8)

                                                                                                      1

                                                                                                      2f48 (102 15) 03 (102thorn 15)g 602 frac14 6885 1000 6885

                                                                                                      Factor of safety

                                                                                                      Fr frac14 6885

                                                                                                      3055frac14 225

                                                                                                      Figure Q69

                                                                                                      Lateral earth pressure 45

                                                                                                      610

                                                                                                      For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                                                      p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                                                      Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                                                      Using the recommendations of Twine and Roscoe

                                                                                                      p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                                                      Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                                                      611

                                                                                                      frac14 18 kN=m3 0 frac14 34

                                                                                                      H frac14 350m nH frac14 335m mH frac14 185m

                                                                                                      Consider a trial value of F frac14 20 Refer to Figure 635

                                                                                                      0m frac14 tan1tan 34

                                                                                                      20

                                                                                                      frac14 186

                                                                                                      Then

                                                                                                      frac14 45 thorn 0m2frac14 543

                                                                                                      W frac14 1

                                                                                                      2 18 3502 cot 543 frac14 792 kN=m

                                                                                                      Figure Q610

                                                                                                      46 Lateral earth pressure

                                                                                                      P frac14 1

                                                                                                      2 s 3352 frac14 561s kN=m

                                                                                                      U frac14 1

                                                                                                      2 98 1852 cosec 543 frac14 206 kN=m

                                                                                                      Equations 630 and 631 then become

                                                                                                      561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                                                      792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                                                      ie

                                                                                                      561s 0616N 405 frac14 0

                                                                                                      792 0857N thorn 563 frac14 0

                                                                                                      N frac14 848

                                                                                                      0857frac14 989 kN=m

                                                                                                      Then

                                                                                                      561s 609 405 frac14 0

                                                                                                      s frac14 649

                                                                                                      561frac14 116 kN=m3

                                                                                                      The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                                                      F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                                                      20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                                                      s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                                                      Figure Q611

                                                                                                      Lateral earth pressure 47

                                                                                                      612

                                                                                                      For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                                      45 thorn 0

                                                                                                      2frac14 63

                                                                                                      For the retained material between the surface and a depth of 36m

                                                                                                      Pa frac14 1

                                                                                                      2 030 18 362 frac14 350 kN=m

                                                                                                      Weight of reinforced fill between the surface and a depth of 36m is

                                                                                                      Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                                      eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                                      Eccentricity of Rv

                                                                                                      e frac14 263 250 frac14 013m

                                                                                                      The average vertical stress at a depth of 36m is

                                                                                                      z frac14 Rv

                                                                                                      L 2efrac14 324

                                                                                                      474frac14 68 kN=m2

                                                                                                      (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                                      Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                                      Tensile stress in the element frac14 138 103

                                                                                                      65 3frac14 71N=mm2

                                                                                                      Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                                      Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                                      Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                                      The weight of ABC is

                                                                                                      W frac14 1

                                                                                                      2 18 52 265 frac14 124 kN=m

                                                                                                      From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                                      48 Lateral earth pressure

                                                                                                      (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                                      Tp frac14 032 68 120 065 frac14 170 kN

                                                                                                      Tr frac14 213 420

                                                                                                      418frac14 214 kN

                                                                                                      Again the tensile failure and slipping limit states are satisfied for this element

                                                                                                      Figure Q612

                                                                                                      Lateral earth pressure 49

                                                                                                      Chapter 7

                                                                                                      Consolidation theory

                                                                                                      71

                                                                                                      Total change in thickness

                                                                                                      H frac14 782 602 frac14 180mm

                                                                                                      Average thickness frac14 1530thorn 180

                                                                                                      2frac14 1620mm

                                                                                                      Length of drainage path d frac14 1620

                                                                                                      2frac14 810mm

                                                                                                      Root time plot (Figure Q71a)

                                                                                                      ffiffiffiffiffiffit90p frac14 33

                                                                                                      t90 frac14 109min

                                                                                                      cv frac14 0848d2

                                                                                                      t90frac14 0848 8102

                                                                                                      109 1440 365

                                                                                                      106frac14 27m2=year

                                                                                                      r0 frac14 782 764

                                                                                                      782 602frac14 018

                                                                                                      180frac14 0100

                                                                                                      rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                                      10 119

                                                                                                      9 180frac14 0735

                                                                                                      rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                                      Log time plot (Figure Q71b)

                                                                                                      t50 frac14 26min

                                                                                                      cv frac14 0196d2

                                                                                                      t50frac14 0196 8102

                                                                                                      26 1440 365

                                                                                                      106frac14 26m2=year

                                                                                                      r0 frac14 782 763

                                                                                                      782 602frac14 019

                                                                                                      180frac14 0106

                                                                                                      rp frac14 763 623

                                                                                                      782 602frac14 140

                                                                                                      180frac14 0778

                                                                                                      rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                                      Figure Q71(a)

                                                                                                      Figure Q71(b)

                                                                                                      Final void ratio

                                                                                                      e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                                      e

                                                                                                      Hfrac14 1thorn e0

                                                                                                      H0frac14 1thorn e1 thorne

                                                                                                      H0

                                                                                                      ie

                                                                                                      e

                                                                                                      180frac14 1631thorne

                                                                                                      1710

                                                                                                      e frac14 2936

                                                                                                      1530frac14 0192

                                                                                                      Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                                      mv frac14 1

                                                                                                      1thorn e0 e0 e101 00

                                                                                                      frac14 1

                                                                                                      1823 0192

                                                                                                      0107frac14 098m2=MN

                                                                                                      k frac14 cvmvw frac14 265 098 98

                                                                                                      60 1440 365 103frac14 81 1010 m=s

                                                                                                      72

                                                                                                      Using Equation 77 (one-dimensional method)

                                                                                                      sc frac14 e0 e11thorn e0 H

                                                                                                      Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                                      Figure Q72

                                                                                                      52 Consolidation theory

                                                                                                      Settlement

                                                                                                      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                                      1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                                      318

                                                                                                      Notes 5 92y 460thorn 84

                                                                                                      Heave

                                                                                                      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                                      1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                                      38

                                                                                                      73

                                                                                                      U frac14 f ethTvTHORN frac14 f cvt

                                                                                                      d2

                                                                                                      Hence if cv is constant

                                                                                                      t1

                                                                                                      t2frac14 d

                                                                                                      21

                                                                                                      d22

                                                                                                      where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                                      d1 frac14 95mm and d2 frac14 2500mm

                                                                                                      for U frac14 050 t2 frac14 t1 d22

                                                                                                      d21

                                                                                                      frac14 20

                                                                                                      60 24 365 25002

                                                                                                      952frac14 263 years

                                                                                                      for U lt 060 Tv frac14

                                                                                                      4U2 (Equation 724(a))

                                                                                                      t030 frac14 t050 0302

                                                                                                      0502

                                                                                                      frac14 263 036 frac14 095 years

                                                                                                      Consolidation theory 53

                                                                                                      74

                                                                                                      The layer is open

                                                                                                      d frac14 8

                                                                                                      2frac14 4m

                                                                                                      Tv frac14 cvtd2frac14 24 3

                                                                                                      42frac14 0450

                                                                                                      ui frac14 frac14 84 kN=m2

                                                                                                      The excess pore water pressure is given by Equation 721

                                                                                                      ue frac14Xmfrac141mfrac140

                                                                                                      2ui

                                                                                                      Msin

                                                                                                      Mz

                                                                                                      d

                                                                                                      expethM2TvTHORN

                                                                                                      In this case z frac14 d

                                                                                                      sinMz

                                                                                                      d

                                                                                                      frac14 sinM

                                                                                                      where

                                                                                                      M frac14

                                                                                                      23

                                                                                                      25

                                                                                                      2

                                                                                                      M sin M M2Tv exp (M2Tv)

                                                                                                      2thorn1 1110 0329

                                                                                                      3

                                                                                                      21 9993 457 105

                                                                                                      ue frac14 2 84 2

                                                                                                      1 0329 ethother terms negligibleTHORN

                                                                                                      frac14 352 kN=m2

                                                                                                      75

                                                                                                      The layer is open

                                                                                                      d frac14 6

                                                                                                      2frac14 3m

                                                                                                      Tv frac14 cvtd2frac14 10 3

                                                                                                      32frac14 0333

                                                                                                      The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                                      54 Consolidation theory

                                                                                                      For an open layer

                                                                                                      Tv frac14 4n

                                                                                                      m2

                                                                                                      n frac14 0333 62

                                                                                                      4frac14 300

                                                                                                      The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                                      ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                                      i j

                                                                                                      0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                                      0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                                      The initial and 3-year isochrones are plotted in Figure Q75

                                                                                                      Area under initial isochrone frac14 180 units

                                                                                                      Area under 3-year isochrone frac14 63 units

                                                                                                      The average degree of consolidation is given by Equation 725Thus

                                                                                                      U frac14 1 63

                                                                                                      180frac14 065

                                                                                                      Figure Q75

                                                                                                      Consolidation theory 55

                                                                                                      76

                                                                                                      At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                      0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                      The final consolidation settlement (one-dimensional method) is

                                                                                                      sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                      Corrected time t frac14 2 1

                                                                                                      2

                                                                                                      40

                                                                                                      52

                                                                                                      frac14 1615 years

                                                                                                      Tv frac14 cvtd2frac14 44 1615

                                                                                                      42frac14 0444

                                                                                                      From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                      77

                                                                                                      The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                      Figure Q77

                                                                                                      56 Consolidation theory

                                                                                                      Point m n Ir (kNm2) sc (mm)

                                                                                                      13020frac14 15 20

                                                                                                      20frac14 10 0194 (4) 113 124

                                                                                                      260

                                                                                                      20frac14 30

                                                                                                      20

                                                                                                      20frac14 10 0204 (2) 59 65

                                                                                                      360

                                                                                                      20frac14 30

                                                                                                      40

                                                                                                      20frac14 20 0238 (1) 35 38

                                                                                                      430

                                                                                                      20frac14 15

                                                                                                      40

                                                                                                      20frac14 20 0224 (2) 65 72

                                                                                                      Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                      78

                                                                                                      Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                      (a) Immediate settlement

                                                                                                      H

                                                                                                      Bfrac14 30

                                                                                                      35frac14 086

                                                                                                      D

                                                                                                      Bfrac14 2

                                                                                                      35frac14 006

                                                                                                      Figure Q78

                                                                                                      Consolidation theory 57

                                                                                                      From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                      si frac14 130131qB

                                                                                                      Eufrac14 10 032 105 35

                                                                                                      40frac14 30mm

                                                                                                      (b) Consolidation settlement

                                                                                                      Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                      1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                      3150

                                                                                                      Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                      Now

                                                                                                      H

                                                                                                      Bfrac14 30

                                                                                                      35frac14 086 and A frac14 065

                                                                                                      from Figure 712 13 frac14 079

                                                                                                      sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                      Total settlement

                                                                                                      s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                      79

                                                                                                      Without sand drains

                                                                                                      Uv frac14 025

                                                                                                      Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                      t frac14 Tvd2

                                                                                                      cvfrac14 0049 82

                                                                                                      cvWith sand drains

                                                                                                      R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                      n frac14 Rrfrac14 169

                                                                                                      015frac14 113

                                                                                                      Tr frac14 cht

                                                                                                      4R2frac14 ch

                                                                                                      4 1692 0049 82

                                                                                                      cvethand ch frac14 cvTHORN

                                                                                                      frac14 0275

                                                                                                      Ur frac14 073 (from Figure 730)

                                                                                                      58 Consolidation theory

                                                                                                      Using Equation 740

                                                                                                      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                      U frac14 080

                                                                                                      710

                                                                                                      Without sand drains

                                                                                                      Uv frac14 090

                                                                                                      Tv frac14 0848

                                                                                                      t frac14 Tvd2

                                                                                                      cvfrac14 0848 102

                                                                                                      96frac14 88 years

                                                                                                      With sand drains

                                                                                                      R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                      n frac14 Rrfrac14 226

                                                                                                      015frac14 15

                                                                                                      Tr

                                                                                                      Tvfrac14 chcv

                                                                                                      d2

                                                                                                      4R2ethsame tTHORN

                                                                                                      Tr

                                                                                                      Tvfrac14 140

                                                                                                      96 102

                                                                                                      4 2262frac14 714 eth1THORN

                                                                                                      Using Equation 740

                                                                                                      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                      An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                      Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                      040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                      Thus

                                                                                                      Uv frac14 0295 and Ur frac14 086

                                                                                                      t frac14 88 00683

                                                                                                      0848frac14 07 years

                                                                                                      Consolidation theory 59

                                                                                                      Chapter 8

                                                                                                      Bearing capacity

                                                                                                      81

                                                                                                      (a) The ultimate bearing capacity is given by Equation 83

                                                                                                      qf frac14 cNc thorn DNq thorn 1

                                                                                                      2BN

                                                                                                      For u frac14 0

                                                                                                      Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                      qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                      The net ultimate bearing capacity is

                                                                                                      qnf frac14 qf D frac14 540 kN=m2

                                                                                                      The net foundation pressure is

                                                                                                      qn frac14 q D frac14 425

                                                                                                      2 eth21 1THORN frac14 192 kN=m2

                                                                                                      The factor of safety (Equation 86) is

                                                                                                      F frac14 qnfqnfrac14 540

                                                                                                      192frac14 28

                                                                                                      (b) For 0 frac14 28

                                                                                                      Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                      qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                      2 112 2 13

                                                                                                      frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                      qnf frac14 574 112 frac14 563 kN=m2

                                                                                                      F frac14 563

                                                                                                      192frac14 29

                                                                                                      (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                      82

                                                                                                      For 0 frac14 38

                                                                                                      Nq frac14 49 N frac14 67

                                                                                                      qnf frac14 DethNq 1THORN thorn 1

                                                                                                      2BN ethfrom Equation 83THORN

                                                                                                      frac14 eth18 075 48THORN thorn 1

                                                                                                      2 18 15 67

                                                                                                      frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                      qn frac14 500

                                                                                                      15 eth18 075THORN frac14 320 kN=m2

                                                                                                      F frac14 qnfqnfrac14 1553

                                                                                                      320frac14 48

                                                                                                      0d frac14 tan1tan 38

                                                                                                      125

                                                                                                      frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                      Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                      2 18 15 25

                                                                                                      frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                      Design load (action) Vd frac14 500 kN=m

                                                                                                      The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                      83

                                                                                                      D

                                                                                                      Bfrac14 350

                                                                                                      225frac14 155

                                                                                                      From Figure 85 for a square foundation

                                                                                                      Nc frac14 81

                                                                                                      Bearing capacity 61

                                                                                                      For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                      Nc frac14 084thorn 016B

                                                                                                      L

                                                                                                      81 frac14 745

                                                                                                      Using Equation 810

                                                                                                      qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                      For F frac14 3

                                                                                                      qn frac14 1006

                                                                                                      3frac14 335 kN=m2

                                                                                                      q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                      Design load frac14 405 450 225 frac14 4100 kN

                                                                                                      Design undrained strength cud frac14 135

                                                                                                      14frac14 96 kN=m2

                                                                                                      Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                      frac14 7241 kN

                                                                                                      Design load Vd frac14 4100 kN

                                                                                                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                      84

                                                                                                      For 0 frac14 40

                                                                                                      Nq frac14 64 N frac14 95

                                                                                                      qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                      (a) Water table 5m below ground level

                                                                                                      qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                      qn frac14 400 17 frac14 383 kN=m2

                                                                                                      F frac14 2686

                                                                                                      383frac14 70

                                                                                                      (b) Water table 1m below ground level (ie at foundation level)

                                                                                                      0 frac14 20 98 frac14 102 kN=m3

                                                                                                      62 Bearing capacity

                                                                                                      qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                      F frac14 2040

                                                                                                      383frac14 53

                                                                                                      (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                      eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                      qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                      F frac14 1296

                                                                                                      392frac14 33

                                                                                                      85

                                                                                                      The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                      Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                      Design value of 0 frac14 tan1tan 39

                                                                                                      125

                                                                                                      frac14 33

                                                                                                      For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                      Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                      86

                                                                                                      (a) Undrained shear for u frac14 0

                                                                                                      Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                      qnf frac14 12cuNc

                                                                                                      frac14 12 100 514 frac14 617 kN=m2

                                                                                                      qn frac14 qnfFfrac14 617

                                                                                                      3frac14 206 kN=m2

                                                                                                      q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                      Bearing capacity 63

                                                                                                      Drained shear for 0 frac14 32

                                                                                                      Nq frac14 23 N frac14 25

                                                                                                      0 frac14 21 98 frac14 112 kN=m3

                                                                                                      qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                      frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                      frac14 694 kN=m2

                                                                                                      q frac14 694

                                                                                                      3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                      Design load frac14 42 227 frac14 3632 kN

                                                                                                      (b) Design undrained strength cud frac14 100

                                                                                                      14frac14 71 kNm2

                                                                                                      Design bearing resistance Rd frac14 12cudNe area

                                                                                                      frac14 12 71 514 42

                                                                                                      frac14 7007 kN

                                                                                                      For drained shear 0d frac14 tan1tan 32

                                                                                                      125

                                                                                                      frac14 26

                                                                                                      Nq frac14 12 N frac14 10

                                                                                                      Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                      (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                      Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                      1 2 100 0175 0700qn 0182qn

                                                                                                      2 6 033 0044 0176qn 0046qn

                                                                                                      3 10 020 0017 0068qn 0018qn

                                                                                                      0246qn

                                                                                                      Diameter of equivalent circle B frac14 45m

                                                                                                      H

                                                                                                      Bfrac14 12

                                                                                                      45frac14 27 and A frac14 042

                                                                                                      13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                      64 Bearing capacity

                                                                                                      For sc frac14 30mm

                                                                                                      qn frac14 30

                                                                                                      0147frac14 204 kN=m2

                                                                                                      q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                      Design load frac14 42 225 frac14 3600 kN

                                                                                                      The design load is 3600 kN settlement being the limiting criterion

                                                                                                      87

                                                                                                      D

                                                                                                      Bfrac14 8

                                                                                                      4frac14 20

                                                                                                      From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                      F frac14 cuNc

                                                                                                      Dfrac14 40 71

                                                                                                      20 8frac14 18

                                                                                                      88

                                                                                                      Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                      Design value of 0 frac14 tan1tan 38

                                                                                                      125

                                                                                                      frac14 32

                                                                                                      Figure Q86

                                                                                                      Bearing capacity 65

                                                                                                      For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                      Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                      The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                      Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                      For B frac14 250m qn frac14 3750

                                                                                                      2502 17 frac14 583 kN=m2

                                                                                                      From Figure 510 m frac14 n frac14 126

                                                                                                      6frac14 021

                                                                                                      Ir frac14 0019

                                                                                                      Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                      Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                      The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                      89

                                                                                                      Depth (m) N 0v (kNm2) CN N1

                                                                                                      070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                      Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                      (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                      Cw frac14 05thorn 0530

                                                                                                      47

                                                                                                      frac14 082

                                                                                                      66 Bearing capacity

                                                                                                      Thus

                                                                                                      qa frac14 150 082 frac14 120 kN=m2

                                                                                                      (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                      Thus

                                                                                                      qa frac14 90 15 frac14 135 kN=m2

                                                                                                      (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                      Ic frac14 171

                                                                                                      1014frac14 0068

                                                                                                      From Equation 819(a) with s frac14 25mm

                                                                                                      q frac14 25

                                                                                                      3507 0068frac14 150 kN=m2

                                                                                                      810

                                                                                                      Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                      Izp frac14 05thorn 01130

                                                                                                      16 27

                                                                                                      05

                                                                                                      frac14 067

                                                                                                      Refer to Figure Q810

                                                                                                      E frac14 25qc

                                                                                                      Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                      Ez (mm3MN)

                                                                                                      1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                      0203

                                                                                                      C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                      130frac14 093

                                                                                                      C2 frac14 1 ethsayTHORN

                                                                                                      s frac14 C1C2qnX Iz

                                                                                                      Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                      Bearing capacity 67

                                                                                                      811

                                                                                                      At pile base level

                                                                                                      cu frac14 220 kN=m2

                                                                                                      qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                      Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                      00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                      qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                      Then

                                                                                                      Qf frac14 Abqb thorn Asqs

                                                                                                      frac14

                                                                                                      4 32 1980

                                                                                                      thorn eth 105 139 86THORN

                                                                                                      frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                      0 01 02 03 04 05 06 07

                                                                                                      0 2 4 6 8 10 12 14

                                                                                                      1

                                                                                                      2

                                                                                                      3

                                                                                                      4

                                                                                                      5

                                                                                                      6

                                                                                                      7

                                                                                                      8

                                                                                                      (1)

                                                                                                      (2)

                                                                                                      (3)

                                                                                                      (4)

                                                                                                      (5)

                                                                                                      qc

                                                                                                      qc

                                                                                                      Iz

                                                                                                      Iz

                                                                                                      (MNm2)

                                                                                                      z (m)

                                                                                                      Figure Q810

                                                                                                      68 Bearing capacity

                                                                                                      Allowable load

                                                                                                      ethaTHORN Qf

                                                                                                      2frac14 17 937

                                                                                                      2frac14 8968 kN

                                                                                                      ethbTHORN Abqb

                                                                                                      3thorn Asqs frac14 13 996

                                                                                                      3thorn 3941 frac14 8606 kN

                                                                                                      ie allowable load frac14 8600 kN

                                                                                                      Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                      According to the limit state method

                                                                                                      Characteristic undrained strength at base level cuk frac14 220

                                                                                                      150kN=m2

                                                                                                      Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                      150frac14 1320 kN=m2

                                                                                                      Characteristic shaft resistance qsk frac14 00150

                                                                                                      frac14 86

                                                                                                      150frac14 57 kN=m2

                                                                                                      Characteristic base and shaft resistances

                                                                                                      Rbk frac14

                                                                                                      4 32 1320 frac14 9330 kN

                                                                                                      Rsk frac14 105 139 86

                                                                                                      150frac14 2629 kN

                                                                                                      For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                      Design bearing resistance Rcd frac14 9330

                                                                                                      160thorn 2629

                                                                                                      130

                                                                                                      frac14 5831thorn 2022

                                                                                                      frac14 7850 kN

                                                                                                      Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                      812

                                                                                                      ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                      qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                      For a single pile

                                                                                                      Qf frac14 Abqb thorn Asqs

                                                                                                      frac14

                                                                                                      4 062 1305

                                                                                                      thorn eth 06 15 42THORN

                                                                                                      frac14 369thorn 1187 frac14 1556 kN

                                                                                                      Bearing capacity 69

                                                                                                      Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                      eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                      (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                      (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                      qbkfrac14 9cuk frac14 9 220

                                                                                                      150frac14 1320 kN=m2

                                                                                                      qskfrac14cuk frac14 040 105

                                                                                                      150frac14 28 kN=m2

                                                                                                      Rbkfrac14

                                                                                                      4 0602 1320 frac14 373 kN

                                                                                                      Rskfrac14 060 15 28 frac14 791 kN

                                                                                                      Rcdfrac14 373

                                                                                                      160thorn 791

                                                                                                      130frac14 233thorn 608 frac14 841 kN

                                                                                                      Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                      Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                      (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                      q frac14 21 000

                                                                                                      1762frac14 68 kN=m2

                                                                                                      Immediate settlement

                                                                                                      H

                                                                                                      Bfrac14 15

                                                                                                      176frac14 085

                                                                                                      D

                                                                                                      Bfrac14 13

                                                                                                      176frac14 074

                                                                                                      L

                                                                                                      Bfrac14 1

                                                                                                      Hence from Figure 515

                                                                                                      130 frac14 078 and 131 frac14 041

                                                                                                      70 Bearing capacity

                                                                                                      Thus using Equation 528

                                                                                                      si frac14 078 041 68 176

                                                                                                      65frac14 6mm

                                                                                                      Consolidation settlement

                                                                                                      Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                      1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                      434 (sod)

                                                                                                      Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                      sc frac14 056 434 frac14 24mm

                                                                                                      The total settlement is (6thorn 24) frac14 30mm

                                                                                                      813

                                                                                                      At base level N frac14 26 Then using Equation 830

                                                                                                      qb frac14 40NDb

                                                                                                      Bfrac14 40 26 2

                                                                                                      025frac14 8320 kN=m2

                                                                                                      ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                      Figure Q812

                                                                                                      Bearing capacity 71

                                                                                                      Over the length embedded in sand

                                                                                                      N frac14 21 ie18thorn 24

                                                                                                      2

                                                                                                      Using Equation 831

                                                                                                      qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                      For a single pile

                                                                                                      Qf frac14 Abqb thorn Asqs

                                                                                                      frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                      For the pile group assuming a group efficiency of 12

                                                                                                      XQf frac14 12 9 604 frac14 6523 kN

                                                                                                      Then the load factor is

                                                                                                      F frac14 6523

                                                                                                      2000thorn 1000frac14 21

                                                                                                      (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                      Characteristic base resistance per unit area qbk frac14 8320

                                                                                                      150frac14 5547 kNm2

                                                                                                      Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                      150frac14 28 kNm2

                                                                                                      Characteristic base and shaft resistances for a single pile

                                                                                                      Rbk frac14 0252 5547 frac14 347 kN

                                                                                                      Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                      For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                      Design bearing resistance Rcd frac14 347

                                                                                                      130thorn 56

                                                                                                      130frac14 310 kN

                                                                                                      For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                      Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                      (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                      From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                      72 Bearing capacity

                                                                                                      N frac14 24thorn 26thorn 34

                                                                                                      3frac14 28

                                                                                                      Ic frac14 171

                                                                                                      2814frac14 0016 ethEquation 818THORN

                                                                                                      s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                      814

                                                                                                      Using Equation 841

                                                                                                      Tf frac14 DLcu thorn

                                                                                                      4ethD2 d2THORNcuNc

                                                                                                      frac14 eth 02 5 06 110THORN thorn

                                                                                                      4eth022 012THORN110 9

                                                                                                      frac14 207thorn 23 frac14 230 kN

                                                                                                      Figure Q813

                                                                                                      Bearing capacity 73

                                                                                                      Chapter 9

                                                                                                      Stability of slopes

                                                                                                      91

                                                                                                      Referring to Figure Q91

                                                                                                      W frac14 417 19 frac14 792 kN=m

                                                                                                      Q frac14 20 28 frac14 56 kN=m

                                                                                                      Arc lengthAB frac14

                                                                                                      180 73 90 frac14 115m

                                                                                                      Arc length BC frac14

                                                                                                      180 28 90 frac14 44m

                                                                                                      The factor of safety is given by

                                                                                                      F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                      90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                      Depth of tension crack z0 frac14 2cu

                                                                                                      frac14 2 20

                                                                                                      19frac14 21m

                                                                                                      Arc length BD frac14

                                                                                                      180 13

                                                                                                      1

                                                                                                      2 90 frac14 21m

                                                                                                      F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                      The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                      Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                      14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                      Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                      The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                      92

                                                                                                      u frac14 0

                                                                                                      Depth factor D frac14 11

                                                                                                      9frac14 122

                                                                                                      Using Equation 92 with F frac14 10

                                                                                                      Ns frac14 cu

                                                                                                      FHfrac14 30

                                                                                                      10 19 9frac14 0175

                                                                                                      Hence from Figure 93

                                                                                                      frac14 50

                                                                                                      For F frac14 12

                                                                                                      Ns frac14 30

                                                                                                      12 19 9frac14 0146

                                                                                                      frac14 27

                                                                                                      93

                                                                                                      Refer to Figure Q93

                                                                                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                      1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                      74 m

                                                                                                      214 1deg

                                                                                                      213 1deg

                                                                                                      39 m

                                                                                                      WB

                                                                                                      D

                                                                                                      C

                                                                                                      28 m

                                                                                                      21 m

                                                                                                      A

                                                                                                      Q

                                                                                                      Soil (1)Soil (2)

                                                                                                      73deg

                                                                                                      Figure Q91

                                                                                                      Stability of slopes 75

                                                                                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                      9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                      599 256 328 1372

                                                                                                      Figure Q93

                                                                                                      76 Stability of slopes

                                                                                                      XW cos frac14 b

                                                                                                      Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                      W sin frac14 bX

                                                                                                      h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                      Arc length La frac14

                                                                                                      180 57

                                                                                                      1

                                                                                                      2 326 frac14 327m

                                                                                                      The factor of safety is given by

                                                                                                      F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                      W sin

                                                                                                      frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                      frac14 091

                                                                                                      According to the limit state method

                                                                                                      0d frac14 tan1tan 32

                                                                                                      125

                                                                                                      frac14 265

                                                                                                      c0 frac14 8

                                                                                                      160frac14 5 kN=m2

                                                                                                      Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                      Design disturbing moment frac14 1075 kN=m

                                                                                                      The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                      94

                                                                                                      F frac14 1

                                                                                                      W sin

                                                                                                      Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                      1thorn ethtan tan0=FTHORN

                                                                                                      c0 frac14 8 kN=m2

                                                                                                      0 frac14 32

                                                                                                      c0b frac14 8 2 frac14 16 kN=m

                                                                                                      W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                      Try F frac14 100

                                                                                                      tan0

                                                                                                      Ffrac14 0625

                                                                                                      Stability of slopes 77

                                                                                                      Values of u are as obtained in Figure Q93

                                                                                                      SliceNo

                                                                                                      h(m)

                                                                                                      W frac14 bh(kNm)

                                                                                                      W sin(kNm)

                                                                                                      ub(kNm)

                                                                                                      c0bthorn (W ub) tan0(kNm)

                                                                                                      sec

                                                                                                      1thorn (tan tan0)FProduct(kNm)

                                                                                                      1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                      23 33 30 1042 31

                                                                                                      3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                      224 92 72 0931 67

                                                                                                      6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                      12 58 112 90 0889 80

                                                                                                      8 60 252 1812

                                                                                                      80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                      10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                      2154 88 116 0853 99

                                                                                                      14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                      1074 1091

                                                                                                      F frac14 1091

                                                                                                      1074frac14 102 (assumed value 100)

                                                                                                      Thus

                                                                                                      F frac14 101

                                                                                                      95

                                                                                                      F frac14 1

                                                                                                      W sin

                                                                                                      XfWeth1 ruTHORN tan0g sec

                                                                                                      1thorn ethtan tan0THORN=F

                                                                                                      0 frac14 33

                                                                                                      ru frac14 020

                                                                                                      W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                      eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                      Try F frac14 110

                                                                                                      tan 0

                                                                                                      Ffrac14 tan 33

                                                                                                      110frac14 0590

                                                                                                      78 Stability of slopes

                                                                                                      Referring to Figure Q95

                                                                                                      SliceNo

                                                                                                      h(m)

                                                                                                      W frac14 bh(kNm)

                                                                                                      W sin(kNm)

                                                                                                      W(1 ru) tan0(kNm)

                                                                                                      sec

                                                                                                      1thorn ( tan tan0)FProduct(kNm)

                                                                                                      1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                      2120 234 0892 209

                                                                                                      4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                      1185 1271

                                                                                                      Figure Q95

                                                                                                      Stability of slopes 79

                                                                                                      F frac14 1271

                                                                                                      1185frac14 107

                                                                                                      The trial value was 110 therefore take F to be 108

                                                                                                      96

                                                                                                      (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                      F frac14 0

                                                                                                      sat

                                                                                                      tan0

                                                                                                      tan

                                                                                                      ptie 15 frac14 92

                                                                                                      19

                                                                                                      tan 36

                                                                                                      tan

                                                                                                      tan frac14 0234

                                                                                                      frac14 13

                                                                                                      Water table well below surface the factor of safety is given by Equation 911

                                                                                                      F frac14 tan0

                                                                                                      tan

                                                                                                      frac14 tan 36

                                                                                                      tan 13

                                                                                                      frac14 31

                                                                                                      (b) 0d frac14 tan1tan 36

                                                                                                      125

                                                                                                      frac14 30

                                                                                                      Depth of potential failure surface frac14 z

                                                                                                      Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                      frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                      frac14 504z kN

                                                                                                      Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                      frac14 19 z sin 13 cos 13

                                                                                                      frac14 416z kN

                                                                                                      Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                      80 Stability of slopes

                                                                                                      • Book Cover
                                                                                                      • Title
                                                                                                      • Contents
                                                                                                      • Basic characteristics of soils
                                                                                                      • Seepage
                                                                                                      • Effective stress
                                                                                                      • Shear strength
                                                                                                      • Stresses and displacements
                                                                                                      • Lateral earth pressure
                                                                                                      • Consolidation theory
                                                                                                      • Bearing capacity
                                                                                                      • Stability of slopes

                                                                                                        610

                                                                                                        For 0 frac14 40 Ka frac14 022The pressure distribution is shown in Figure Q610

                                                                                                        p frac14 065KaH frac14 065 022 19 9 frac14 245 kN=m2

                                                                                                        Strut load frac14 245 15 3 frac14 110 kN (a load factor of at least 20 would be applied tothis value)

                                                                                                        Using the recommendations of Twine and Roscoe

                                                                                                        p frac14 02H frac14 02 19 9 frac14 342 kN=m2

                                                                                                        Strut load frac14 342 15 3 frac14 154 kN (this value would be multiplied by a partialfactor of 135)

                                                                                                        611

                                                                                                        frac14 18 kN=m3 0 frac14 34

                                                                                                        H frac14 350m nH frac14 335m mH frac14 185m

                                                                                                        Consider a trial value of F frac14 20 Refer to Figure 635

                                                                                                        0m frac14 tan1tan 34

                                                                                                        20

                                                                                                        frac14 186

                                                                                                        Then

                                                                                                        frac14 45 thorn 0m2frac14 543

                                                                                                        W frac14 1

                                                                                                        2 18 3502 cot 543 frac14 792 kN=m

                                                                                                        Figure Q610

                                                                                                        46 Lateral earth pressure

                                                                                                        P frac14 1

                                                                                                        2 s 3352 frac14 561s kN=m

                                                                                                        U frac14 1

                                                                                                        2 98 1852 cosec 543 frac14 206 kN=m

                                                                                                        Equations 630 and 631 then become

                                                                                                        561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                                                        792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                                                        ie

                                                                                                        561s 0616N 405 frac14 0

                                                                                                        792 0857N thorn 563 frac14 0

                                                                                                        N frac14 848

                                                                                                        0857frac14 989 kN=m

                                                                                                        Then

                                                                                                        561s 609 405 frac14 0

                                                                                                        s frac14 649

                                                                                                        561frac14 116 kN=m3

                                                                                                        The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                                                        F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                                                        20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                                                        s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                                                        Figure Q611

                                                                                                        Lateral earth pressure 47

                                                                                                        612

                                                                                                        For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                                        45 thorn 0

                                                                                                        2frac14 63

                                                                                                        For the retained material between the surface and a depth of 36m

                                                                                                        Pa frac14 1

                                                                                                        2 030 18 362 frac14 350 kN=m

                                                                                                        Weight of reinforced fill between the surface and a depth of 36m is

                                                                                                        Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                                        eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                                        Eccentricity of Rv

                                                                                                        e frac14 263 250 frac14 013m

                                                                                                        The average vertical stress at a depth of 36m is

                                                                                                        z frac14 Rv

                                                                                                        L 2efrac14 324

                                                                                                        474frac14 68 kN=m2

                                                                                                        (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                                        Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                                        Tensile stress in the element frac14 138 103

                                                                                                        65 3frac14 71N=mm2

                                                                                                        Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                                        Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                                        Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                                        The weight of ABC is

                                                                                                        W frac14 1

                                                                                                        2 18 52 265 frac14 124 kN=m

                                                                                                        From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                                        48 Lateral earth pressure

                                                                                                        (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                                        Tp frac14 032 68 120 065 frac14 170 kN

                                                                                                        Tr frac14 213 420

                                                                                                        418frac14 214 kN

                                                                                                        Again the tensile failure and slipping limit states are satisfied for this element

                                                                                                        Figure Q612

                                                                                                        Lateral earth pressure 49

                                                                                                        Chapter 7

                                                                                                        Consolidation theory

                                                                                                        71

                                                                                                        Total change in thickness

                                                                                                        H frac14 782 602 frac14 180mm

                                                                                                        Average thickness frac14 1530thorn 180

                                                                                                        2frac14 1620mm

                                                                                                        Length of drainage path d frac14 1620

                                                                                                        2frac14 810mm

                                                                                                        Root time plot (Figure Q71a)

                                                                                                        ffiffiffiffiffiffit90p frac14 33

                                                                                                        t90 frac14 109min

                                                                                                        cv frac14 0848d2

                                                                                                        t90frac14 0848 8102

                                                                                                        109 1440 365

                                                                                                        106frac14 27m2=year

                                                                                                        r0 frac14 782 764

                                                                                                        782 602frac14 018

                                                                                                        180frac14 0100

                                                                                                        rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                                        10 119

                                                                                                        9 180frac14 0735

                                                                                                        rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                                        Log time plot (Figure Q71b)

                                                                                                        t50 frac14 26min

                                                                                                        cv frac14 0196d2

                                                                                                        t50frac14 0196 8102

                                                                                                        26 1440 365

                                                                                                        106frac14 26m2=year

                                                                                                        r0 frac14 782 763

                                                                                                        782 602frac14 019

                                                                                                        180frac14 0106

                                                                                                        rp frac14 763 623

                                                                                                        782 602frac14 140

                                                                                                        180frac14 0778

                                                                                                        rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                                        Figure Q71(a)

                                                                                                        Figure Q71(b)

                                                                                                        Final void ratio

                                                                                                        e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                                        e

                                                                                                        Hfrac14 1thorn e0

                                                                                                        H0frac14 1thorn e1 thorne

                                                                                                        H0

                                                                                                        ie

                                                                                                        e

                                                                                                        180frac14 1631thorne

                                                                                                        1710

                                                                                                        e frac14 2936

                                                                                                        1530frac14 0192

                                                                                                        Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                                        mv frac14 1

                                                                                                        1thorn e0 e0 e101 00

                                                                                                        frac14 1

                                                                                                        1823 0192

                                                                                                        0107frac14 098m2=MN

                                                                                                        k frac14 cvmvw frac14 265 098 98

                                                                                                        60 1440 365 103frac14 81 1010 m=s

                                                                                                        72

                                                                                                        Using Equation 77 (one-dimensional method)

                                                                                                        sc frac14 e0 e11thorn e0 H

                                                                                                        Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                                        Figure Q72

                                                                                                        52 Consolidation theory

                                                                                                        Settlement

                                                                                                        Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                                        1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                                        318

                                                                                                        Notes 5 92y 460thorn 84

                                                                                                        Heave

                                                                                                        Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                                        1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                                        38

                                                                                                        73

                                                                                                        U frac14 f ethTvTHORN frac14 f cvt

                                                                                                        d2

                                                                                                        Hence if cv is constant

                                                                                                        t1

                                                                                                        t2frac14 d

                                                                                                        21

                                                                                                        d22

                                                                                                        where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                                        d1 frac14 95mm and d2 frac14 2500mm

                                                                                                        for U frac14 050 t2 frac14 t1 d22

                                                                                                        d21

                                                                                                        frac14 20

                                                                                                        60 24 365 25002

                                                                                                        952frac14 263 years

                                                                                                        for U lt 060 Tv frac14

                                                                                                        4U2 (Equation 724(a))

                                                                                                        t030 frac14 t050 0302

                                                                                                        0502

                                                                                                        frac14 263 036 frac14 095 years

                                                                                                        Consolidation theory 53

                                                                                                        74

                                                                                                        The layer is open

                                                                                                        d frac14 8

                                                                                                        2frac14 4m

                                                                                                        Tv frac14 cvtd2frac14 24 3

                                                                                                        42frac14 0450

                                                                                                        ui frac14 frac14 84 kN=m2

                                                                                                        The excess pore water pressure is given by Equation 721

                                                                                                        ue frac14Xmfrac141mfrac140

                                                                                                        2ui

                                                                                                        Msin

                                                                                                        Mz

                                                                                                        d

                                                                                                        expethM2TvTHORN

                                                                                                        In this case z frac14 d

                                                                                                        sinMz

                                                                                                        d

                                                                                                        frac14 sinM

                                                                                                        where

                                                                                                        M frac14

                                                                                                        23

                                                                                                        25

                                                                                                        2

                                                                                                        M sin M M2Tv exp (M2Tv)

                                                                                                        2thorn1 1110 0329

                                                                                                        3

                                                                                                        21 9993 457 105

                                                                                                        ue frac14 2 84 2

                                                                                                        1 0329 ethother terms negligibleTHORN

                                                                                                        frac14 352 kN=m2

                                                                                                        75

                                                                                                        The layer is open

                                                                                                        d frac14 6

                                                                                                        2frac14 3m

                                                                                                        Tv frac14 cvtd2frac14 10 3

                                                                                                        32frac14 0333

                                                                                                        The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                                        54 Consolidation theory

                                                                                                        For an open layer

                                                                                                        Tv frac14 4n

                                                                                                        m2

                                                                                                        n frac14 0333 62

                                                                                                        4frac14 300

                                                                                                        The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                                        ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                                        i j

                                                                                                        0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                                        0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                                        The initial and 3-year isochrones are plotted in Figure Q75

                                                                                                        Area under initial isochrone frac14 180 units

                                                                                                        Area under 3-year isochrone frac14 63 units

                                                                                                        The average degree of consolidation is given by Equation 725Thus

                                                                                                        U frac14 1 63

                                                                                                        180frac14 065

                                                                                                        Figure Q75

                                                                                                        Consolidation theory 55

                                                                                                        76

                                                                                                        At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                        0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                        The final consolidation settlement (one-dimensional method) is

                                                                                                        sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                        Corrected time t frac14 2 1

                                                                                                        2

                                                                                                        40

                                                                                                        52

                                                                                                        frac14 1615 years

                                                                                                        Tv frac14 cvtd2frac14 44 1615

                                                                                                        42frac14 0444

                                                                                                        From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                        77

                                                                                                        The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                        Figure Q77

                                                                                                        56 Consolidation theory

                                                                                                        Point m n Ir (kNm2) sc (mm)

                                                                                                        13020frac14 15 20

                                                                                                        20frac14 10 0194 (4) 113 124

                                                                                                        260

                                                                                                        20frac14 30

                                                                                                        20

                                                                                                        20frac14 10 0204 (2) 59 65

                                                                                                        360

                                                                                                        20frac14 30

                                                                                                        40

                                                                                                        20frac14 20 0238 (1) 35 38

                                                                                                        430

                                                                                                        20frac14 15

                                                                                                        40

                                                                                                        20frac14 20 0224 (2) 65 72

                                                                                                        Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                        78

                                                                                                        Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                        (a) Immediate settlement

                                                                                                        H

                                                                                                        Bfrac14 30

                                                                                                        35frac14 086

                                                                                                        D

                                                                                                        Bfrac14 2

                                                                                                        35frac14 006

                                                                                                        Figure Q78

                                                                                                        Consolidation theory 57

                                                                                                        From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                        si frac14 130131qB

                                                                                                        Eufrac14 10 032 105 35

                                                                                                        40frac14 30mm

                                                                                                        (b) Consolidation settlement

                                                                                                        Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                        1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                        3150

                                                                                                        Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                        Now

                                                                                                        H

                                                                                                        Bfrac14 30

                                                                                                        35frac14 086 and A frac14 065

                                                                                                        from Figure 712 13 frac14 079

                                                                                                        sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                        Total settlement

                                                                                                        s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                        79

                                                                                                        Without sand drains

                                                                                                        Uv frac14 025

                                                                                                        Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                        t frac14 Tvd2

                                                                                                        cvfrac14 0049 82

                                                                                                        cvWith sand drains

                                                                                                        R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                        n frac14 Rrfrac14 169

                                                                                                        015frac14 113

                                                                                                        Tr frac14 cht

                                                                                                        4R2frac14 ch

                                                                                                        4 1692 0049 82

                                                                                                        cvethand ch frac14 cvTHORN

                                                                                                        frac14 0275

                                                                                                        Ur frac14 073 (from Figure 730)

                                                                                                        58 Consolidation theory

                                                                                                        Using Equation 740

                                                                                                        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                        U frac14 080

                                                                                                        710

                                                                                                        Without sand drains

                                                                                                        Uv frac14 090

                                                                                                        Tv frac14 0848

                                                                                                        t frac14 Tvd2

                                                                                                        cvfrac14 0848 102

                                                                                                        96frac14 88 years

                                                                                                        With sand drains

                                                                                                        R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                        n frac14 Rrfrac14 226

                                                                                                        015frac14 15

                                                                                                        Tr

                                                                                                        Tvfrac14 chcv

                                                                                                        d2

                                                                                                        4R2ethsame tTHORN

                                                                                                        Tr

                                                                                                        Tvfrac14 140

                                                                                                        96 102

                                                                                                        4 2262frac14 714 eth1THORN

                                                                                                        Using Equation 740

                                                                                                        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                        An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                        Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                        040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                        Thus

                                                                                                        Uv frac14 0295 and Ur frac14 086

                                                                                                        t frac14 88 00683

                                                                                                        0848frac14 07 years

                                                                                                        Consolidation theory 59

                                                                                                        Chapter 8

                                                                                                        Bearing capacity

                                                                                                        81

                                                                                                        (a) The ultimate bearing capacity is given by Equation 83

                                                                                                        qf frac14 cNc thorn DNq thorn 1

                                                                                                        2BN

                                                                                                        For u frac14 0

                                                                                                        Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                        qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                        The net ultimate bearing capacity is

                                                                                                        qnf frac14 qf D frac14 540 kN=m2

                                                                                                        The net foundation pressure is

                                                                                                        qn frac14 q D frac14 425

                                                                                                        2 eth21 1THORN frac14 192 kN=m2

                                                                                                        The factor of safety (Equation 86) is

                                                                                                        F frac14 qnfqnfrac14 540

                                                                                                        192frac14 28

                                                                                                        (b) For 0 frac14 28

                                                                                                        Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                        qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                        2 112 2 13

                                                                                                        frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                        qnf frac14 574 112 frac14 563 kN=m2

                                                                                                        F frac14 563

                                                                                                        192frac14 29

                                                                                                        (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                        82

                                                                                                        For 0 frac14 38

                                                                                                        Nq frac14 49 N frac14 67

                                                                                                        qnf frac14 DethNq 1THORN thorn 1

                                                                                                        2BN ethfrom Equation 83THORN

                                                                                                        frac14 eth18 075 48THORN thorn 1

                                                                                                        2 18 15 67

                                                                                                        frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                        qn frac14 500

                                                                                                        15 eth18 075THORN frac14 320 kN=m2

                                                                                                        F frac14 qnfqnfrac14 1553

                                                                                                        320frac14 48

                                                                                                        0d frac14 tan1tan 38

                                                                                                        125

                                                                                                        frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                        Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                        2 18 15 25

                                                                                                        frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                        Design load (action) Vd frac14 500 kN=m

                                                                                                        The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                        83

                                                                                                        D

                                                                                                        Bfrac14 350

                                                                                                        225frac14 155

                                                                                                        From Figure 85 for a square foundation

                                                                                                        Nc frac14 81

                                                                                                        Bearing capacity 61

                                                                                                        For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                        Nc frac14 084thorn 016B

                                                                                                        L

                                                                                                        81 frac14 745

                                                                                                        Using Equation 810

                                                                                                        qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                        For F frac14 3

                                                                                                        qn frac14 1006

                                                                                                        3frac14 335 kN=m2

                                                                                                        q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                        Design load frac14 405 450 225 frac14 4100 kN

                                                                                                        Design undrained strength cud frac14 135

                                                                                                        14frac14 96 kN=m2

                                                                                                        Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                        frac14 7241 kN

                                                                                                        Design load Vd frac14 4100 kN

                                                                                                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                        84

                                                                                                        For 0 frac14 40

                                                                                                        Nq frac14 64 N frac14 95

                                                                                                        qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                        (a) Water table 5m below ground level

                                                                                                        qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                        qn frac14 400 17 frac14 383 kN=m2

                                                                                                        F frac14 2686

                                                                                                        383frac14 70

                                                                                                        (b) Water table 1m below ground level (ie at foundation level)

                                                                                                        0 frac14 20 98 frac14 102 kN=m3

                                                                                                        62 Bearing capacity

                                                                                                        qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                        F frac14 2040

                                                                                                        383frac14 53

                                                                                                        (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                        eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                        qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                        F frac14 1296

                                                                                                        392frac14 33

                                                                                                        85

                                                                                                        The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                        Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                        Design value of 0 frac14 tan1tan 39

                                                                                                        125

                                                                                                        frac14 33

                                                                                                        For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                        Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                        86

                                                                                                        (a) Undrained shear for u frac14 0

                                                                                                        Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                        qnf frac14 12cuNc

                                                                                                        frac14 12 100 514 frac14 617 kN=m2

                                                                                                        qn frac14 qnfFfrac14 617

                                                                                                        3frac14 206 kN=m2

                                                                                                        q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                        Bearing capacity 63

                                                                                                        Drained shear for 0 frac14 32

                                                                                                        Nq frac14 23 N frac14 25

                                                                                                        0 frac14 21 98 frac14 112 kN=m3

                                                                                                        qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                        frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                        frac14 694 kN=m2

                                                                                                        q frac14 694

                                                                                                        3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                        Design load frac14 42 227 frac14 3632 kN

                                                                                                        (b) Design undrained strength cud frac14 100

                                                                                                        14frac14 71 kNm2

                                                                                                        Design bearing resistance Rd frac14 12cudNe area

                                                                                                        frac14 12 71 514 42

                                                                                                        frac14 7007 kN

                                                                                                        For drained shear 0d frac14 tan1tan 32

                                                                                                        125

                                                                                                        frac14 26

                                                                                                        Nq frac14 12 N frac14 10

                                                                                                        Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                        (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                        Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                        1 2 100 0175 0700qn 0182qn

                                                                                                        2 6 033 0044 0176qn 0046qn

                                                                                                        3 10 020 0017 0068qn 0018qn

                                                                                                        0246qn

                                                                                                        Diameter of equivalent circle B frac14 45m

                                                                                                        H

                                                                                                        Bfrac14 12

                                                                                                        45frac14 27 and A frac14 042

                                                                                                        13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                        64 Bearing capacity

                                                                                                        For sc frac14 30mm

                                                                                                        qn frac14 30

                                                                                                        0147frac14 204 kN=m2

                                                                                                        q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                        Design load frac14 42 225 frac14 3600 kN

                                                                                                        The design load is 3600 kN settlement being the limiting criterion

                                                                                                        87

                                                                                                        D

                                                                                                        Bfrac14 8

                                                                                                        4frac14 20

                                                                                                        From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                        F frac14 cuNc

                                                                                                        Dfrac14 40 71

                                                                                                        20 8frac14 18

                                                                                                        88

                                                                                                        Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                        Design value of 0 frac14 tan1tan 38

                                                                                                        125

                                                                                                        frac14 32

                                                                                                        Figure Q86

                                                                                                        Bearing capacity 65

                                                                                                        For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                        Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                        The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                        Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                        For B frac14 250m qn frac14 3750

                                                                                                        2502 17 frac14 583 kN=m2

                                                                                                        From Figure 510 m frac14 n frac14 126

                                                                                                        6frac14 021

                                                                                                        Ir frac14 0019

                                                                                                        Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                        Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                        The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                        89

                                                                                                        Depth (m) N 0v (kNm2) CN N1

                                                                                                        070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                        Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                        (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                        Cw frac14 05thorn 0530

                                                                                                        47

                                                                                                        frac14 082

                                                                                                        66 Bearing capacity

                                                                                                        Thus

                                                                                                        qa frac14 150 082 frac14 120 kN=m2

                                                                                                        (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                        Thus

                                                                                                        qa frac14 90 15 frac14 135 kN=m2

                                                                                                        (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                        Ic frac14 171

                                                                                                        1014frac14 0068

                                                                                                        From Equation 819(a) with s frac14 25mm

                                                                                                        q frac14 25

                                                                                                        3507 0068frac14 150 kN=m2

                                                                                                        810

                                                                                                        Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                        Izp frac14 05thorn 01130

                                                                                                        16 27

                                                                                                        05

                                                                                                        frac14 067

                                                                                                        Refer to Figure Q810

                                                                                                        E frac14 25qc

                                                                                                        Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                        Ez (mm3MN)

                                                                                                        1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                        0203

                                                                                                        C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                        130frac14 093

                                                                                                        C2 frac14 1 ethsayTHORN

                                                                                                        s frac14 C1C2qnX Iz

                                                                                                        Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                        Bearing capacity 67

                                                                                                        811

                                                                                                        At pile base level

                                                                                                        cu frac14 220 kN=m2

                                                                                                        qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                        Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                        00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                        qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                        Then

                                                                                                        Qf frac14 Abqb thorn Asqs

                                                                                                        frac14

                                                                                                        4 32 1980

                                                                                                        thorn eth 105 139 86THORN

                                                                                                        frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                        0 01 02 03 04 05 06 07

                                                                                                        0 2 4 6 8 10 12 14

                                                                                                        1

                                                                                                        2

                                                                                                        3

                                                                                                        4

                                                                                                        5

                                                                                                        6

                                                                                                        7

                                                                                                        8

                                                                                                        (1)

                                                                                                        (2)

                                                                                                        (3)

                                                                                                        (4)

                                                                                                        (5)

                                                                                                        qc

                                                                                                        qc

                                                                                                        Iz

                                                                                                        Iz

                                                                                                        (MNm2)

                                                                                                        z (m)

                                                                                                        Figure Q810

                                                                                                        68 Bearing capacity

                                                                                                        Allowable load

                                                                                                        ethaTHORN Qf

                                                                                                        2frac14 17 937

                                                                                                        2frac14 8968 kN

                                                                                                        ethbTHORN Abqb

                                                                                                        3thorn Asqs frac14 13 996

                                                                                                        3thorn 3941 frac14 8606 kN

                                                                                                        ie allowable load frac14 8600 kN

                                                                                                        Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                        According to the limit state method

                                                                                                        Characteristic undrained strength at base level cuk frac14 220

                                                                                                        150kN=m2

                                                                                                        Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                        150frac14 1320 kN=m2

                                                                                                        Characteristic shaft resistance qsk frac14 00150

                                                                                                        frac14 86

                                                                                                        150frac14 57 kN=m2

                                                                                                        Characteristic base and shaft resistances

                                                                                                        Rbk frac14

                                                                                                        4 32 1320 frac14 9330 kN

                                                                                                        Rsk frac14 105 139 86

                                                                                                        150frac14 2629 kN

                                                                                                        For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                        Design bearing resistance Rcd frac14 9330

                                                                                                        160thorn 2629

                                                                                                        130

                                                                                                        frac14 5831thorn 2022

                                                                                                        frac14 7850 kN

                                                                                                        Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                        812

                                                                                                        ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                        qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                        For a single pile

                                                                                                        Qf frac14 Abqb thorn Asqs

                                                                                                        frac14

                                                                                                        4 062 1305

                                                                                                        thorn eth 06 15 42THORN

                                                                                                        frac14 369thorn 1187 frac14 1556 kN

                                                                                                        Bearing capacity 69

                                                                                                        Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                        eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                        (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                        (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                        qbkfrac14 9cuk frac14 9 220

                                                                                                        150frac14 1320 kN=m2

                                                                                                        qskfrac14cuk frac14 040 105

                                                                                                        150frac14 28 kN=m2

                                                                                                        Rbkfrac14

                                                                                                        4 0602 1320 frac14 373 kN

                                                                                                        Rskfrac14 060 15 28 frac14 791 kN

                                                                                                        Rcdfrac14 373

                                                                                                        160thorn 791

                                                                                                        130frac14 233thorn 608 frac14 841 kN

                                                                                                        Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                        Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                        (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                        q frac14 21 000

                                                                                                        1762frac14 68 kN=m2

                                                                                                        Immediate settlement

                                                                                                        H

                                                                                                        Bfrac14 15

                                                                                                        176frac14 085

                                                                                                        D

                                                                                                        Bfrac14 13

                                                                                                        176frac14 074

                                                                                                        L

                                                                                                        Bfrac14 1

                                                                                                        Hence from Figure 515

                                                                                                        130 frac14 078 and 131 frac14 041

                                                                                                        70 Bearing capacity

                                                                                                        Thus using Equation 528

                                                                                                        si frac14 078 041 68 176

                                                                                                        65frac14 6mm

                                                                                                        Consolidation settlement

                                                                                                        Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                        1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                        434 (sod)

                                                                                                        Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                        sc frac14 056 434 frac14 24mm

                                                                                                        The total settlement is (6thorn 24) frac14 30mm

                                                                                                        813

                                                                                                        At base level N frac14 26 Then using Equation 830

                                                                                                        qb frac14 40NDb

                                                                                                        Bfrac14 40 26 2

                                                                                                        025frac14 8320 kN=m2

                                                                                                        ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                        Figure Q812

                                                                                                        Bearing capacity 71

                                                                                                        Over the length embedded in sand

                                                                                                        N frac14 21 ie18thorn 24

                                                                                                        2

                                                                                                        Using Equation 831

                                                                                                        qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                        For a single pile

                                                                                                        Qf frac14 Abqb thorn Asqs

                                                                                                        frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                        For the pile group assuming a group efficiency of 12

                                                                                                        XQf frac14 12 9 604 frac14 6523 kN

                                                                                                        Then the load factor is

                                                                                                        F frac14 6523

                                                                                                        2000thorn 1000frac14 21

                                                                                                        (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                        Characteristic base resistance per unit area qbk frac14 8320

                                                                                                        150frac14 5547 kNm2

                                                                                                        Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                        150frac14 28 kNm2

                                                                                                        Characteristic base and shaft resistances for a single pile

                                                                                                        Rbk frac14 0252 5547 frac14 347 kN

                                                                                                        Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                        For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                        Design bearing resistance Rcd frac14 347

                                                                                                        130thorn 56

                                                                                                        130frac14 310 kN

                                                                                                        For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                        Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                        (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                        From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                        72 Bearing capacity

                                                                                                        N frac14 24thorn 26thorn 34

                                                                                                        3frac14 28

                                                                                                        Ic frac14 171

                                                                                                        2814frac14 0016 ethEquation 818THORN

                                                                                                        s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                        814

                                                                                                        Using Equation 841

                                                                                                        Tf frac14 DLcu thorn

                                                                                                        4ethD2 d2THORNcuNc

                                                                                                        frac14 eth 02 5 06 110THORN thorn

                                                                                                        4eth022 012THORN110 9

                                                                                                        frac14 207thorn 23 frac14 230 kN

                                                                                                        Figure Q813

                                                                                                        Bearing capacity 73

                                                                                                        Chapter 9

                                                                                                        Stability of slopes

                                                                                                        91

                                                                                                        Referring to Figure Q91

                                                                                                        W frac14 417 19 frac14 792 kN=m

                                                                                                        Q frac14 20 28 frac14 56 kN=m

                                                                                                        Arc lengthAB frac14

                                                                                                        180 73 90 frac14 115m

                                                                                                        Arc length BC frac14

                                                                                                        180 28 90 frac14 44m

                                                                                                        The factor of safety is given by

                                                                                                        F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                        90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                        Depth of tension crack z0 frac14 2cu

                                                                                                        frac14 2 20

                                                                                                        19frac14 21m

                                                                                                        Arc length BD frac14

                                                                                                        180 13

                                                                                                        1

                                                                                                        2 90 frac14 21m

                                                                                                        F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                        The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                        Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                        14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                        Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                        The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                        92

                                                                                                        u frac14 0

                                                                                                        Depth factor D frac14 11

                                                                                                        9frac14 122

                                                                                                        Using Equation 92 with F frac14 10

                                                                                                        Ns frac14 cu

                                                                                                        FHfrac14 30

                                                                                                        10 19 9frac14 0175

                                                                                                        Hence from Figure 93

                                                                                                        frac14 50

                                                                                                        For F frac14 12

                                                                                                        Ns frac14 30

                                                                                                        12 19 9frac14 0146

                                                                                                        frac14 27

                                                                                                        93

                                                                                                        Refer to Figure Q93

                                                                                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                        1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                        74 m

                                                                                                        214 1deg

                                                                                                        213 1deg

                                                                                                        39 m

                                                                                                        WB

                                                                                                        D

                                                                                                        C

                                                                                                        28 m

                                                                                                        21 m

                                                                                                        A

                                                                                                        Q

                                                                                                        Soil (1)Soil (2)

                                                                                                        73deg

                                                                                                        Figure Q91

                                                                                                        Stability of slopes 75

                                                                                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                        9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                        599 256 328 1372

                                                                                                        Figure Q93

                                                                                                        76 Stability of slopes

                                                                                                        XW cos frac14 b

                                                                                                        Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                        W sin frac14 bX

                                                                                                        h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                        Arc length La frac14

                                                                                                        180 57

                                                                                                        1

                                                                                                        2 326 frac14 327m

                                                                                                        The factor of safety is given by

                                                                                                        F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                        W sin

                                                                                                        frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                        frac14 091

                                                                                                        According to the limit state method

                                                                                                        0d frac14 tan1tan 32

                                                                                                        125

                                                                                                        frac14 265

                                                                                                        c0 frac14 8

                                                                                                        160frac14 5 kN=m2

                                                                                                        Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                        Design disturbing moment frac14 1075 kN=m

                                                                                                        The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                        94

                                                                                                        F frac14 1

                                                                                                        W sin

                                                                                                        Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                        1thorn ethtan tan0=FTHORN

                                                                                                        c0 frac14 8 kN=m2

                                                                                                        0 frac14 32

                                                                                                        c0b frac14 8 2 frac14 16 kN=m

                                                                                                        W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                        Try F frac14 100

                                                                                                        tan0

                                                                                                        Ffrac14 0625

                                                                                                        Stability of slopes 77

                                                                                                        Values of u are as obtained in Figure Q93

                                                                                                        SliceNo

                                                                                                        h(m)

                                                                                                        W frac14 bh(kNm)

                                                                                                        W sin(kNm)

                                                                                                        ub(kNm)

                                                                                                        c0bthorn (W ub) tan0(kNm)

                                                                                                        sec

                                                                                                        1thorn (tan tan0)FProduct(kNm)

                                                                                                        1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                        23 33 30 1042 31

                                                                                                        3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                        224 92 72 0931 67

                                                                                                        6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                        12 58 112 90 0889 80

                                                                                                        8 60 252 1812

                                                                                                        80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                        10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                        2154 88 116 0853 99

                                                                                                        14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                        1074 1091

                                                                                                        F frac14 1091

                                                                                                        1074frac14 102 (assumed value 100)

                                                                                                        Thus

                                                                                                        F frac14 101

                                                                                                        95

                                                                                                        F frac14 1

                                                                                                        W sin

                                                                                                        XfWeth1 ruTHORN tan0g sec

                                                                                                        1thorn ethtan tan0THORN=F

                                                                                                        0 frac14 33

                                                                                                        ru frac14 020

                                                                                                        W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                        eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                        Try F frac14 110

                                                                                                        tan 0

                                                                                                        Ffrac14 tan 33

                                                                                                        110frac14 0590

                                                                                                        78 Stability of slopes

                                                                                                        Referring to Figure Q95

                                                                                                        SliceNo

                                                                                                        h(m)

                                                                                                        W frac14 bh(kNm)

                                                                                                        W sin(kNm)

                                                                                                        W(1 ru) tan0(kNm)

                                                                                                        sec

                                                                                                        1thorn ( tan tan0)FProduct(kNm)

                                                                                                        1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                        2120 234 0892 209

                                                                                                        4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                        1185 1271

                                                                                                        Figure Q95

                                                                                                        Stability of slopes 79

                                                                                                        F frac14 1271

                                                                                                        1185frac14 107

                                                                                                        The trial value was 110 therefore take F to be 108

                                                                                                        96

                                                                                                        (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                        F frac14 0

                                                                                                        sat

                                                                                                        tan0

                                                                                                        tan

                                                                                                        ptie 15 frac14 92

                                                                                                        19

                                                                                                        tan 36

                                                                                                        tan

                                                                                                        tan frac14 0234

                                                                                                        frac14 13

                                                                                                        Water table well below surface the factor of safety is given by Equation 911

                                                                                                        F frac14 tan0

                                                                                                        tan

                                                                                                        frac14 tan 36

                                                                                                        tan 13

                                                                                                        frac14 31

                                                                                                        (b) 0d frac14 tan1tan 36

                                                                                                        125

                                                                                                        frac14 30

                                                                                                        Depth of potential failure surface frac14 z

                                                                                                        Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                        frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                        frac14 504z kN

                                                                                                        Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                        frac14 19 z sin 13 cos 13

                                                                                                        frac14 416z kN

                                                                                                        Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                        80 Stability of slopes

                                                                                                        • Book Cover
                                                                                                        • Title
                                                                                                        • Contents
                                                                                                        • Basic characteristics of soils
                                                                                                        • Seepage
                                                                                                        • Effective stress
                                                                                                        • Shear strength
                                                                                                        • Stresses and displacements
                                                                                                        • Lateral earth pressure
                                                                                                        • Consolidation theory
                                                                                                        • Bearing capacity
                                                                                                        • Stability of slopes

                                                                                                          P frac14 1

                                                                                                          2 s 3352 frac14 561s kN=m

                                                                                                          U frac14 1

                                                                                                          2 98 1852 cosec 543 frac14 206 kN=m

                                                                                                          Equations 630 and 631 then become

                                                                                                          561s thorn ethN 206THORN tan 186 cos 543 N sin 543 frac14 0

                                                                                                          792 ethN 206THORN tan 186 sin 543 N cos 543 frac14 0

                                                                                                          ie

                                                                                                          561s 0616N 405 frac14 0

                                                                                                          792 0857N thorn 563 frac14 0

                                                                                                          N frac14 848

                                                                                                          0857frac14 989 kN=m

                                                                                                          Then

                                                                                                          561s 609 405 frac14 0

                                                                                                          s frac14 649

                                                                                                          561frac14 116 kN=m3

                                                                                                          The calculations for trial values of F of 20 15 and 10 are summarized below

                                                                                                          F 0m W (kNm) U (kNm) N (kNm) s (kNm3)

                                                                                                          20 186 543 792 206 989 11615 242 571 713 199 856 9910 34 62 586 191 657 77

                                                                                                          s is plotted against F in Figure Q611From Figure Q611 for s frac14 106 kNm3 F frac14 17

                                                                                                          Figure Q611

                                                                                                          Lateral earth pressure 47

                                                                                                          612

                                                                                                          For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                                          45 thorn 0

                                                                                                          2frac14 63

                                                                                                          For the retained material between the surface and a depth of 36m

                                                                                                          Pa frac14 1

                                                                                                          2 030 18 362 frac14 350 kN=m

                                                                                                          Weight of reinforced fill between the surface and a depth of 36m is

                                                                                                          Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                                          eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                                          Eccentricity of Rv

                                                                                                          e frac14 263 250 frac14 013m

                                                                                                          The average vertical stress at a depth of 36m is

                                                                                                          z frac14 Rv

                                                                                                          L 2efrac14 324

                                                                                                          474frac14 68 kN=m2

                                                                                                          (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                                          Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                                          Tensile stress in the element frac14 138 103

                                                                                                          65 3frac14 71N=mm2

                                                                                                          Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                                          Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                                          Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                                          The weight of ABC is

                                                                                                          W frac14 1

                                                                                                          2 18 52 265 frac14 124 kN=m

                                                                                                          From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                                          48 Lateral earth pressure

                                                                                                          (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                                          Tp frac14 032 68 120 065 frac14 170 kN

                                                                                                          Tr frac14 213 420

                                                                                                          418frac14 214 kN

                                                                                                          Again the tensile failure and slipping limit states are satisfied for this element

                                                                                                          Figure Q612

                                                                                                          Lateral earth pressure 49

                                                                                                          Chapter 7

                                                                                                          Consolidation theory

                                                                                                          71

                                                                                                          Total change in thickness

                                                                                                          H frac14 782 602 frac14 180mm

                                                                                                          Average thickness frac14 1530thorn 180

                                                                                                          2frac14 1620mm

                                                                                                          Length of drainage path d frac14 1620

                                                                                                          2frac14 810mm

                                                                                                          Root time plot (Figure Q71a)

                                                                                                          ffiffiffiffiffiffit90p frac14 33

                                                                                                          t90 frac14 109min

                                                                                                          cv frac14 0848d2

                                                                                                          t90frac14 0848 8102

                                                                                                          109 1440 365

                                                                                                          106frac14 27m2=year

                                                                                                          r0 frac14 782 764

                                                                                                          782 602frac14 018

                                                                                                          180frac14 0100

                                                                                                          rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                                          10 119

                                                                                                          9 180frac14 0735

                                                                                                          rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                                          Log time plot (Figure Q71b)

                                                                                                          t50 frac14 26min

                                                                                                          cv frac14 0196d2

                                                                                                          t50frac14 0196 8102

                                                                                                          26 1440 365

                                                                                                          106frac14 26m2=year

                                                                                                          r0 frac14 782 763

                                                                                                          782 602frac14 019

                                                                                                          180frac14 0106

                                                                                                          rp frac14 763 623

                                                                                                          782 602frac14 140

                                                                                                          180frac14 0778

                                                                                                          rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                                          Figure Q71(a)

                                                                                                          Figure Q71(b)

                                                                                                          Final void ratio

                                                                                                          e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                                          e

                                                                                                          Hfrac14 1thorn e0

                                                                                                          H0frac14 1thorn e1 thorne

                                                                                                          H0

                                                                                                          ie

                                                                                                          e

                                                                                                          180frac14 1631thorne

                                                                                                          1710

                                                                                                          e frac14 2936

                                                                                                          1530frac14 0192

                                                                                                          Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                                          mv frac14 1

                                                                                                          1thorn e0 e0 e101 00

                                                                                                          frac14 1

                                                                                                          1823 0192

                                                                                                          0107frac14 098m2=MN

                                                                                                          k frac14 cvmvw frac14 265 098 98

                                                                                                          60 1440 365 103frac14 81 1010 m=s

                                                                                                          72

                                                                                                          Using Equation 77 (one-dimensional method)

                                                                                                          sc frac14 e0 e11thorn e0 H

                                                                                                          Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                                          Figure Q72

                                                                                                          52 Consolidation theory

                                                                                                          Settlement

                                                                                                          Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                                          1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                                          318

                                                                                                          Notes 5 92y 460thorn 84

                                                                                                          Heave

                                                                                                          Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                                          1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                                          38

                                                                                                          73

                                                                                                          U frac14 f ethTvTHORN frac14 f cvt

                                                                                                          d2

                                                                                                          Hence if cv is constant

                                                                                                          t1

                                                                                                          t2frac14 d

                                                                                                          21

                                                                                                          d22

                                                                                                          where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                                          d1 frac14 95mm and d2 frac14 2500mm

                                                                                                          for U frac14 050 t2 frac14 t1 d22

                                                                                                          d21

                                                                                                          frac14 20

                                                                                                          60 24 365 25002

                                                                                                          952frac14 263 years

                                                                                                          for U lt 060 Tv frac14

                                                                                                          4U2 (Equation 724(a))

                                                                                                          t030 frac14 t050 0302

                                                                                                          0502

                                                                                                          frac14 263 036 frac14 095 years

                                                                                                          Consolidation theory 53

                                                                                                          74

                                                                                                          The layer is open

                                                                                                          d frac14 8

                                                                                                          2frac14 4m

                                                                                                          Tv frac14 cvtd2frac14 24 3

                                                                                                          42frac14 0450

                                                                                                          ui frac14 frac14 84 kN=m2

                                                                                                          The excess pore water pressure is given by Equation 721

                                                                                                          ue frac14Xmfrac141mfrac140

                                                                                                          2ui

                                                                                                          Msin

                                                                                                          Mz

                                                                                                          d

                                                                                                          expethM2TvTHORN

                                                                                                          In this case z frac14 d

                                                                                                          sinMz

                                                                                                          d

                                                                                                          frac14 sinM

                                                                                                          where

                                                                                                          M frac14

                                                                                                          23

                                                                                                          25

                                                                                                          2

                                                                                                          M sin M M2Tv exp (M2Tv)

                                                                                                          2thorn1 1110 0329

                                                                                                          3

                                                                                                          21 9993 457 105

                                                                                                          ue frac14 2 84 2

                                                                                                          1 0329 ethother terms negligibleTHORN

                                                                                                          frac14 352 kN=m2

                                                                                                          75

                                                                                                          The layer is open

                                                                                                          d frac14 6

                                                                                                          2frac14 3m

                                                                                                          Tv frac14 cvtd2frac14 10 3

                                                                                                          32frac14 0333

                                                                                                          The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                                          54 Consolidation theory

                                                                                                          For an open layer

                                                                                                          Tv frac14 4n

                                                                                                          m2

                                                                                                          n frac14 0333 62

                                                                                                          4frac14 300

                                                                                                          The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                                          ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                                          i j

                                                                                                          0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                                          0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                                          The initial and 3-year isochrones are plotted in Figure Q75

                                                                                                          Area under initial isochrone frac14 180 units

                                                                                                          Area under 3-year isochrone frac14 63 units

                                                                                                          The average degree of consolidation is given by Equation 725Thus

                                                                                                          U frac14 1 63

                                                                                                          180frac14 065

                                                                                                          Figure Q75

                                                                                                          Consolidation theory 55

                                                                                                          76

                                                                                                          At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                          0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                          The final consolidation settlement (one-dimensional method) is

                                                                                                          sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                          Corrected time t frac14 2 1

                                                                                                          2

                                                                                                          40

                                                                                                          52

                                                                                                          frac14 1615 years

                                                                                                          Tv frac14 cvtd2frac14 44 1615

                                                                                                          42frac14 0444

                                                                                                          From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                          77

                                                                                                          The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                          Figure Q77

                                                                                                          56 Consolidation theory

                                                                                                          Point m n Ir (kNm2) sc (mm)

                                                                                                          13020frac14 15 20

                                                                                                          20frac14 10 0194 (4) 113 124

                                                                                                          260

                                                                                                          20frac14 30

                                                                                                          20

                                                                                                          20frac14 10 0204 (2) 59 65

                                                                                                          360

                                                                                                          20frac14 30

                                                                                                          40

                                                                                                          20frac14 20 0238 (1) 35 38

                                                                                                          430

                                                                                                          20frac14 15

                                                                                                          40

                                                                                                          20frac14 20 0224 (2) 65 72

                                                                                                          Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                          78

                                                                                                          Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                          (a) Immediate settlement

                                                                                                          H

                                                                                                          Bfrac14 30

                                                                                                          35frac14 086

                                                                                                          D

                                                                                                          Bfrac14 2

                                                                                                          35frac14 006

                                                                                                          Figure Q78

                                                                                                          Consolidation theory 57

                                                                                                          From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                          si frac14 130131qB

                                                                                                          Eufrac14 10 032 105 35

                                                                                                          40frac14 30mm

                                                                                                          (b) Consolidation settlement

                                                                                                          Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                          1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                          3150

                                                                                                          Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                          Now

                                                                                                          H

                                                                                                          Bfrac14 30

                                                                                                          35frac14 086 and A frac14 065

                                                                                                          from Figure 712 13 frac14 079

                                                                                                          sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                          Total settlement

                                                                                                          s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                          79

                                                                                                          Without sand drains

                                                                                                          Uv frac14 025

                                                                                                          Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                          t frac14 Tvd2

                                                                                                          cvfrac14 0049 82

                                                                                                          cvWith sand drains

                                                                                                          R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                          n frac14 Rrfrac14 169

                                                                                                          015frac14 113

                                                                                                          Tr frac14 cht

                                                                                                          4R2frac14 ch

                                                                                                          4 1692 0049 82

                                                                                                          cvethand ch frac14 cvTHORN

                                                                                                          frac14 0275

                                                                                                          Ur frac14 073 (from Figure 730)

                                                                                                          58 Consolidation theory

                                                                                                          Using Equation 740

                                                                                                          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                          U frac14 080

                                                                                                          710

                                                                                                          Without sand drains

                                                                                                          Uv frac14 090

                                                                                                          Tv frac14 0848

                                                                                                          t frac14 Tvd2

                                                                                                          cvfrac14 0848 102

                                                                                                          96frac14 88 years

                                                                                                          With sand drains

                                                                                                          R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                          n frac14 Rrfrac14 226

                                                                                                          015frac14 15

                                                                                                          Tr

                                                                                                          Tvfrac14 chcv

                                                                                                          d2

                                                                                                          4R2ethsame tTHORN

                                                                                                          Tr

                                                                                                          Tvfrac14 140

                                                                                                          96 102

                                                                                                          4 2262frac14 714 eth1THORN

                                                                                                          Using Equation 740

                                                                                                          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                          An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                          Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                          040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                          Thus

                                                                                                          Uv frac14 0295 and Ur frac14 086

                                                                                                          t frac14 88 00683

                                                                                                          0848frac14 07 years

                                                                                                          Consolidation theory 59

                                                                                                          Chapter 8

                                                                                                          Bearing capacity

                                                                                                          81

                                                                                                          (a) The ultimate bearing capacity is given by Equation 83

                                                                                                          qf frac14 cNc thorn DNq thorn 1

                                                                                                          2BN

                                                                                                          For u frac14 0

                                                                                                          Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                          qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                          The net ultimate bearing capacity is

                                                                                                          qnf frac14 qf D frac14 540 kN=m2

                                                                                                          The net foundation pressure is

                                                                                                          qn frac14 q D frac14 425

                                                                                                          2 eth21 1THORN frac14 192 kN=m2

                                                                                                          The factor of safety (Equation 86) is

                                                                                                          F frac14 qnfqnfrac14 540

                                                                                                          192frac14 28

                                                                                                          (b) For 0 frac14 28

                                                                                                          Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                          qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                          2 112 2 13

                                                                                                          frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                          qnf frac14 574 112 frac14 563 kN=m2

                                                                                                          F frac14 563

                                                                                                          192frac14 29

                                                                                                          (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                          82

                                                                                                          For 0 frac14 38

                                                                                                          Nq frac14 49 N frac14 67

                                                                                                          qnf frac14 DethNq 1THORN thorn 1

                                                                                                          2BN ethfrom Equation 83THORN

                                                                                                          frac14 eth18 075 48THORN thorn 1

                                                                                                          2 18 15 67

                                                                                                          frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                          qn frac14 500

                                                                                                          15 eth18 075THORN frac14 320 kN=m2

                                                                                                          F frac14 qnfqnfrac14 1553

                                                                                                          320frac14 48

                                                                                                          0d frac14 tan1tan 38

                                                                                                          125

                                                                                                          frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                          Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                          2 18 15 25

                                                                                                          frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                          Design load (action) Vd frac14 500 kN=m

                                                                                                          The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                          83

                                                                                                          D

                                                                                                          Bfrac14 350

                                                                                                          225frac14 155

                                                                                                          From Figure 85 for a square foundation

                                                                                                          Nc frac14 81

                                                                                                          Bearing capacity 61

                                                                                                          For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                          Nc frac14 084thorn 016B

                                                                                                          L

                                                                                                          81 frac14 745

                                                                                                          Using Equation 810

                                                                                                          qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                          For F frac14 3

                                                                                                          qn frac14 1006

                                                                                                          3frac14 335 kN=m2

                                                                                                          q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                          Design load frac14 405 450 225 frac14 4100 kN

                                                                                                          Design undrained strength cud frac14 135

                                                                                                          14frac14 96 kN=m2

                                                                                                          Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                          frac14 7241 kN

                                                                                                          Design load Vd frac14 4100 kN

                                                                                                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                          84

                                                                                                          For 0 frac14 40

                                                                                                          Nq frac14 64 N frac14 95

                                                                                                          qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                          (a) Water table 5m below ground level

                                                                                                          qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                          qn frac14 400 17 frac14 383 kN=m2

                                                                                                          F frac14 2686

                                                                                                          383frac14 70

                                                                                                          (b) Water table 1m below ground level (ie at foundation level)

                                                                                                          0 frac14 20 98 frac14 102 kN=m3

                                                                                                          62 Bearing capacity

                                                                                                          qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                          F frac14 2040

                                                                                                          383frac14 53

                                                                                                          (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                          eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                          qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                          F frac14 1296

                                                                                                          392frac14 33

                                                                                                          85

                                                                                                          The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                          Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                          Design value of 0 frac14 tan1tan 39

                                                                                                          125

                                                                                                          frac14 33

                                                                                                          For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                          Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                          86

                                                                                                          (a) Undrained shear for u frac14 0

                                                                                                          Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                          qnf frac14 12cuNc

                                                                                                          frac14 12 100 514 frac14 617 kN=m2

                                                                                                          qn frac14 qnfFfrac14 617

                                                                                                          3frac14 206 kN=m2

                                                                                                          q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                          Bearing capacity 63

                                                                                                          Drained shear for 0 frac14 32

                                                                                                          Nq frac14 23 N frac14 25

                                                                                                          0 frac14 21 98 frac14 112 kN=m3

                                                                                                          qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                          frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                          frac14 694 kN=m2

                                                                                                          q frac14 694

                                                                                                          3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                          Design load frac14 42 227 frac14 3632 kN

                                                                                                          (b) Design undrained strength cud frac14 100

                                                                                                          14frac14 71 kNm2

                                                                                                          Design bearing resistance Rd frac14 12cudNe area

                                                                                                          frac14 12 71 514 42

                                                                                                          frac14 7007 kN

                                                                                                          For drained shear 0d frac14 tan1tan 32

                                                                                                          125

                                                                                                          frac14 26

                                                                                                          Nq frac14 12 N frac14 10

                                                                                                          Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                          (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                          Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                          1 2 100 0175 0700qn 0182qn

                                                                                                          2 6 033 0044 0176qn 0046qn

                                                                                                          3 10 020 0017 0068qn 0018qn

                                                                                                          0246qn

                                                                                                          Diameter of equivalent circle B frac14 45m

                                                                                                          H

                                                                                                          Bfrac14 12

                                                                                                          45frac14 27 and A frac14 042

                                                                                                          13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                          64 Bearing capacity

                                                                                                          For sc frac14 30mm

                                                                                                          qn frac14 30

                                                                                                          0147frac14 204 kN=m2

                                                                                                          q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                          Design load frac14 42 225 frac14 3600 kN

                                                                                                          The design load is 3600 kN settlement being the limiting criterion

                                                                                                          87

                                                                                                          D

                                                                                                          Bfrac14 8

                                                                                                          4frac14 20

                                                                                                          From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                          F frac14 cuNc

                                                                                                          Dfrac14 40 71

                                                                                                          20 8frac14 18

                                                                                                          88

                                                                                                          Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                          Design value of 0 frac14 tan1tan 38

                                                                                                          125

                                                                                                          frac14 32

                                                                                                          Figure Q86

                                                                                                          Bearing capacity 65

                                                                                                          For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                          Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                          The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                          Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                          For B frac14 250m qn frac14 3750

                                                                                                          2502 17 frac14 583 kN=m2

                                                                                                          From Figure 510 m frac14 n frac14 126

                                                                                                          6frac14 021

                                                                                                          Ir frac14 0019

                                                                                                          Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                          Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                          The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                          89

                                                                                                          Depth (m) N 0v (kNm2) CN N1

                                                                                                          070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                          Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                          (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                          Cw frac14 05thorn 0530

                                                                                                          47

                                                                                                          frac14 082

                                                                                                          66 Bearing capacity

                                                                                                          Thus

                                                                                                          qa frac14 150 082 frac14 120 kN=m2

                                                                                                          (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                          Thus

                                                                                                          qa frac14 90 15 frac14 135 kN=m2

                                                                                                          (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                          Ic frac14 171

                                                                                                          1014frac14 0068

                                                                                                          From Equation 819(a) with s frac14 25mm

                                                                                                          q frac14 25

                                                                                                          3507 0068frac14 150 kN=m2

                                                                                                          810

                                                                                                          Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                          Izp frac14 05thorn 01130

                                                                                                          16 27

                                                                                                          05

                                                                                                          frac14 067

                                                                                                          Refer to Figure Q810

                                                                                                          E frac14 25qc

                                                                                                          Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                          Ez (mm3MN)

                                                                                                          1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                          0203

                                                                                                          C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                          130frac14 093

                                                                                                          C2 frac14 1 ethsayTHORN

                                                                                                          s frac14 C1C2qnX Iz

                                                                                                          Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                          Bearing capacity 67

                                                                                                          811

                                                                                                          At pile base level

                                                                                                          cu frac14 220 kN=m2

                                                                                                          qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                          Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                          00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                          qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                          Then

                                                                                                          Qf frac14 Abqb thorn Asqs

                                                                                                          frac14

                                                                                                          4 32 1980

                                                                                                          thorn eth 105 139 86THORN

                                                                                                          frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                          0 01 02 03 04 05 06 07

                                                                                                          0 2 4 6 8 10 12 14

                                                                                                          1

                                                                                                          2

                                                                                                          3

                                                                                                          4

                                                                                                          5

                                                                                                          6

                                                                                                          7

                                                                                                          8

                                                                                                          (1)

                                                                                                          (2)

                                                                                                          (3)

                                                                                                          (4)

                                                                                                          (5)

                                                                                                          qc

                                                                                                          qc

                                                                                                          Iz

                                                                                                          Iz

                                                                                                          (MNm2)

                                                                                                          z (m)

                                                                                                          Figure Q810

                                                                                                          68 Bearing capacity

                                                                                                          Allowable load

                                                                                                          ethaTHORN Qf

                                                                                                          2frac14 17 937

                                                                                                          2frac14 8968 kN

                                                                                                          ethbTHORN Abqb

                                                                                                          3thorn Asqs frac14 13 996

                                                                                                          3thorn 3941 frac14 8606 kN

                                                                                                          ie allowable load frac14 8600 kN

                                                                                                          Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                          According to the limit state method

                                                                                                          Characteristic undrained strength at base level cuk frac14 220

                                                                                                          150kN=m2

                                                                                                          Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                          150frac14 1320 kN=m2

                                                                                                          Characteristic shaft resistance qsk frac14 00150

                                                                                                          frac14 86

                                                                                                          150frac14 57 kN=m2

                                                                                                          Characteristic base and shaft resistances

                                                                                                          Rbk frac14

                                                                                                          4 32 1320 frac14 9330 kN

                                                                                                          Rsk frac14 105 139 86

                                                                                                          150frac14 2629 kN

                                                                                                          For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                          Design bearing resistance Rcd frac14 9330

                                                                                                          160thorn 2629

                                                                                                          130

                                                                                                          frac14 5831thorn 2022

                                                                                                          frac14 7850 kN

                                                                                                          Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                          812

                                                                                                          ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                          qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                          For a single pile

                                                                                                          Qf frac14 Abqb thorn Asqs

                                                                                                          frac14

                                                                                                          4 062 1305

                                                                                                          thorn eth 06 15 42THORN

                                                                                                          frac14 369thorn 1187 frac14 1556 kN

                                                                                                          Bearing capacity 69

                                                                                                          Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                          eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                          (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                          (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                          qbkfrac14 9cuk frac14 9 220

                                                                                                          150frac14 1320 kN=m2

                                                                                                          qskfrac14cuk frac14 040 105

                                                                                                          150frac14 28 kN=m2

                                                                                                          Rbkfrac14

                                                                                                          4 0602 1320 frac14 373 kN

                                                                                                          Rskfrac14 060 15 28 frac14 791 kN

                                                                                                          Rcdfrac14 373

                                                                                                          160thorn 791

                                                                                                          130frac14 233thorn 608 frac14 841 kN

                                                                                                          Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                          Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                          (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                          q frac14 21 000

                                                                                                          1762frac14 68 kN=m2

                                                                                                          Immediate settlement

                                                                                                          H

                                                                                                          Bfrac14 15

                                                                                                          176frac14 085

                                                                                                          D

                                                                                                          Bfrac14 13

                                                                                                          176frac14 074

                                                                                                          L

                                                                                                          Bfrac14 1

                                                                                                          Hence from Figure 515

                                                                                                          130 frac14 078 and 131 frac14 041

                                                                                                          70 Bearing capacity

                                                                                                          Thus using Equation 528

                                                                                                          si frac14 078 041 68 176

                                                                                                          65frac14 6mm

                                                                                                          Consolidation settlement

                                                                                                          Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                          1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                          434 (sod)

                                                                                                          Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                          sc frac14 056 434 frac14 24mm

                                                                                                          The total settlement is (6thorn 24) frac14 30mm

                                                                                                          813

                                                                                                          At base level N frac14 26 Then using Equation 830

                                                                                                          qb frac14 40NDb

                                                                                                          Bfrac14 40 26 2

                                                                                                          025frac14 8320 kN=m2

                                                                                                          ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                          Figure Q812

                                                                                                          Bearing capacity 71

                                                                                                          Over the length embedded in sand

                                                                                                          N frac14 21 ie18thorn 24

                                                                                                          2

                                                                                                          Using Equation 831

                                                                                                          qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                          For a single pile

                                                                                                          Qf frac14 Abqb thorn Asqs

                                                                                                          frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                          For the pile group assuming a group efficiency of 12

                                                                                                          XQf frac14 12 9 604 frac14 6523 kN

                                                                                                          Then the load factor is

                                                                                                          F frac14 6523

                                                                                                          2000thorn 1000frac14 21

                                                                                                          (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                          Characteristic base resistance per unit area qbk frac14 8320

                                                                                                          150frac14 5547 kNm2

                                                                                                          Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                          150frac14 28 kNm2

                                                                                                          Characteristic base and shaft resistances for a single pile

                                                                                                          Rbk frac14 0252 5547 frac14 347 kN

                                                                                                          Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                          For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                          Design bearing resistance Rcd frac14 347

                                                                                                          130thorn 56

                                                                                                          130frac14 310 kN

                                                                                                          For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                          Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                          (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                          From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                          72 Bearing capacity

                                                                                                          N frac14 24thorn 26thorn 34

                                                                                                          3frac14 28

                                                                                                          Ic frac14 171

                                                                                                          2814frac14 0016 ethEquation 818THORN

                                                                                                          s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                          814

                                                                                                          Using Equation 841

                                                                                                          Tf frac14 DLcu thorn

                                                                                                          4ethD2 d2THORNcuNc

                                                                                                          frac14 eth 02 5 06 110THORN thorn

                                                                                                          4eth022 012THORN110 9

                                                                                                          frac14 207thorn 23 frac14 230 kN

                                                                                                          Figure Q813

                                                                                                          Bearing capacity 73

                                                                                                          Chapter 9

                                                                                                          Stability of slopes

                                                                                                          91

                                                                                                          Referring to Figure Q91

                                                                                                          W frac14 417 19 frac14 792 kN=m

                                                                                                          Q frac14 20 28 frac14 56 kN=m

                                                                                                          Arc lengthAB frac14

                                                                                                          180 73 90 frac14 115m

                                                                                                          Arc length BC frac14

                                                                                                          180 28 90 frac14 44m

                                                                                                          The factor of safety is given by

                                                                                                          F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                          90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                          Depth of tension crack z0 frac14 2cu

                                                                                                          frac14 2 20

                                                                                                          19frac14 21m

                                                                                                          Arc length BD frac14

                                                                                                          180 13

                                                                                                          1

                                                                                                          2 90 frac14 21m

                                                                                                          F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                          The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                          Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                          14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                          Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                          The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                          92

                                                                                                          u frac14 0

                                                                                                          Depth factor D frac14 11

                                                                                                          9frac14 122

                                                                                                          Using Equation 92 with F frac14 10

                                                                                                          Ns frac14 cu

                                                                                                          FHfrac14 30

                                                                                                          10 19 9frac14 0175

                                                                                                          Hence from Figure 93

                                                                                                          frac14 50

                                                                                                          For F frac14 12

                                                                                                          Ns frac14 30

                                                                                                          12 19 9frac14 0146

                                                                                                          frac14 27

                                                                                                          93

                                                                                                          Refer to Figure Q93

                                                                                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                          1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                          74 m

                                                                                                          214 1deg

                                                                                                          213 1deg

                                                                                                          39 m

                                                                                                          WB

                                                                                                          D

                                                                                                          C

                                                                                                          28 m

                                                                                                          21 m

                                                                                                          A

                                                                                                          Q

                                                                                                          Soil (1)Soil (2)

                                                                                                          73deg

                                                                                                          Figure Q91

                                                                                                          Stability of slopes 75

                                                                                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                          9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                          599 256 328 1372

                                                                                                          Figure Q93

                                                                                                          76 Stability of slopes

                                                                                                          XW cos frac14 b

                                                                                                          Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                          W sin frac14 bX

                                                                                                          h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                          Arc length La frac14

                                                                                                          180 57

                                                                                                          1

                                                                                                          2 326 frac14 327m

                                                                                                          The factor of safety is given by

                                                                                                          F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                          W sin

                                                                                                          frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                          frac14 091

                                                                                                          According to the limit state method

                                                                                                          0d frac14 tan1tan 32

                                                                                                          125

                                                                                                          frac14 265

                                                                                                          c0 frac14 8

                                                                                                          160frac14 5 kN=m2

                                                                                                          Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                          Design disturbing moment frac14 1075 kN=m

                                                                                                          The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                          94

                                                                                                          F frac14 1

                                                                                                          W sin

                                                                                                          Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                          1thorn ethtan tan0=FTHORN

                                                                                                          c0 frac14 8 kN=m2

                                                                                                          0 frac14 32

                                                                                                          c0b frac14 8 2 frac14 16 kN=m

                                                                                                          W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                          Try F frac14 100

                                                                                                          tan0

                                                                                                          Ffrac14 0625

                                                                                                          Stability of slopes 77

                                                                                                          Values of u are as obtained in Figure Q93

                                                                                                          SliceNo

                                                                                                          h(m)

                                                                                                          W frac14 bh(kNm)

                                                                                                          W sin(kNm)

                                                                                                          ub(kNm)

                                                                                                          c0bthorn (W ub) tan0(kNm)

                                                                                                          sec

                                                                                                          1thorn (tan tan0)FProduct(kNm)

                                                                                                          1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                          23 33 30 1042 31

                                                                                                          3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                          224 92 72 0931 67

                                                                                                          6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                          12 58 112 90 0889 80

                                                                                                          8 60 252 1812

                                                                                                          80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                          10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                          2154 88 116 0853 99

                                                                                                          14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                          1074 1091

                                                                                                          F frac14 1091

                                                                                                          1074frac14 102 (assumed value 100)

                                                                                                          Thus

                                                                                                          F frac14 101

                                                                                                          95

                                                                                                          F frac14 1

                                                                                                          W sin

                                                                                                          XfWeth1 ruTHORN tan0g sec

                                                                                                          1thorn ethtan tan0THORN=F

                                                                                                          0 frac14 33

                                                                                                          ru frac14 020

                                                                                                          W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                          eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                          Try F frac14 110

                                                                                                          tan 0

                                                                                                          Ffrac14 tan 33

                                                                                                          110frac14 0590

                                                                                                          78 Stability of slopes

                                                                                                          Referring to Figure Q95

                                                                                                          SliceNo

                                                                                                          h(m)

                                                                                                          W frac14 bh(kNm)

                                                                                                          W sin(kNm)

                                                                                                          W(1 ru) tan0(kNm)

                                                                                                          sec

                                                                                                          1thorn ( tan tan0)FProduct(kNm)

                                                                                                          1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                          2120 234 0892 209

                                                                                                          4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                          1185 1271

                                                                                                          Figure Q95

                                                                                                          Stability of slopes 79

                                                                                                          F frac14 1271

                                                                                                          1185frac14 107

                                                                                                          The trial value was 110 therefore take F to be 108

                                                                                                          96

                                                                                                          (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                          F frac14 0

                                                                                                          sat

                                                                                                          tan0

                                                                                                          tan

                                                                                                          ptie 15 frac14 92

                                                                                                          19

                                                                                                          tan 36

                                                                                                          tan

                                                                                                          tan frac14 0234

                                                                                                          frac14 13

                                                                                                          Water table well below surface the factor of safety is given by Equation 911

                                                                                                          F frac14 tan0

                                                                                                          tan

                                                                                                          frac14 tan 36

                                                                                                          tan 13

                                                                                                          frac14 31

                                                                                                          (b) 0d frac14 tan1tan 36

                                                                                                          125

                                                                                                          frac14 30

                                                                                                          Depth of potential failure surface frac14 z

                                                                                                          Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                          frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                          frac14 504z kN

                                                                                                          Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                          frac14 19 z sin 13 cos 13

                                                                                                          frac14 416z kN

                                                                                                          Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                          80 Stability of slopes

                                                                                                          • Book Cover
                                                                                                          • Title
                                                                                                          • Contents
                                                                                                          • Basic characteristics of soils
                                                                                                          • Seepage
                                                                                                          • Effective stress
                                                                                                          • Shear strength
                                                                                                          • Stresses and displacements
                                                                                                          • Lateral earth pressure
                                                                                                          • Consolidation theory
                                                                                                          • Bearing capacity
                                                                                                          • Stability of slopes

                                                                                                            612

                                                                                                            For 0 frac14 36Ka frac14 026 and K0 frac14 1 sin 36 frac14 041

                                                                                                            45 thorn 0

                                                                                                            2frac14 63

                                                                                                            For the retained material between the surface and a depth of 36m

                                                                                                            Pa frac14 1

                                                                                                            2 030 18 362 frac14 350 kN=m

                                                                                                            Weight of reinforced fill between the surface and a depth of 36m is

                                                                                                            Wf frac14 18 36 50 frac14 324 kN=m frac14Rveth or VTHORNConsidering moments about X

                                                                                                            eth35 12THORN thorn eth324 25THORN frac14 Rva etha frac14 lever armTHORN a frac14 263m

                                                                                                            Eccentricity of Rv

                                                                                                            e frac14 263 250 frac14 013m

                                                                                                            The average vertical stress at a depth of 36m is

                                                                                                            z frac14 Rv

                                                                                                            L 2efrac14 324

                                                                                                            474frac14 68 kN=m2

                                                                                                            (a) In the tie back wedge method K frac14 Ka and Le frac14 418m

                                                                                                            Tp frac14 026 68 120 065 frac14 138 kN ethEquation 632THORN

                                                                                                            Tensile stress in the element frac14 138 103

                                                                                                            65 3frac14 71N=mm2

                                                                                                            Tensile failure of the element will not occur the ultimate tensile strength being 48times the tensile stress

                                                                                                            Tr frac14 2 0065 418 68 tan 30 frac14 213 kN

                                                                                                            Slipping between element and soil will not occur Tr being greater than Tp by a factorof 15 The stability of wedge ABC would also be checked after the above calculationshad been performed for all elements

                                                                                                            The weight of ABC is

                                                                                                            W frac14 1

                                                                                                            2 18 52 265 frac14 124 kN=m

                                                                                                            From the force diagram (Figure Q612) Tw frac14 124 tan 27 frac14 632 kNThe factored sum of the forces Tr in all elements (Tr) must be greater than Tw

                                                                                                            48 Lateral earth pressure

                                                                                                            (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                                            Tp frac14 032 68 120 065 frac14 170 kN

                                                                                                            Tr frac14 213 420

                                                                                                            418frac14 214 kN

                                                                                                            Again the tensile failure and slipping limit states are satisfied for this element

                                                                                                            Figure Q612

                                                                                                            Lateral earth pressure 49

                                                                                                            Chapter 7

                                                                                                            Consolidation theory

                                                                                                            71

                                                                                                            Total change in thickness

                                                                                                            H frac14 782 602 frac14 180mm

                                                                                                            Average thickness frac14 1530thorn 180

                                                                                                            2frac14 1620mm

                                                                                                            Length of drainage path d frac14 1620

                                                                                                            2frac14 810mm

                                                                                                            Root time plot (Figure Q71a)

                                                                                                            ffiffiffiffiffiffit90p frac14 33

                                                                                                            t90 frac14 109min

                                                                                                            cv frac14 0848d2

                                                                                                            t90frac14 0848 8102

                                                                                                            109 1440 365

                                                                                                            106frac14 27m2=year

                                                                                                            r0 frac14 782 764

                                                                                                            782 602frac14 018

                                                                                                            180frac14 0100

                                                                                                            rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                                            10 119

                                                                                                            9 180frac14 0735

                                                                                                            rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                                            Log time plot (Figure Q71b)

                                                                                                            t50 frac14 26min

                                                                                                            cv frac14 0196d2

                                                                                                            t50frac14 0196 8102

                                                                                                            26 1440 365

                                                                                                            106frac14 26m2=year

                                                                                                            r0 frac14 782 763

                                                                                                            782 602frac14 019

                                                                                                            180frac14 0106

                                                                                                            rp frac14 763 623

                                                                                                            782 602frac14 140

                                                                                                            180frac14 0778

                                                                                                            rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                                            Figure Q71(a)

                                                                                                            Figure Q71(b)

                                                                                                            Final void ratio

                                                                                                            e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                                            e

                                                                                                            Hfrac14 1thorn e0

                                                                                                            H0frac14 1thorn e1 thorne

                                                                                                            H0

                                                                                                            ie

                                                                                                            e

                                                                                                            180frac14 1631thorne

                                                                                                            1710

                                                                                                            e frac14 2936

                                                                                                            1530frac14 0192

                                                                                                            Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                                            mv frac14 1

                                                                                                            1thorn e0 e0 e101 00

                                                                                                            frac14 1

                                                                                                            1823 0192

                                                                                                            0107frac14 098m2=MN

                                                                                                            k frac14 cvmvw frac14 265 098 98

                                                                                                            60 1440 365 103frac14 81 1010 m=s

                                                                                                            72

                                                                                                            Using Equation 77 (one-dimensional method)

                                                                                                            sc frac14 e0 e11thorn e0 H

                                                                                                            Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                                            Figure Q72

                                                                                                            52 Consolidation theory

                                                                                                            Settlement

                                                                                                            Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                                            1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                                            318

                                                                                                            Notes 5 92y 460thorn 84

                                                                                                            Heave

                                                                                                            Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                                            1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                                            38

                                                                                                            73

                                                                                                            U frac14 f ethTvTHORN frac14 f cvt

                                                                                                            d2

                                                                                                            Hence if cv is constant

                                                                                                            t1

                                                                                                            t2frac14 d

                                                                                                            21

                                                                                                            d22

                                                                                                            where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                                            d1 frac14 95mm and d2 frac14 2500mm

                                                                                                            for U frac14 050 t2 frac14 t1 d22

                                                                                                            d21

                                                                                                            frac14 20

                                                                                                            60 24 365 25002

                                                                                                            952frac14 263 years

                                                                                                            for U lt 060 Tv frac14

                                                                                                            4U2 (Equation 724(a))

                                                                                                            t030 frac14 t050 0302

                                                                                                            0502

                                                                                                            frac14 263 036 frac14 095 years

                                                                                                            Consolidation theory 53

                                                                                                            74

                                                                                                            The layer is open

                                                                                                            d frac14 8

                                                                                                            2frac14 4m

                                                                                                            Tv frac14 cvtd2frac14 24 3

                                                                                                            42frac14 0450

                                                                                                            ui frac14 frac14 84 kN=m2

                                                                                                            The excess pore water pressure is given by Equation 721

                                                                                                            ue frac14Xmfrac141mfrac140

                                                                                                            2ui

                                                                                                            Msin

                                                                                                            Mz

                                                                                                            d

                                                                                                            expethM2TvTHORN

                                                                                                            In this case z frac14 d

                                                                                                            sinMz

                                                                                                            d

                                                                                                            frac14 sinM

                                                                                                            where

                                                                                                            M frac14

                                                                                                            23

                                                                                                            25

                                                                                                            2

                                                                                                            M sin M M2Tv exp (M2Tv)

                                                                                                            2thorn1 1110 0329

                                                                                                            3

                                                                                                            21 9993 457 105

                                                                                                            ue frac14 2 84 2

                                                                                                            1 0329 ethother terms negligibleTHORN

                                                                                                            frac14 352 kN=m2

                                                                                                            75

                                                                                                            The layer is open

                                                                                                            d frac14 6

                                                                                                            2frac14 3m

                                                                                                            Tv frac14 cvtd2frac14 10 3

                                                                                                            32frac14 0333

                                                                                                            The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                                            54 Consolidation theory

                                                                                                            For an open layer

                                                                                                            Tv frac14 4n

                                                                                                            m2

                                                                                                            n frac14 0333 62

                                                                                                            4frac14 300

                                                                                                            The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                                            ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                                            i j

                                                                                                            0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                                            0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                                            The initial and 3-year isochrones are plotted in Figure Q75

                                                                                                            Area under initial isochrone frac14 180 units

                                                                                                            Area under 3-year isochrone frac14 63 units

                                                                                                            The average degree of consolidation is given by Equation 725Thus

                                                                                                            U frac14 1 63

                                                                                                            180frac14 065

                                                                                                            Figure Q75

                                                                                                            Consolidation theory 55

                                                                                                            76

                                                                                                            At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                            0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                            The final consolidation settlement (one-dimensional method) is

                                                                                                            sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                            Corrected time t frac14 2 1

                                                                                                            2

                                                                                                            40

                                                                                                            52

                                                                                                            frac14 1615 years

                                                                                                            Tv frac14 cvtd2frac14 44 1615

                                                                                                            42frac14 0444

                                                                                                            From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                            77

                                                                                                            The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                            Figure Q77

                                                                                                            56 Consolidation theory

                                                                                                            Point m n Ir (kNm2) sc (mm)

                                                                                                            13020frac14 15 20

                                                                                                            20frac14 10 0194 (4) 113 124

                                                                                                            260

                                                                                                            20frac14 30

                                                                                                            20

                                                                                                            20frac14 10 0204 (2) 59 65

                                                                                                            360

                                                                                                            20frac14 30

                                                                                                            40

                                                                                                            20frac14 20 0238 (1) 35 38

                                                                                                            430

                                                                                                            20frac14 15

                                                                                                            40

                                                                                                            20frac14 20 0224 (2) 65 72

                                                                                                            Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                            78

                                                                                                            Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                            (a) Immediate settlement

                                                                                                            H

                                                                                                            Bfrac14 30

                                                                                                            35frac14 086

                                                                                                            D

                                                                                                            Bfrac14 2

                                                                                                            35frac14 006

                                                                                                            Figure Q78

                                                                                                            Consolidation theory 57

                                                                                                            From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                            si frac14 130131qB

                                                                                                            Eufrac14 10 032 105 35

                                                                                                            40frac14 30mm

                                                                                                            (b) Consolidation settlement

                                                                                                            Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                            1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                            3150

                                                                                                            Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                            Now

                                                                                                            H

                                                                                                            Bfrac14 30

                                                                                                            35frac14 086 and A frac14 065

                                                                                                            from Figure 712 13 frac14 079

                                                                                                            sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                            Total settlement

                                                                                                            s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                            79

                                                                                                            Without sand drains

                                                                                                            Uv frac14 025

                                                                                                            Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                            t frac14 Tvd2

                                                                                                            cvfrac14 0049 82

                                                                                                            cvWith sand drains

                                                                                                            R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                            n frac14 Rrfrac14 169

                                                                                                            015frac14 113

                                                                                                            Tr frac14 cht

                                                                                                            4R2frac14 ch

                                                                                                            4 1692 0049 82

                                                                                                            cvethand ch frac14 cvTHORN

                                                                                                            frac14 0275

                                                                                                            Ur frac14 073 (from Figure 730)

                                                                                                            58 Consolidation theory

                                                                                                            Using Equation 740

                                                                                                            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                            U frac14 080

                                                                                                            710

                                                                                                            Without sand drains

                                                                                                            Uv frac14 090

                                                                                                            Tv frac14 0848

                                                                                                            t frac14 Tvd2

                                                                                                            cvfrac14 0848 102

                                                                                                            96frac14 88 years

                                                                                                            With sand drains

                                                                                                            R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                            n frac14 Rrfrac14 226

                                                                                                            015frac14 15

                                                                                                            Tr

                                                                                                            Tvfrac14 chcv

                                                                                                            d2

                                                                                                            4R2ethsame tTHORN

                                                                                                            Tr

                                                                                                            Tvfrac14 140

                                                                                                            96 102

                                                                                                            4 2262frac14 714 eth1THORN

                                                                                                            Using Equation 740

                                                                                                            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                            An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                            Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                            040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                            Thus

                                                                                                            Uv frac14 0295 and Ur frac14 086

                                                                                                            t frac14 88 00683

                                                                                                            0848frac14 07 years

                                                                                                            Consolidation theory 59

                                                                                                            Chapter 8

                                                                                                            Bearing capacity

                                                                                                            81

                                                                                                            (a) The ultimate bearing capacity is given by Equation 83

                                                                                                            qf frac14 cNc thorn DNq thorn 1

                                                                                                            2BN

                                                                                                            For u frac14 0

                                                                                                            Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                            qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                            The net ultimate bearing capacity is

                                                                                                            qnf frac14 qf D frac14 540 kN=m2

                                                                                                            The net foundation pressure is

                                                                                                            qn frac14 q D frac14 425

                                                                                                            2 eth21 1THORN frac14 192 kN=m2

                                                                                                            The factor of safety (Equation 86) is

                                                                                                            F frac14 qnfqnfrac14 540

                                                                                                            192frac14 28

                                                                                                            (b) For 0 frac14 28

                                                                                                            Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                            qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                            2 112 2 13

                                                                                                            frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                            qnf frac14 574 112 frac14 563 kN=m2

                                                                                                            F frac14 563

                                                                                                            192frac14 29

                                                                                                            (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                            82

                                                                                                            For 0 frac14 38

                                                                                                            Nq frac14 49 N frac14 67

                                                                                                            qnf frac14 DethNq 1THORN thorn 1

                                                                                                            2BN ethfrom Equation 83THORN

                                                                                                            frac14 eth18 075 48THORN thorn 1

                                                                                                            2 18 15 67

                                                                                                            frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                            qn frac14 500

                                                                                                            15 eth18 075THORN frac14 320 kN=m2

                                                                                                            F frac14 qnfqnfrac14 1553

                                                                                                            320frac14 48

                                                                                                            0d frac14 tan1tan 38

                                                                                                            125

                                                                                                            frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                            Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                            2 18 15 25

                                                                                                            frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                            Design load (action) Vd frac14 500 kN=m

                                                                                                            The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                            83

                                                                                                            D

                                                                                                            Bfrac14 350

                                                                                                            225frac14 155

                                                                                                            From Figure 85 for a square foundation

                                                                                                            Nc frac14 81

                                                                                                            Bearing capacity 61

                                                                                                            For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                            Nc frac14 084thorn 016B

                                                                                                            L

                                                                                                            81 frac14 745

                                                                                                            Using Equation 810

                                                                                                            qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                            For F frac14 3

                                                                                                            qn frac14 1006

                                                                                                            3frac14 335 kN=m2

                                                                                                            q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                            Design load frac14 405 450 225 frac14 4100 kN

                                                                                                            Design undrained strength cud frac14 135

                                                                                                            14frac14 96 kN=m2

                                                                                                            Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                            frac14 7241 kN

                                                                                                            Design load Vd frac14 4100 kN

                                                                                                            Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                            84

                                                                                                            For 0 frac14 40

                                                                                                            Nq frac14 64 N frac14 95

                                                                                                            qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                            (a) Water table 5m below ground level

                                                                                                            qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                            qn frac14 400 17 frac14 383 kN=m2

                                                                                                            F frac14 2686

                                                                                                            383frac14 70

                                                                                                            (b) Water table 1m below ground level (ie at foundation level)

                                                                                                            0 frac14 20 98 frac14 102 kN=m3

                                                                                                            62 Bearing capacity

                                                                                                            qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                            F frac14 2040

                                                                                                            383frac14 53

                                                                                                            (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                            eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                            qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                            F frac14 1296

                                                                                                            392frac14 33

                                                                                                            85

                                                                                                            The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                            Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                            Design value of 0 frac14 tan1tan 39

                                                                                                            125

                                                                                                            frac14 33

                                                                                                            For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                            Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                            Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                            86

                                                                                                            (a) Undrained shear for u frac14 0

                                                                                                            Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                            qnf frac14 12cuNc

                                                                                                            frac14 12 100 514 frac14 617 kN=m2

                                                                                                            qn frac14 qnfFfrac14 617

                                                                                                            3frac14 206 kN=m2

                                                                                                            q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                            Bearing capacity 63

                                                                                                            Drained shear for 0 frac14 32

                                                                                                            Nq frac14 23 N frac14 25

                                                                                                            0 frac14 21 98 frac14 112 kN=m3

                                                                                                            qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                            frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                            frac14 694 kN=m2

                                                                                                            q frac14 694

                                                                                                            3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                            Design load frac14 42 227 frac14 3632 kN

                                                                                                            (b) Design undrained strength cud frac14 100

                                                                                                            14frac14 71 kNm2

                                                                                                            Design bearing resistance Rd frac14 12cudNe area

                                                                                                            frac14 12 71 514 42

                                                                                                            frac14 7007 kN

                                                                                                            For drained shear 0d frac14 tan1tan 32

                                                                                                            125

                                                                                                            frac14 26

                                                                                                            Nq frac14 12 N frac14 10

                                                                                                            Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                            (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                            Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                            1 2 100 0175 0700qn 0182qn

                                                                                                            2 6 033 0044 0176qn 0046qn

                                                                                                            3 10 020 0017 0068qn 0018qn

                                                                                                            0246qn

                                                                                                            Diameter of equivalent circle B frac14 45m

                                                                                                            H

                                                                                                            Bfrac14 12

                                                                                                            45frac14 27 and A frac14 042

                                                                                                            13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                            64 Bearing capacity

                                                                                                            For sc frac14 30mm

                                                                                                            qn frac14 30

                                                                                                            0147frac14 204 kN=m2

                                                                                                            q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                            Design load frac14 42 225 frac14 3600 kN

                                                                                                            The design load is 3600 kN settlement being the limiting criterion

                                                                                                            87

                                                                                                            D

                                                                                                            Bfrac14 8

                                                                                                            4frac14 20

                                                                                                            From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                            F frac14 cuNc

                                                                                                            Dfrac14 40 71

                                                                                                            20 8frac14 18

                                                                                                            88

                                                                                                            Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                            Design value of 0 frac14 tan1tan 38

                                                                                                            125

                                                                                                            frac14 32

                                                                                                            Figure Q86

                                                                                                            Bearing capacity 65

                                                                                                            For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                            Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                            The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                            Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                            For B frac14 250m qn frac14 3750

                                                                                                            2502 17 frac14 583 kN=m2

                                                                                                            From Figure 510 m frac14 n frac14 126

                                                                                                            6frac14 021

                                                                                                            Ir frac14 0019

                                                                                                            Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                            Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                            The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                            89

                                                                                                            Depth (m) N 0v (kNm2) CN N1

                                                                                                            070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                            Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                            (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                            Cw frac14 05thorn 0530

                                                                                                            47

                                                                                                            frac14 082

                                                                                                            66 Bearing capacity

                                                                                                            Thus

                                                                                                            qa frac14 150 082 frac14 120 kN=m2

                                                                                                            (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                            Thus

                                                                                                            qa frac14 90 15 frac14 135 kN=m2

                                                                                                            (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                            Ic frac14 171

                                                                                                            1014frac14 0068

                                                                                                            From Equation 819(a) with s frac14 25mm

                                                                                                            q frac14 25

                                                                                                            3507 0068frac14 150 kN=m2

                                                                                                            810

                                                                                                            Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                            Izp frac14 05thorn 01130

                                                                                                            16 27

                                                                                                            05

                                                                                                            frac14 067

                                                                                                            Refer to Figure Q810

                                                                                                            E frac14 25qc

                                                                                                            Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                            Ez (mm3MN)

                                                                                                            1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                            0203

                                                                                                            C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                            130frac14 093

                                                                                                            C2 frac14 1 ethsayTHORN

                                                                                                            s frac14 C1C2qnX Iz

                                                                                                            Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                            Bearing capacity 67

                                                                                                            811

                                                                                                            At pile base level

                                                                                                            cu frac14 220 kN=m2

                                                                                                            qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                            Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                            00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                            qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                            Then

                                                                                                            Qf frac14 Abqb thorn Asqs

                                                                                                            frac14

                                                                                                            4 32 1980

                                                                                                            thorn eth 105 139 86THORN

                                                                                                            frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                            0 01 02 03 04 05 06 07

                                                                                                            0 2 4 6 8 10 12 14

                                                                                                            1

                                                                                                            2

                                                                                                            3

                                                                                                            4

                                                                                                            5

                                                                                                            6

                                                                                                            7

                                                                                                            8

                                                                                                            (1)

                                                                                                            (2)

                                                                                                            (3)

                                                                                                            (4)

                                                                                                            (5)

                                                                                                            qc

                                                                                                            qc

                                                                                                            Iz

                                                                                                            Iz

                                                                                                            (MNm2)

                                                                                                            z (m)

                                                                                                            Figure Q810

                                                                                                            68 Bearing capacity

                                                                                                            Allowable load

                                                                                                            ethaTHORN Qf

                                                                                                            2frac14 17 937

                                                                                                            2frac14 8968 kN

                                                                                                            ethbTHORN Abqb

                                                                                                            3thorn Asqs frac14 13 996

                                                                                                            3thorn 3941 frac14 8606 kN

                                                                                                            ie allowable load frac14 8600 kN

                                                                                                            Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                            According to the limit state method

                                                                                                            Characteristic undrained strength at base level cuk frac14 220

                                                                                                            150kN=m2

                                                                                                            Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                            150frac14 1320 kN=m2

                                                                                                            Characteristic shaft resistance qsk frac14 00150

                                                                                                            frac14 86

                                                                                                            150frac14 57 kN=m2

                                                                                                            Characteristic base and shaft resistances

                                                                                                            Rbk frac14

                                                                                                            4 32 1320 frac14 9330 kN

                                                                                                            Rsk frac14 105 139 86

                                                                                                            150frac14 2629 kN

                                                                                                            For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                            Design bearing resistance Rcd frac14 9330

                                                                                                            160thorn 2629

                                                                                                            130

                                                                                                            frac14 5831thorn 2022

                                                                                                            frac14 7850 kN

                                                                                                            Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                            812

                                                                                                            ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                            qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                            For a single pile

                                                                                                            Qf frac14 Abqb thorn Asqs

                                                                                                            frac14

                                                                                                            4 062 1305

                                                                                                            thorn eth 06 15 42THORN

                                                                                                            frac14 369thorn 1187 frac14 1556 kN

                                                                                                            Bearing capacity 69

                                                                                                            Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                            eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                            (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                            (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                            qbkfrac14 9cuk frac14 9 220

                                                                                                            150frac14 1320 kN=m2

                                                                                                            qskfrac14cuk frac14 040 105

                                                                                                            150frac14 28 kN=m2

                                                                                                            Rbkfrac14

                                                                                                            4 0602 1320 frac14 373 kN

                                                                                                            Rskfrac14 060 15 28 frac14 791 kN

                                                                                                            Rcdfrac14 373

                                                                                                            160thorn 791

                                                                                                            130frac14 233thorn 608 frac14 841 kN

                                                                                                            Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                            Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                            (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                            q frac14 21 000

                                                                                                            1762frac14 68 kN=m2

                                                                                                            Immediate settlement

                                                                                                            H

                                                                                                            Bfrac14 15

                                                                                                            176frac14 085

                                                                                                            D

                                                                                                            Bfrac14 13

                                                                                                            176frac14 074

                                                                                                            L

                                                                                                            Bfrac14 1

                                                                                                            Hence from Figure 515

                                                                                                            130 frac14 078 and 131 frac14 041

                                                                                                            70 Bearing capacity

                                                                                                            Thus using Equation 528

                                                                                                            si frac14 078 041 68 176

                                                                                                            65frac14 6mm

                                                                                                            Consolidation settlement

                                                                                                            Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                            1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                            434 (sod)

                                                                                                            Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                            sc frac14 056 434 frac14 24mm

                                                                                                            The total settlement is (6thorn 24) frac14 30mm

                                                                                                            813

                                                                                                            At base level N frac14 26 Then using Equation 830

                                                                                                            qb frac14 40NDb

                                                                                                            Bfrac14 40 26 2

                                                                                                            025frac14 8320 kN=m2

                                                                                                            ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                            Figure Q812

                                                                                                            Bearing capacity 71

                                                                                                            Over the length embedded in sand

                                                                                                            N frac14 21 ie18thorn 24

                                                                                                            2

                                                                                                            Using Equation 831

                                                                                                            qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                            For a single pile

                                                                                                            Qf frac14 Abqb thorn Asqs

                                                                                                            frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                            For the pile group assuming a group efficiency of 12

                                                                                                            XQf frac14 12 9 604 frac14 6523 kN

                                                                                                            Then the load factor is

                                                                                                            F frac14 6523

                                                                                                            2000thorn 1000frac14 21

                                                                                                            (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                            Characteristic base resistance per unit area qbk frac14 8320

                                                                                                            150frac14 5547 kNm2

                                                                                                            Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                            150frac14 28 kNm2

                                                                                                            Characteristic base and shaft resistances for a single pile

                                                                                                            Rbk frac14 0252 5547 frac14 347 kN

                                                                                                            Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                            For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                            Design bearing resistance Rcd frac14 347

                                                                                                            130thorn 56

                                                                                                            130frac14 310 kN

                                                                                                            For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                            Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                            (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                            From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                            72 Bearing capacity

                                                                                                            N frac14 24thorn 26thorn 34

                                                                                                            3frac14 28

                                                                                                            Ic frac14 171

                                                                                                            2814frac14 0016 ethEquation 818THORN

                                                                                                            s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                            814

                                                                                                            Using Equation 841

                                                                                                            Tf frac14 DLcu thorn

                                                                                                            4ethD2 d2THORNcuNc

                                                                                                            frac14 eth 02 5 06 110THORN thorn

                                                                                                            4eth022 012THORN110 9

                                                                                                            frac14 207thorn 23 frac14 230 kN

                                                                                                            Figure Q813

                                                                                                            Bearing capacity 73

                                                                                                            Chapter 9

                                                                                                            Stability of slopes

                                                                                                            91

                                                                                                            Referring to Figure Q91

                                                                                                            W frac14 417 19 frac14 792 kN=m

                                                                                                            Q frac14 20 28 frac14 56 kN=m

                                                                                                            Arc lengthAB frac14

                                                                                                            180 73 90 frac14 115m

                                                                                                            Arc length BC frac14

                                                                                                            180 28 90 frac14 44m

                                                                                                            The factor of safety is given by

                                                                                                            F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                            90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                            Depth of tension crack z0 frac14 2cu

                                                                                                            frac14 2 20

                                                                                                            19frac14 21m

                                                                                                            Arc length BD frac14

                                                                                                            180 13

                                                                                                            1

                                                                                                            2 90 frac14 21m

                                                                                                            F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                            The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                            Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                            14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                            Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                            The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                            92

                                                                                                            u frac14 0

                                                                                                            Depth factor D frac14 11

                                                                                                            9frac14 122

                                                                                                            Using Equation 92 with F frac14 10

                                                                                                            Ns frac14 cu

                                                                                                            FHfrac14 30

                                                                                                            10 19 9frac14 0175

                                                                                                            Hence from Figure 93

                                                                                                            frac14 50

                                                                                                            For F frac14 12

                                                                                                            Ns frac14 30

                                                                                                            12 19 9frac14 0146

                                                                                                            frac14 27

                                                                                                            93

                                                                                                            Refer to Figure Q93

                                                                                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                            1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                            74 m

                                                                                                            214 1deg

                                                                                                            213 1deg

                                                                                                            39 m

                                                                                                            WB

                                                                                                            D

                                                                                                            C

                                                                                                            28 m

                                                                                                            21 m

                                                                                                            A

                                                                                                            Q

                                                                                                            Soil (1)Soil (2)

                                                                                                            73deg

                                                                                                            Figure Q91

                                                                                                            Stability of slopes 75

                                                                                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                            9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                            599 256 328 1372

                                                                                                            Figure Q93

                                                                                                            76 Stability of slopes

                                                                                                            XW cos frac14 b

                                                                                                            Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                            W sin frac14 bX

                                                                                                            h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                            Arc length La frac14

                                                                                                            180 57

                                                                                                            1

                                                                                                            2 326 frac14 327m

                                                                                                            The factor of safety is given by

                                                                                                            F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                            W sin

                                                                                                            frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                            frac14 091

                                                                                                            According to the limit state method

                                                                                                            0d frac14 tan1tan 32

                                                                                                            125

                                                                                                            frac14 265

                                                                                                            c0 frac14 8

                                                                                                            160frac14 5 kN=m2

                                                                                                            Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                            Design disturbing moment frac14 1075 kN=m

                                                                                                            The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                            94

                                                                                                            F frac14 1

                                                                                                            W sin

                                                                                                            Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                            1thorn ethtan tan0=FTHORN

                                                                                                            c0 frac14 8 kN=m2

                                                                                                            0 frac14 32

                                                                                                            c0b frac14 8 2 frac14 16 kN=m

                                                                                                            W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                            Try F frac14 100

                                                                                                            tan0

                                                                                                            Ffrac14 0625

                                                                                                            Stability of slopes 77

                                                                                                            Values of u are as obtained in Figure Q93

                                                                                                            SliceNo

                                                                                                            h(m)

                                                                                                            W frac14 bh(kNm)

                                                                                                            W sin(kNm)

                                                                                                            ub(kNm)

                                                                                                            c0bthorn (W ub) tan0(kNm)

                                                                                                            sec

                                                                                                            1thorn (tan tan0)FProduct(kNm)

                                                                                                            1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                            23 33 30 1042 31

                                                                                                            3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                            224 92 72 0931 67

                                                                                                            6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                            12 58 112 90 0889 80

                                                                                                            8 60 252 1812

                                                                                                            80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                            10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                            2154 88 116 0853 99

                                                                                                            14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                            1074 1091

                                                                                                            F frac14 1091

                                                                                                            1074frac14 102 (assumed value 100)

                                                                                                            Thus

                                                                                                            F frac14 101

                                                                                                            95

                                                                                                            F frac14 1

                                                                                                            W sin

                                                                                                            XfWeth1 ruTHORN tan0g sec

                                                                                                            1thorn ethtan tan0THORN=F

                                                                                                            0 frac14 33

                                                                                                            ru frac14 020

                                                                                                            W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                            eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                            Try F frac14 110

                                                                                                            tan 0

                                                                                                            Ffrac14 tan 33

                                                                                                            110frac14 0590

                                                                                                            78 Stability of slopes

                                                                                                            Referring to Figure Q95

                                                                                                            SliceNo

                                                                                                            h(m)

                                                                                                            W frac14 bh(kNm)

                                                                                                            W sin(kNm)

                                                                                                            W(1 ru) tan0(kNm)

                                                                                                            sec

                                                                                                            1thorn ( tan tan0)FProduct(kNm)

                                                                                                            1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                            2120 234 0892 209

                                                                                                            4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                            1185 1271

                                                                                                            Figure Q95

                                                                                                            Stability of slopes 79

                                                                                                            F frac14 1271

                                                                                                            1185frac14 107

                                                                                                            The trial value was 110 therefore take F to be 108

                                                                                                            96

                                                                                                            (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                            F frac14 0

                                                                                                            sat

                                                                                                            tan0

                                                                                                            tan

                                                                                                            ptie 15 frac14 92

                                                                                                            19

                                                                                                            tan 36

                                                                                                            tan

                                                                                                            tan frac14 0234

                                                                                                            frac14 13

                                                                                                            Water table well below surface the factor of safety is given by Equation 911

                                                                                                            F frac14 tan0

                                                                                                            tan

                                                                                                            frac14 tan 36

                                                                                                            tan 13

                                                                                                            frac14 31

                                                                                                            (b) 0d frac14 tan1tan 36

                                                                                                            125

                                                                                                            frac14 30

                                                                                                            Depth of potential failure surface frac14 z

                                                                                                            Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                            frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                            frac14 504z kN

                                                                                                            Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                            frac14 19 z sin 13 cos 13

                                                                                                            frac14 416z kN

                                                                                                            Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                            80 Stability of slopes

                                                                                                            • Book Cover
                                                                                                            • Title
                                                                                                            • Contents
                                                                                                            • Basic characteristics of soils
                                                                                                            • Seepage
                                                                                                            • Effective stress
                                                                                                            • Shear strength
                                                                                                            • Stresses and displacements
                                                                                                            • Lateral earth pressure
                                                                                                            • Consolidation theory
                                                                                                            • Bearing capacity
                                                                                                            • Stability of slopes

                                                                                                              (b) In the coherent gravity method K frac14 032 (K varying linearly between 041 at thesurface and 026 at a depth of 6m) and Le frac14 420m

                                                                                                              Tp frac14 032 68 120 065 frac14 170 kN

                                                                                                              Tr frac14 213 420

                                                                                                              418frac14 214 kN

                                                                                                              Again the tensile failure and slipping limit states are satisfied for this element

                                                                                                              Figure Q612

                                                                                                              Lateral earth pressure 49

                                                                                                              Chapter 7

                                                                                                              Consolidation theory

                                                                                                              71

                                                                                                              Total change in thickness

                                                                                                              H frac14 782 602 frac14 180mm

                                                                                                              Average thickness frac14 1530thorn 180

                                                                                                              2frac14 1620mm

                                                                                                              Length of drainage path d frac14 1620

                                                                                                              2frac14 810mm

                                                                                                              Root time plot (Figure Q71a)

                                                                                                              ffiffiffiffiffiffit90p frac14 33

                                                                                                              t90 frac14 109min

                                                                                                              cv frac14 0848d2

                                                                                                              t90frac14 0848 8102

                                                                                                              109 1440 365

                                                                                                              106frac14 27m2=year

                                                                                                              r0 frac14 782 764

                                                                                                              782 602frac14 018

                                                                                                              180frac14 0100

                                                                                                              rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                                              10 119

                                                                                                              9 180frac14 0735

                                                                                                              rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                                              Log time plot (Figure Q71b)

                                                                                                              t50 frac14 26min

                                                                                                              cv frac14 0196d2

                                                                                                              t50frac14 0196 8102

                                                                                                              26 1440 365

                                                                                                              106frac14 26m2=year

                                                                                                              r0 frac14 782 763

                                                                                                              782 602frac14 019

                                                                                                              180frac14 0106

                                                                                                              rp frac14 763 623

                                                                                                              782 602frac14 140

                                                                                                              180frac14 0778

                                                                                                              rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                                              Figure Q71(a)

                                                                                                              Figure Q71(b)

                                                                                                              Final void ratio

                                                                                                              e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                                              e

                                                                                                              Hfrac14 1thorn e0

                                                                                                              H0frac14 1thorn e1 thorne

                                                                                                              H0

                                                                                                              ie

                                                                                                              e

                                                                                                              180frac14 1631thorne

                                                                                                              1710

                                                                                                              e frac14 2936

                                                                                                              1530frac14 0192

                                                                                                              Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                                              mv frac14 1

                                                                                                              1thorn e0 e0 e101 00

                                                                                                              frac14 1

                                                                                                              1823 0192

                                                                                                              0107frac14 098m2=MN

                                                                                                              k frac14 cvmvw frac14 265 098 98

                                                                                                              60 1440 365 103frac14 81 1010 m=s

                                                                                                              72

                                                                                                              Using Equation 77 (one-dimensional method)

                                                                                                              sc frac14 e0 e11thorn e0 H

                                                                                                              Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                                              Figure Q72

                                                                                                              52 Consolidation theory

                                                                                                              Settlement

                                                                                                              Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                                              1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                                              318

                                                                                                              Notes 5 92y 460thorn 84

                                                                                                              Heave

                                                                                                              Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                                              1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                                              38

                                                                                                              73

                                                                                                              U frac14 f ethTvTHORN frac14 f cvt

                                                                                                              d2

                                                                                                              Hence if cv is constant

                                                                                                              t1

                                                                                                              t2frac14 d

                                                                                                              21

                                                                                                              d22

                                                                                                              where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                                              d1 frac14 95mm and d2 frac14 2500mm

                                                                                                              for U frac14 050 t2 frac14 t1 d22

                                                                                                              d21

                                                                                                              frac14 20

                                                                                                              60 24 365 25002

                                                                                                              952frac14 263 years

                                                                                                              for U lt 060 Tv frac14

                                                                                                              4U2 (Equation 724(a))

                                                                                                              t030 frac14 t050 0302

                                                                                                              0502

                                                                                                              frac14 263 036 frac14 095 years

                                                                                                              Consolidation theory 53

                                                                                                              74

                                                                                                              The layer is open

                                                                                                              d frac14 8

                                                                                                              2frac14 4m

                                                                                                              Tv frac14 cvtd2frac14 24 3

                                                                                                              42frac14 0450

                                                                                                              ui frac14 frac14 84 kN=m2

                                                                                                              The excess pore water pressure is given by Equation 721

                                                                                                              ue frac14Xmfrac141mfrac140

                                                                                                              2ui

                                                                                                              Msin

                                                                                                              Mz

                                                                                                              d

                                                                                                              expethM2TvTHORN

                                                                                                              In this case z frac14 d

                                                                                                              sinMz

                                                                                                              d

                                                                                                              frac14 sinM

                                                                                                              where

                                                                                                              M frac14

                                                                                                              23

                                                                                                              25

                                                                                                              2

                                                                                                              M sin M M2Tv exp (M2Tv)

                                                                                                              2thorn1 1110 0329

                                                                                                              3

                                                                                                              21 9993 457 105

                                                                                                              ue frac14 2 84 2

                                                                                                              1 0329 ethother terms negligibleTHORN

                                                                                                              frac14 352 kN=m2

                                                                                                              75

                                                                                                              The layer is open

                                                                                                              d frac14 6

                                                                                                              2frac14 3m

                                                                                                              Tv frac14 cvtd2frac14 10 3

                                                                                                              32frac14 0333

                                                                                                              The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                                              54 Consolidation theory

                                                                                                              For an open layer

                                                                                                              Tv frac14 4n

                                                                                                              m2

                                                                                                              n frac14 0333 62

                                                                                                              4frac14 300

                                                                                                              The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                                              ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                                              i j

                                                                                                              0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                                              0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                                              The initial and 3-year isochrones are plotted in Figure Q75

                                                                                                              Area under initial isochrone frac14 180 units

                                                                                                              Area under 3-year isochrone frac14 63 units

                                                                                                              The average degree of consolidation is given by Equation 725Thus

                                                                                                              U frac14 1 63

                                                                                                              180frac14 065

                                                                                                              Figure Q75

                                                                                                              Consolidation theory 55

                                                                                                              76

                                                                                                              At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                              0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                              The final consolidation settlement (one-dimensional method) is

                                                                                                              sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                              Corrected time t frac14 2 1

                                                                                                              2

                                                                                                              40

                                                                                                              52

                                                                                                              frac14 1615 years

                                                                                                              Tv frac14 cvtd2frac14 44 1615

                                                                                                              42frac14 0444

                                                                                                              From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                              77

                                                                                                              The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                              Figure Q77

                                                                                                              56 Consolidation theory

                                                                                                              Point m n Ir (kNm2) sc (mm)

                                                                                                              13020frac14 15 20

                                                                                                              20frac14 10 0194 (4) 113 124

                                                                                                              260

                                                                                                              20frac14 30

                                                                                                              20

                                                                                                              20frac14 10 0204 (2) 59 65

                                                                                                              360

                                                                                                              20frac14 30

                                                                                                              40

                                                                                                              20frac14 20 0238 (1) 35 38

                                                                                                              430

                                                                                                              20frac14 15

                                                                                                              40

                                                                                                              20frac14 20 0224 (2) 65 72

                                                                                                              Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                              78

                                                                                                              Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                              (a) Immediate settlement

                                                                                                              H

                                                                                                              Bfrac14 30

                                                                                                              35frac14 086

                                                                                                              D

                                                                                                              Bfrac14 2

                                                                                                              35frac14 006

                                                                                                              Figure Q78

                                                                                                              Consolidation theory 57

                                                                                                              From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                              si frac14 130131qB

                                                                                                              Eufrac14 10 032 105 35

                                                                                                              40frac14 30mm

                                                                                                              (b) Consolidation settlement

                                                                                                              Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                              1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                              3150

                                                                                                              Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                              Now

                                                                                                              H

                                                                                                              Bfrac14 30

                                                                                                              35frac14 086 and A frac14 065

                                                                                                              from Figure 712 13 frac14 079

                                                                                                              sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                              Total settlement

                                                                                                              s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                              79

                                                                                                              Without sand drains

                                                                                                              Uv frac14 025

                                                                                                              Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                              t frac14 Tvd2

                                                                                                              cvfrac14 0049 82

                                                                                                              cvWith sand drains

                                                                                                              R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                              n frac14 Rrfrac14 169

                                                                                                              015frac14 113

                                                                                                              Tr frac14 cht

                                                                                                              4R2frac14 ch

                                                                                                              4 1692 0049 82

                                                                                                              cvethand ch frac14 cvTHORN

                                                                                                              frac14 0275

                                                                                                              Ur frac14 073 (from Figure 730)

                                                                                                              58 Consolidation theory

                                                                                                              Using Equation 740

                                                                                                              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                              U frac14 080

                                                                                                              710

                                                                                                              Without sand drains

                                                                                                              Uv frac14 090

                                                                                                              Tv frac14 0848

                                                                                                              t frac14 Tvd2

                                                                                                              cvfrac14 0848 102

                                                                                                              96frac14 88 years

                                                                                                              With sand drains

                                                                                                              R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                              n frac14 Rrfrac14 226

                                                                                                              015frac14 15

                                                                                                              Tr

                                                                                                              Tvfrac14 chcv

                                                                                                              d2

                                                                                                              4R2ethsame tTHORN

                                                                                                              Tr

                                                                                                              Tvfrac14 140

                                                                                                              96 102

                                                                                                              4 2262frac14 714 eth1THORN

                                                                                                              Using Equation 740

                                                                                                              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                              An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                              Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                              040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                              Thus

                                                                                                              Uv frac14 0295 and Ur frac14 086

                                                                                                              t frac14 88 00683

                                                                                                              0848frac14 07 years

                                                                                                              Consolidation theory 59

                                                                                                              Chapter 8

                                                                                                              Bearing capacity

                                                                                                              81

                                                                                                              (a) The ultimate bearing capacity is given by Equation 83

                                                                                                              qf frac14 cNc thorn DNq thorn 1

                                                                                                              2BN

                                                                                                              For u frac14 0

                                                                                                              Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                              qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                              The net ultimate bearing capacity is

                                                                                                              qnf frac14 qf D frac14 540 kN=m2

                                                                                                              The net foundation pressure is

                                                                                                              qn frac14 q D frac14 425

                                                                                                              2 eth21 1THORN frac14 192 kN=m2

                                                                                                              The factor of safety (Equation 86) is

                                                                                                              F frac14 qnfqnfrac14 540

                                                                                                              192frac14 28

                                                                                                              (b) For 0 frac14 28

                                                                                                              Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                              qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                              2 112 2 13

                                                                                                              frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                              qnf frac14 574 112 frac14 563 kN=m2

                                                                                                              F frac14 563

                                                                                                              192frac14 29

                                                                                                              (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                              82

                                                                                                              For 0 frac14 38

                                                                                                              Nq frac14 49 N frac14 67

                                                                                                              qnf frac14 DethNq 1THORN thorn 1

                                                                                                              2BN ethfrom Equation 83THORN

                                                                                                              frac14 eth18 075 48THORN thorn 1

                                                                                                              2 18 15 67

                                                                                                              frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                              qn frac14 500

                                                                                                              15 eth18 075THORN frac14 320 kN=m2

                                                                                                              F frac14 qnfqnfrac14 1553

                                                                                                              320frac14 48

                                                                                                              0d frac14 tan1tan 38

                                                                                                              125

                                                                                                              frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                              Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                              2 18 15 25

                                                                                                              frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                              Design load (action) Vd frac14 500 kN=m

                                                                                                              The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                              83

                                                                                                              D

                                                                                                              Bfrac14 350

                                                                                                              225frac14 155

                                                                                                              From Figure 85 for a square foundation

                                                                                                              Nc frac14 81

                                                                                                              Bearing capacity 61

                                                                                                              For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                              Nc frac14 084thorn 016B

                                                                                                              L

                                                                                                              81 frac14 745

                                                                                                              Using Equation 810

                                                                                                              qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                              For F frac14 3

                                                                                                              qn frac14 1006

                                                                                                              3frac14 335 kN=m2

                                                                                                              q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                              Design load frac14 405 450 225 frac14 4100 kN

                                                                                                              Design undrained strength cud frac14 135

                                                                                                              14frac14 96 kN=m2

                                                                                                              Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                              frac14 7241 kN

                                                                                                              Design load Vd frac14 4100 kN

                                                                                                              Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                              84

                                                                                                              For 0 frac14 40

                                                                                                              Nq frac14 64 N frac14 95

                                                                                                              qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                              (a) Water table 5m below ground level

                                                                                                              qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                              qn frac14 400 17 frac14 383 kN=m2

                                                                                                              F frac14 2686

                                                                                                              383frac14 70

                                                                                                              (b) Water table 1m below ground level (ie at foundation level)

                                                                                                              0 frac14 20 98 frac14 102 kN=m3

                                                                                                              62 Bearing capacity

                                                                                                              qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                              F frac14 2040

                                                                                                              383frac14 53

                                                                                                              (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                              eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                              qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                              F frac14 1296

                                                                                                              392frac14 33

                                                                                                              85

                                                                                                              The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                              Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                              Design value of 0 frac14 tan1tan 39

                                                                                                              125

                                                                                                              frac14 33

                                                                                                              For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                              Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                              Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                              86

                                                                                                              (a) Undrained shear for u frac14 0

                                                                                                              Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                              qnf frac14 12cuNc

                                                                                                              frac14 12 100 514 frac14 617 kN=m2

                                                                                                              qn frac14 qnfFfrac14 617

                                                                                                              3frac14 206 kN=m2

                                                                                                              q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                              Bearing capacity 63

                                                                                                              Drained shear for 0 frac14 32

                                                                                                              Nq frac14 23 N frac14 25

                                                                                                              0 frac14 21 98 frac14 112 kN=m3

                                                                                                              qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                              frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                              frac14 694 kN=m2

                                                                                                              q frac14 694

                                                                                                              3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                              Design load frac14 42 227 frac14 3632 kN

                                                                                                              (b) Design undrained strength cud frac14 100

                                                                                                              14frac14 71 kNm2

                                                                                                              Design bearing resistance Rd frac14 12cudNe area

                                                                                                              frac14 12 71 514 42

                                                                                                              frac14 7007 kN

                                                                                                              For drained shear 0d frac14 tan1tan 32

                                                                                                              125

                                                                                                              frac14 26

                                                                                                              Nq frac14 12 N frac14 10

                                                                                                              Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                              (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                              Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                              1 2 100 0175 0700qn 0182qn

                                                                                                              2 6 033 0044 0176qn 0046qn

                                                                                                              3 10 020 0017 0068qn 0018qn

                                                                                                              0246qn

                                                                                                              Diameter of equivalent circle B frac14 45m

                                                                                                              H

                                                                                                              Bfrac14 12

                                                                                                              45frac14 27 and A frac14 042

                                                                                                              13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                              64 Bearing capacity

                                                                                                              For sc frac14 30mm

                                                                                                              qn frac14 30

                                                                                                              0147frac14 204 kN=m2

                                                                                                              q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                              Design load frac14 42 225 frac14 3600 kN

                                                                                                              The design load is 3600 kN settlement being the limiting criterion

                                                                                                              87

                                                                                                              D

                                                                                                              Bfrac14 8

                                                                                                              4frac14 20

                                                                                                              From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                              F frac14 cuNc

                                                                                                              Dfrac14 40 71

                                                                                                              20 8frac14 18

                                                                                                              88

                                                                                                              Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                              Design value of 0 frac14 tan1tan 38

                                                                                                              125

                                                                                                              frac14 32

                                                                                                              Figure Q86

                                                                                                              Bearing capacity 65

                                                                                                              For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                              Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                              The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                              Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                              For B frac14 250m qn frac14 3750

                                                                                                              2502 17 frac14 583 kN=m2

                                                                                                              From Figure 510 m frac14 n frac14 126

                                                                                                              6frac14 021

                                                                                                              Ir frac14 0019

                                                                                                              Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                              Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                              The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                              89

                                                                                                              Depth (m) N 0v (kNm2) CN N1

                                                                                                              070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                              Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                              (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                              Cw frac14 05thorn 0530

                                                                                                              47

                                                                                                              frac14 082

                                                                                                              66 Bearing capacity

                                                                                                              Thus

                                                                                                              qa frac14 150 082 frac14 120 kN=m2

                                                                                                              (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                              Thus

                                                                                                              qa frac14 90 15 frac14 135 kN=m2

                                                                                                              (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                              Ic frac14 171

                                                                                                              1014frac14 0068

                                                                                                              From Equation 819(a) with s frac14 25mm

                                                                                                              q frac14 25

                                                                                                              3507 0068frac14 150 kN=m2

                                                                                                              810

                                                                                                              Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                              Izp frac14 05thorn 01130

                                                                                                              16 27

                                                                                                              05

                                                                                                              frac14 067

                                                                                                              Refer to Figure Q810

                                                                                                              E frac14 25qc

                                                                                                              Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                              Ez (mm3MN)

                                                                                                              1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                              0203

                                                                                                              C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                              130frac14 093

                                                                                                              C2 frac14 1 ethsayTHORN

                                                                                                              s frac14 C1C2qnX Iz

                                                                                                              Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                              Bearing capacity 67

                                                                                                              811

                                                                                                              At pile base level

                                                                                                              cu frac14 220 kN=m2

                                                                                                              qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                              Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                              00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                              qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                              Then

                                                                                                              Qf frac14 Abqb thorn Asqs

                                                                                                              frac14

                                                                                                              4 32 1980

                                                                                                              thorn eth 105 139 86THORN

                                                                                                              frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                              0 01 02 03 04 05 06 07

                                                                                                              0 2 4 6 8 10 12 14

                                                                                                              1

                                                                                                              2

                                                                                                              3

                                                                                                              4

                                                                                                              5

                                                                                                              6

                                                                                                              7

                                                                                                              8

                                                                                                              (1)

                                                                                                              (2)

                                                                                                              (3)

                                                                                                              (4)

                                                                                                              (5)

                                                                                                              qc

                                                                                                              qc

                                                                                                              Iz

                                                                                                              Iz

                                                                                                              (MNm2)

                                                                                                              z (m)

                                                                                                              Figure Q810

                                                                                                              68 Bearing capacity

                                                                                                              Allowable load

                                                                                                              ethaTHORN Qf

                                                                                                              2frac14 17 937

                                                                                                              2frac14 8968 kN

                                                                                                              ethbTHORN Abqb

                                                                                                              3thorn Asqs frac14 13 996

                                                                                                              3thorn 3941 frac14 8606 kN

                                                                                                              ie allowable load frac14 8600 kN

                                                                                                              Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                              According to the limit state method

                                                                                                              Characteristic undrained strength at base level cuk frac14 220

                                                                                                              150kN=m2

                                                                                                              Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                              150frac14 1320 kN=m2

                                                                                                              Characteristic shaft resistance qsk frac14 00150

                                                                                                              frac14 86

                                                                                                              150frac14 57 kN=m2

                                                                                                              Characteristic base and shaft resistances

                                                                                                              Rbk frac14

                                                                                                              4 32 1320 frac14 9330 kN

                                                                                                              Rsk frac14 105 139 86

                                                                                                              150frac14 2629 kN

                                                                                                              For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                              Design bearing resistance Rcd frac14 9330

                                                                                                              160thorn 2629

                                                                                                              130

                                                                                                              frac14 5831thorn 2022

                                                                                                              frac14 7850 kN

                                                                                                              Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                              812

                                                                                                              ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                              qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                              For a single pile

                                                                                                              Qf frac14 Abqb thorn Asqs

                                                                                                              frac14

                                                                                                              4 062 1305

                                                                                                              thorn eth 06 15 42THORN

                                                                                                              frac14 369thorn 1187 frac14 1556 kN

                                                                                                              Bearing capacity 69

                                                                                                              Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                              eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                              (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                              (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                              qbkfrac14 9cuk frac14 9 220

                                                                                                              150frac14 1320 kN=m2

                                                                                                              qskfrac14cuk frac14 040 105

                                                                                                              150frac14 28 kN=m2

                                                                                                              Rbkfrac14

                                                                                                              4 0602 1320 frac14 373 kN

                                                                                                              Rskfrac14 060 15 28 frac14 791 kN

                                                                                                              Rcdfrac14 373

                                                                                                              160thorn 791

                                                                                                              130frac14 233thorn 608 frac14 841 kN

                                                                                                              Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                              Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                              (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                              q frac14 21 000

                                                                                                              1762frac14 68 kN=m2

                                                                                                              Immediate settlement

                                                                                                              H

                                                                                                              Bfrac14 15

                                                                                                              176frac14 085

                                                                                                              D

                                                                                                              Bfrac14 13

                                                                                                              176frac14 074

                                                                                                              L

                                                                                                              Bfrac14 1

                                                                                                              Hence from Figure 515

                                                                                                              130 frac14 078 and 131 frac14 041

                                                                                                              70 Bearing capacity

                                                                                                              Thus using Equation 528

                                                                                                              si frac14 078 041 68 176

                                                                                                              65frac14 6mm

                                                                                                              Consolidation settlement

                                                                                                              Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                              1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                              434 (sod)

                                                                                                              Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                              sc frac14 056 434 frac14 24mm

                                                                                                              The total settlement is (6thorn 24) frac14 30mm

                                                                                                              813

                                                                                                              At base level N frac14 26 Then using Equation 830

                                                                                                              qb frac14 40NDb

                                                                                                              Bfrac14 40 26 2

                                                                                                              025frac14 8320 kN=m2

                                                                                                              ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                              Figure Q812

                                                                                                              Bearing capacity 71

                                                                                                              Over the length embedded in sand

                                                                                                              N frac14 21 ie18thorn 24

                                                                                                              2

                                                                                                              Using Equation 831

                                                                                                              qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                              For a single pile

                                                                                                              Qf frac14 Abqb thorn Asqs

                                                                                                              frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                              For the pile group assuming a group efficiency of 12

                                                                                                              XQf frac14 12 9 604 frac14 6523 kN

                                                                                                              Then the load factor is

                                                                                                              F frac14 6523

                                                                                                              2000thorn 1000frac14 21

                                                                                                              (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                              Characteristic base resistance per unit area qbk frac14 8320

                                                                                                              150frac14 5547 kNm2

                                                                                                              Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                              150frac14 28 kNm2

                                                                                                              Characteristic base and shaft resistances for a single pile

                                                                                                              Rbk frac14 0252 5547 frac14 347 kN

                                                                                                              Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                              For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                              Design bearing resistance Rcd frac14 347

                                                                                                              130thorn 56

                                                                                                              130frac14 310 kN

                                                                                                              For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                              Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                              (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                              From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                              72 Bearing capacity

                                                                                                              N frac14 24thorn 26thorn 34

                                                                                                              3frac14 28

                                                                                                              Ic frac14 171

                                                                                                              2814frac14 0016 ethEquation 818THORN

                                                                                                              s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                              814

                                                                                                              Using Equation 841

                                                                                                              Tf frac14 DLcu thorn

                                                                                                              4ethD2 d2THORNcuNc

                                                                                                              frac14 eth 02 5 06 110THORN thorn

                                                                                                              4eth022 012THORN110 9

                                                                                                              frac14 207thorn 23 frac14 230 kN

                                                                                                              Figure Q813

                                                                                                              Bearing capacity 73

                                                                                                              Chapter 9

                                                                                                              Stability of slopes

                                                                                                              91

                                                                                                              Referring to Figure Q91

                                                                                                              W frac14 417 19 frac14 792 kN=m

                                                                                                              Q frac14 20 28 frac14 56 kN=m

                                                                                                              Arc lengthAB frac14

                                                                                                              180 73 90 frac14 115m

                                                                                                              Arc length BC frac14

                                                                                                              180 28 90 frac14 44m

                                                                                                              The factor of safety is given by

                                                                                                              F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                              90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                              Depth of tension crack z0 frac14 2cu

                                                                                                              frac14 2 20

                                                                                                              19frac14 21m

                                                                                                              Arc length BD frac14

                                                                                                              180 13

                                                                                                              1

                                                                                                              2 90 frac14 21m

                                                                                                              F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                              The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                              Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                              14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                              Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                              The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                              92

                                                                                                              u frac14 0

                                                                                                              Depth factor D frac14 11

                                                                                                              9frac14 122

                                                                                                              Using Equation 92 with F frac14 10

                                                                                                              Ns frac14 cu

                                                                                                              FHfrac14 30

                                                                                                              10 19 9frac14 0175

                                                                                                              Hence from Figure 93

                                                                                                              frac14 50

                                                                                                              For F frac14 12

                                                                                                              Ns frac14 30

                                                                                                              12 19 9frac14 0146

                                                                                                              frac14 27

                                                                                                              93

                                                                                                              Refer to Figure Q93

                                                                                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                              1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                              74 m

                                                                                                              214 1deg

                                                                                                              213 1deg

                                                                                                              39 m

                                                                                                              WB

                                                                                                              D

                                                                                                              C

                                                                                                              28 m

                                                                                                              21 m

                                                                                                              A

                                                                                                              Q

                                                                                                              Soil (1)Soil (2)

                                                                                                              73deg

                                                                                                              Figure Q91

                                                                                                              Stability of slopes 75

                                                                                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                              9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                              599 256 328 1372

                                                                                                              Figure Q93

                                                                                                              76 Stability of slopes

                                                                                                              XW cos frac14 b

                                                                                                              Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                              W sin frac14 bX

                                                                                                              h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                              Arc length La frac14

                                                                                                              180 57

                                                                                                              1

                                                                                                              2 326 frac14 327m

                                                                                                              The factor of safety is given by

                                                                                                              F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                              W sin

                                                                                                              frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                              frac14 091

                                                                                                              According to the limit state method

                                                                                                              0d frac14 tan1tan 32

                                                                                                              125

                                                                                                              frac14 265

                                                                                                              c0 frac14 8

                                                                                                              160frac14 5 kN=m2

                                                                                                              Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                              Design disturbing moment frac14 1075 kN=m

                                                                                                              The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                              94

                                                                                                              F frac14 1

                                                                                                              W sin

                                                                                                              Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                              1thorn ethtan tan0=FTHORN

                                                                                                              c0 frac14 8 kN=m2

                                                                                                              0 frac14 32

                                                                                                              c0b frac14 8 2 frac14 16 kN=m

                                                                                                              W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                              Try F frac14 100

                                                                                                              tan0

                                                                                                              Ffrac14 0625

                                                                                                              Stability of slopes 77

                                                                                                              Values of u are as obtained in Figure Q93

                                                                                                              SliceNo

                                                                                                              h(m)

                                                                                                              W frac14 bh(kNm)

                                                                                                              W sin(kNm)

                                                                                                              ub(kNm)

                                                                                                              c0bthorn (W ub) tan0(kNm)

                                                                                                              sec

                                                                                                              1thorn (tan tan0)FProduct(kNm)

                                                                                                              1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                              23 33 30 1042 31

                                                                                                              3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                              224 92 72 0931 67

                                                                                                              6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                              12 58 112 90 0889 80

                                                                                                              8 60 252 1812

                                                                                                              80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                              10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                              2154 88 116 0853 99

                                                                                                              14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                              1074 1091

                                                                                                              F frac14 1091

                                                                                                              1074frac14 102 (assumed value 100)

                                                                                                              Thus

                                                                                                              F frac14 101

                                                                                                              95

                                                                                                              F frac14 1

                                                                                                              W sin

                                                                                                              XfWeth1 ruTHORN tan0g sec

                                                                                                              1thorn ethtan tan0THORN=F

                                                                                                              0 frac14 33

                                                                                                              ru frac14 020

                                                                                                              W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                              eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                              Try F frac14 110

                                                                                                              tan 0

                                                                                                              Ffrac14 tan 33

                                                                                                              110frac14 0590

                                                                                                              78 Stability of slopes

                                                                                                              Referring to Figure Q95

                                                                                                              SliceNo

                                                                                                              h(m)

                                                                                                              W frac14 bh(kNm)

                                                                                                              W sin(kNm)

                                                                                                              W(1 ru) tan0(kNm)

                                                                                                              sec

                                                                                                              1thorn ( tan tan0)FProduct(kNm)

                                                                                                              1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                              2120 234 0892 209

                                                                                                              4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                              1185 1271

                                                                                                              Figure Q95

                                                                                                              Stability of slopes 79

                                                                                                              F frac14 1271

                                                                                                              1185frac14 107

                                                                                                              The trial value was 110 therefore take F to be 108

                                                                                                              96

                                                                                                              (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                              F frac14 0

                                                                                                              sat

                                                                                                              tan0

                                                                                                              tan

                                                                                                              ptie 15 frac14 92

                                                                                                              19

                                                                                                              tan 36

                                                                                                              tan

                                                                                                              tan frac14 0234

                                                                                                              frac14 13

                                                                                                              Water table well below surface the factor of safety is given by Equation 911

                                                                                                              F frac14 tan0

                                                                                                              tan

                                                                                                              frac14 tan 36

                                                                                                              tan 13

                                                                                                              frac14 31

                                                                                                              (b) 0d frac14 tan1tan 36

                                                                                                              125

                                                                                                              frac14 30

                                                                                                              Depth of potential failure surface frac14 z

                                                                                                              Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                              frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                              frac14 504z kN

                                                                                                              Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                              frac14 19 z sin 13 cos 13

                                                                                                              frac14 416z kN

                                                                                                              Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                              80 Stability of slopes

                                                                                                              • Book Cover
                                                                                                              • Title
                                                                                                              • Contents
                                                                                                              • Basic characteristics of soils
                                                                                                              • Seepage
                                                                                                              • Effective stress
                                                                                                              • Shear strength
                                                                                                              • Stresses and displacements
                                                                                                              • Lateral earth pressure
                                                                                                              • Consolidation theory
                                                                                                              • Bearing capacity
                                                                                                              • Stability of slopes

                                                                                                                Chapter 7

                                                                                                                Consolidation theory

                                                                                                                71

                                                                                                                Total change in thickness

                                                                                                                H frac14 782 602 frac14 180mm

                                                                                                                Average thickness frac14 1530thorn 180

                                                                                                                2frac14 1620mm

                                                                                                                Length of drainage path d frac14 1620

                                                                                                                2frac14 810mm

                                                                                                                Root time plot (Figure Q71a)

                                                                                                                ffiffiffiffiffiffit90p frac14 33

                                                                                                                t90 frac14 109min

                                                                                                                cv frac14 0848d2

                                                                                                                t90frac14 0848 8102

                                                                                                                109 1440 365

                                                                                                                106frac14 27m2=year

                                                                                                                r0 frac14 782 764

                                                                                                                782 602frac14 018

                                                                                                                180frac14 0100

                                                                                                                rp frac14 10eth764 645THORN9eth782 602THORN frac14

                                                                                                                10 119

                                                                                                                9 180frac14 0735

                                                                                                                rs frac14 1 eth0100thorn 0735THORN frac14 0165

                                                                                                                Log time plot (Figure Q71b)

                                                                                                                t50 frac14 26min

                                                                                                                cv frac14 0196d2

                                                                                                                t50frac14 0196 8102

                                                                                                                26 1440 365

                                                                                                                106frac14 26m2=year

                                                                                                                r0 frac14 782 763

                                                                                                                782 602frac14 019

                                                                                                                180frac14 0106

                                                                                                                rp frac14 763 623

                                                                                                                782 602frac14 140

                                                                                                                180frac14 0778

                                                                                                                rs frac14 1 eth0106thorn 0778THORN frac14 0116

                                                                                                                Figure Q71(a)

                                                                                                                Figure Q71(b)

                                                                                                                Final void ratio

                                                                                                                e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                                                e

                                                                                                                Hfrac14 1thorn e0

                                                                                                                H0frac14 1thorn e1 thorne

                                                                                                                H0

                                                                                                                ie

                                                                                                                e

                                                                                                                180frac14 1631thorne

                                                                                                                1710

                                                                                                                e frac14 2936

                                                                                                                1530frac14 0192

                                                                                                                Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                                                mv frac14 1

                                                                                                                1thorn e0 e0 e101 00

                                                                                                                frac14 1

                                                                                                                1823 0192

                                                                                                                0107frac14 098m2=MN

                                                                                                                k frac14 cvmvw frac14 265 098 98

                                                                                                                60 1440 365 103frac14 81 1010 m=s

                                                                                                                72

                                                                                                                Using Equation 77 (one-dimensional method)

                                                                                                                sc frac14 e0 e11thorn e0 H

                                                                                                                Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                                                Figure Q72

                                                                                                                52 Consolidation theory

                                                                                                                Settlement

                                                                                                                Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                                                1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                                                318

                                                                                                                Notes 5 92y 460thorn 84

                                                                                                                Heave

                                                                                                                Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                                                1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                                                38

                                                                                                                73

                                                                                                                U frac14 f ethTvTHORN frac14 f cvt

                                                                                                                d2

                                                                                                                Hence if cv is constant

                                                                                                                t1

                                                                                                                t2frac14 d

                                                                                                                21

                                                                                                                d22

                                                                                                                where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                                                d1 frac14 95mm and d2 frac14 2500mm

                                                                                                                for U frac14 050 t2 frac14 t1 d22

                                                                                                                d21

                                                                                                                frac14 20

                                                                                                                60 24 365 25002

                                                                                                                952frac14 263 years

                                                                                                                for U lt 060 Tv frac14

                                                                                                                4U2 (Equation 724(a))

                                                                                                                t030 frac14 t050 0302

                                                                                                                0502

                                                                                                                frac14 263 036 frac14 095 years

                                                                                                                Consolidation theory 53

                                                                                                                74

                                                                                                                The layer is open

                                                                                                                d frac14 8

                                                                                                                2frac14 4m

                                                                                                                Tv frac14 cvtd2frac14 24 3

                                                                                                                42frac14 0450

                                                                                                                ui frac14 frac14 84 kN=m2

                                                                                                                The excess pore water pressure is given by Equation 721

                                                                                                                ue frac14Xmfrac141mfrac140

                                                                                                                2ui

                                                                                                                Msin

                                                                                                                Mz

                                                                                                                d

                                                                                                                expethM2TvTHORN

                                                                                                                In this case z frac14 d

                                                                                                                sinMz

                                                                                                                d

                                                                                                                frac14 sinM

                                                                                                                where

                                                                                                                M frac14

                                                                                                                23

                                                                                                                25

                                                                                                                2

                                                                                                                M sin M M2Tv exp (M2Tv)

                                                                                                                2thorn1 1110 0329

                                                                                                                3

                                                                                                                21 9993 457 105

                                                                                                                ue frac14 2 84 2

                                                                                                                1 0329 ethother terms negligibleTHORN

                                                                                                                frac14 352 kN=m2

                                                                                                                75

                                                                                                                The layer is open

                                                                                                                d frac14 6

                                                                                                                2frac14 3m

                                                                                                                Tv frac14 cvtd2frac14 10 3

                                                                                                                32frac14 0333

                                                                                                                The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                                                54 Consolidation theory

                                                                                                                For an open layer

                                                                                                                Tv frac14 4n

                                                                                                                m2

                                                                                                                n frac14 0333 62

                                                                                                                4frac14 300

                                                                                                                The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                                                ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                                                i j

                                                                                                                0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                                                0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                                                The initial and 3-year isochrones are plotted in Figure Q75

                                                                                                                Area under initial isochrone frac14 180 units

                                                                                                                Area under 3-year isochrone frac14 63 units

                                                                                                                The average degree of consolidation is given by Equation 725Thus

                                                                                                                U frac14 1 63

                                                                                                                180frac14 065

                                                                                                                Figure Q75

                                                                                                                Consolidation theory 55

                                                                                                                76

                                                                                                                At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                                0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                                The final consolidation settlement (one-dimensional method) is

                                                                                                                sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                                Corrected time t frac14 2 1

                                                                                                                2

                                                                                                                40

                                                                                                                52

                                                                                                                frac14 1615 years

                                                                                                                Tv frac14 cvtd2frac14 44 1615

                                                                                                                42frac14 0444

                                                                                                                From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                                77

                                                                                                                The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                                Figure Q77

                                                                                                                56 Consolidation theory

                                                                                                                Point m n Ir (kNm2) sc (mm)

                                                                                                                13020frac14 15 20

                                                                                                                20frac14 10 0194 (4) 113 124

                                                                                                                260

                                                                                                                20frac14 30

                                                                                                                20

                                                                                                                20frac14 10 0204 (2) 59 65

                                                                                                                360

                                                                                                                20frac14 30

                                                                                                                40

                                                                                                                20frac14 20 0238 (1) 35 38

                                                                                                                430

                                                                                                                20frac14 15

                                                                                                                40

                                                                                                                20frac14 20 0224 (2) 65 72

                                                                                                                Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                                78

                                                                                                                Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                                (a) Immediate settlement

                                                                                                                H

                                                                                                                Bfrac14 30

                                                                                                                35frac14 086

                                                                                                                D

                                                                                                                Bfrac14 2

                                                                                                                35frac14 006

                                                                                                                Figure Q78

                                                                                                                Consolidation theory 57

                                                                                                                From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                                si frac14 130131qB

                                                                                                                Eufrac14 10 032 105 35

                                                                                                                40frac14 30mm

                                                                                                                (b) Consolidation settlement

                                                                                                                Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                                1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                                3150

                                                                                                                Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                                Now

                                                                                                                H

                                                                                                                Bfrac14 30

                                                                                                                35frac14 086 and A frac14 065

                                                                                                                from Figure 712 13 frac14 079

                                                                                                                sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                                Total settlement

                                                                                                                s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                                79

                                                                                                                Without sand drains

                                                                                                                Uv frac14 025

                                                                                                                Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                                t frac14 Tvd2

                                                                                                                cvfrac14 0049 82

                                                                                                                cvWith sand drains

                                                                                                                R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                                n frac14 Rrfrac14 169

                                                                                                                015frac14 113

                                                                                                                Tr frac14 cht

                                                                                                                4R2frac14 ch

                                                                                                                4 1692 0049 82

                                                                                                                cvethand ch frac14 cvTHORN

                                                                                                                frac14 0275

                                                                                                                Ur frac14 073 (from Figure 730)

                                                                                                                58 Consolidation theory

                                                                                                                Using Equation 740

                                                                                                                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                                U frac14 080

                                                                                                                710

                                                                                                                Without sand drains

                                                                                                                Uv frac14 090

                                                                                                                Tv frac14 0848

                                                                                                                t frac14 Tvd2

                                                                                                                cvfrac14 0848 102

                                                                                                                96frac14 88 years

                                                                                                                With sand drains

                                                                                                                R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                                n frac14 Rrfrac14 226

                                                                                                                015frac14 15

                                                                                                                Tr

                                                                                                                Tvfrac14 chcv

                                                                                                                d2

                                                                                                                4R2ethsame tTHORN

                                                                                                                Tr

                                                                                                                Tvfrac14 140

                                                                                                                96 102

                                                                                                                4 2262frac14 714 eth1THORN

                                                                                                                Using Equation 740

                                                                                                                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                                An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                                Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                                040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                                Thus

                                                                                                                Uv frac14 0295 and Ur frac14 086

                                                                                                                t frac14 88 00683

                                                                                                                0848frac14 07 years

                                                                                                                Consolidation theory 59

                                                                                                                Chapter 8

                                                                                                                Bearing capacity

                                                                                                                81

                                                                                                                (a) The ultimate bearing capacity is given by Equation 83

                                                                                                                qf frac14 cNc thorn DNq thorn 1

                                                                                                                2BN

                                                                                                                For u frac14 0

                                                                                                                Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                                The net ultimate bearing capacity is

                                                                                                                qnf frac14 qf D frac14 540 kN=m2

                                                                                                                The net foundation pressure is

                                                                                                                qn frac14 q D frac14 425

                                                                                                                2 eth21 1THORN frac14 192 kN=m2

                                                                                                                The factor of safety (Equation 86) is

                                                                                                                F frac14 qnfqnfrac14 540

                                                                                                                192frac14 28

                                                                                                                (b) For 0 frac14 28

                                                                                                                Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                                qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                                2 112 2 13

                                                                                                                frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                                qnf frac14 574 112 frac14 563 kN=m2

                                                                                                                F frac14 563

                                                                                                                192frac14 29

                                                                                                                (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                                82

                                                                                                                For 0 frac14 38

                                                                                                                Nq frac14 49 N frac14 67

                                                                                                                qnf frac14 DethNq 1THORN thorn 1

                                                                                                                2BN ethfrom Equation 83THORN

                                                                                                                frac14 eth18 075 48THORN thorn 1

                                                                                                                2 18 15 67

                                                                                                                frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                                qn frac14 500

                                                                                                                15 eth18 075THORN frac14 320 kN=m2

                                                                                                                F frac14 qnfqnfrac14 1553

                                                                                                                320frac14 48

                                                                                                                0d frac14 tan1tan 38

                                                                                                                125

                                                                                                                frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                                Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                                2 18 15 25

                                                                                                                frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                                Design load (action) Vd frac14 500 kN=m

                                                                                                                The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                                83

                                                                                                                D

                                                                                                                Bfrac14 350

                                                                                                                225frac14 155

                                                                                                                From Figure 85 for a square foundation

                                                                                                                Nc frac14 81

                                                                                                                Bearing capacity 61

                                                                                                                For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                                Nc frac14 084thorn 016B

                                                                                                                L

                                                                                                                81 frac14 745

                                                                                                                Using Equation 810

                                                                                                                qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                                For F frac14 3

                                                                                                                qn frac14 1006

                                                                                                                3frac14 335 kN=m2

                                                                                                                q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                                Design load frac14 405 450 225 frac14 4100 kN

                                                                                                                Design undrained strength cud frac14 135

                                                                                                                14frac14 96 kN=m2

                                                                                                                Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                                frac14 7241 kN

                                                                                                                Design load Vd frac14 4100 kN

                                                                                                                Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                84

                                                                                                                For 0 frac14 40

                                                                                                                Nq frac14 64 N frac14 95

                                                                                                                qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                                (a) Water table 5m below ground level

                                                                                                                qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                                qn frac14 400 17 frac14 383 kN=m2

                                                                                                                F frac14 2686

                                                                                                                383frac14 70

                                                                                                                (b) Water table 1m below ground level (ie at foundation level)

                                                                                                                0 frac14 20 98 frac14 102 kN=m3

                                                                                                                62 Bearing capacity

                                                                                                                qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                                F frac14 2040

                                                                                                                383frac14 53

                                                                                                                (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                                eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                                qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                                F frac14 1296

                                                                                                                392frac14 33

                                                                                                                85

                                                                                                                The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                                Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                                Design value of 0 frac14 tan1tan 39

                                                                                                                125

                                                                                                                frac14 33

                                                                                                                For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                                Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                                Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                86

                                                                                                                (a) Undrained shear for u frac14 0

                                                                                                                Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                qnf frac14 12cuNc

                                                                                                                frac14 12 100 514 frac14 617 kN=m2

                                                                                                                qn frac14 qnfFfrac14 617

                                                                                                                3frac14 206 kN=m2

                                                                                                                q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                                Bearing capacity 63

                                                                                                                Drained shear for 0 frac14 32

                                                                                                                Nq frac14 23 N frac14 25

                                                                                                                0 frac14 21 98 frac14 112 kN=m3

                                                                                                                qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                frac14 694 kN=m2

                                                                                                                q frac14 694

                                                                                                                3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                Design load frac14 42 227 frac14 3632 kN

                                                                                                                (b) Design undrained strength cud frac14 100

                                                                                                                14frac14 71 kNm2

                                                                                                                Design bearing resistance Rd frac14 12cudNe area

                                                                                                                frac14 12 71 514 42

                                                                                                                frac14 7007 kN

                                                                                                                For drained shear 0d frac14 tan1tan 32

                                                                                                                125

                                                                                                                frac14 26

                                                                                                                Nq frac14 12 N frac14 10

                                                                                                                Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                1 2 100 0175 0700qn 0182qn

                                                                                                                2 6 033 0044 0176qn 0046qn

                                                                                                                3 10 020 0017 0068qn 0018qn

                                                                                                                0246qn

                                                                                                                Diameter of equivalent circle B frac14 45m

                                                                                                                H

                                                                                                                Bfrac14 12

                                                                                                                45frac14 27 and A frac14 042

                                                                                                                13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                64 Bearing capacity

                                                                                                                For sc frac14 30mm

                                                                                                                qn frac14 30

                                                                                                                0147frac14 204 kN=m2

                                                                                                                q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                Design load frac14 42 225 frac14 3600 kN

                                                                                                                The design load is 3600 kN settlement being the limiting criterion

                                                                                                                87

                                                                                                                D

                                                                                                                Bfrac14 8

                                                                                                                4frac14 20

                                                                                                                From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                F frac14 cuNc

                                                                                                                Dfrac14 40 71

                                                                                                                20 8frac14 18

                                                                                                                88

                                                                                                                Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                Design value of 0 frac14 tan1tan 38

                                                                                                                125

                                                                                                                frac14 32

                                                                                                                Figure Q86

                                                                                                                Bearing capacity 65

                                                                                                                For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                For B frac14 250m qn frac14 3750

                                                                                                                2502 17 frac14 583 kN=m2

                                                                                                                From Figure 510 m frac14 n frac14 126

                                                                                                                6frac14 021

                                                                                                                Ir frac14 0019

                                                                                                                Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                89

                                                                                                                Depth (m) N 0v (kNm2) CN N1

                                                                                                                070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                Cw frac14 05thorn 0530

                                                                                                                47

                                                                                                                frac14 082

                                                                                                                66 Bearing capacity

                                                                                                                Thus

                                                                                                                qa frac14 150 082 frac14 120 kN=m2

                                                                                                                (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                Thus

                                                                                                                qa frac14 90 15 frac14 135 kN=m2

                                                                                                                (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                Ic frac14 171

                                                                                                                1014frac14 0068

                                                                                                                From Equation 819(a) with s frac14 25mm

                                                                                                                q frac14 25

                                                                                                                3507 0068frac14 150 kN=m2

                                                                                                                810

                                                                                                                Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                Izp frac14 05thorn 01130

                                                                                                                16 27

                                                                                                                05

                                                                                                                frac14 067

                                                                                                                Refer to Figure Q810

                                                                                                                E frac14 25qc

                                                                                                                Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                Ez (mm3MN)

                                                                                                                1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                0203

                                                                                                                C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                130frac14 093

                                                                                                                C2 frac14 1 ethsayTHORN

                                                                                                                s frac14 C1C2qnX Iz

                                                                                                                Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                Bearing capacity 67

                                                                                                                811

                                                                                                                At pile base level

                                                                                                                cu frac14 220 kN=m2

                                                                                                                qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                Then

                                                                                                                Qf frac14 Abqb thorn Asqs

                                                                                                                frac14

                                                                                                                4 32 1980

                                                                                                                thorn eth 105 139 86THORN

                                                                                                                frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                0 01 02 03 04 05 06 07

                                                                                                                0 2 4 6 8 10 12 14

                                                                                                                1

                                                                                                                2

                                                                                                                3

                                                                                                                4

                                                                                                                5

                                                                                                                6

                                                                                                                7

                                                                                                                8

                                                                                                                (1)

                                                                                                                (2)

                                                                                                                (3)

                                                                                                                (4)

                                                                                                                (5)

                                                                                                                qc

                                                                                                                qc

                                                                                                                Iz

                                                                                                                Iz

                                                                                                                (MNm2)

                                                                                                                z (m)

                                                                                                                Figure Q810

                                                                                                                68 Bearing capacity

                                                                                                                Allowable load

                                                                                                                ethaTHORN Qf

                                                                                                                2frac14 17 937

                                                                                                                2frac14 8968 kN

                                                                                                                ethbTHORN Abqb

                                                                                                                3thorn Asqs frac14 13 996

                                                                                                                3thorn 3941 frac14 8606 kN

                                                                                                                ie allowable load frac14 8600 kN

                                                                                                                Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                According to the limit state method

                                                                                                                Characteristic undrained strength at base level cuk frac14 220

                                                                                                                150kN=m2

                                                                                                                Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                150frac14 1320 kN=m2

                                                                                                                Characteristic shaft resistance qsk frac14 00150

                                                                                                                frac14 86

                                                                                                                150frac14 57 kN=m2

                                                                                                                Characteristic base and shaft resistances

                                                                                                                Rbk frac14

                                                                                                                4 32 1320 frac14 9330 kN

                                                                                                                Rsk frac14 105 139 86

                                                                                                                150frac14 2629 kN

                                                                                                                For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                Design bearing resistance Rcd frac14 9330

                                                                                                                160thorn 2629

                                                                                                                130

                                                                                                                frac14 5831thorn 2022

                                                                                                                frac14 7850 kN

                                                                                                                Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                812

                                                                                                                ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                For a single pile

                                                                                                                Qf frac14 Abqb thorn Asqs

                                                                                                                frac14

                                                                                                                4 062 1305

                                                                                                                thorn eth 06 15 42THORN

                                                                                                                frac14 369thorn 1187 frac14 1556 kN

                                                                                                                Bearing capacity 69

                                                                                                                Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                qbkfrac14 9cuk frac14 9 220

                                                                                                                150frac14 1320 kN=m2

                                                                                                                qskfrac14cuk frac14 040 105

                                                                                                                150frac14 28 kN=m2

                                                                                                                Rbkfrac14

                                                                                                                4 0602 1320 frac14 373 kN

                                                                                                                Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                Rcdfrac14 373

                                                                                                                160thorn 791

                                                                                                                130frac14 233thorn 608 frac14 841 kN

                                                                                                                Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                q frac14 21 000

                                                                                                                1762frac14 68 kN=m2

                                                                                                                Immediate settlement

                                                                                                                H

                                                                                                                Bfrac14 15

                                                                                                                176frac14 085

                                                                                                                D

                                                                                                                Bfrac14 13

                                                                                                                176frac14 074

                                                                                                                L

                                                                                                                Bfrac14 1

                                                                                                                Hence from Figure 515

                                                                                                                130 frac14 078 and 131 frac14 041

                                                                                                                70 Bearing capacity

                                                                                                                Thus using Equation 528

                                                                                                                si frac14 078 041 68 176

                                                                                                                65frac14 6mm

                                                                                                                Consolidation settlement

                                                                                                                Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                434 (sod)

                                                                                                                Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                sc frac14 056 434 frac14 24mm

                                                                                                                The total settlement is (6thorn 24) frac14 30mm

                                                                                                                813

                                                                                                                At base level N frac14 26 Then using Equation 830

                                                                                                                qb frac14 40NDb

                                                                                                                Bfrac14 40 26 2

                                                                                                                025frac14 8320 kN=m2

                                                                                                                ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                Figure Q812

                                                                                                                Bearing capacity 71

                                                                                                                Over the length embedded in sand

                                                                                                                N frac14 21 ie18thorn 24

                                                                                                                2

                                                                                                                Using Equation 831

                                                                                                                qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                For a single pile

                                                                                                                Qf frac14 Abqb thorn Asqs

                                                                                                                frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                For the pile group assuming a group efficiency of 12

                                                                                                                XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                Then the load factor is

                                                                                                                F frac14 6523

                                                                                                                2000thorn 1000frac14 21

                                                                                                                (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                150frac14 5547 kNm2

                                                                                                                Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                150frac14 28 kNm2

                                                                                                                Characteristic base and shaft resistances for a single pile

                                                                                                                Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                Design bearing resistance Rcd frac14 347

                                                                                                                130thorn 56

                                                                                                                130frac14 310 kN

                                                                                                                For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                72 Bearing capacity

                                                                                                                N frac14 24thorn 26thorn 34

                                                                                                                3frac14 28

                                                                                                                Ic frac14 171

                                                                                                                2814frac14 0016 ethEquation 818THORN

                                                                                                                s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                814

                                                                                                                Using Equation 841

                                                                                                                Tf frac14 DLcu thorn

                                                                                                                4ethD2 d2THORNcuNc

                                                                                                                frac14 eth 02 5 06 110THORN thorn

                                                                                                                4eth022 012THORN110 9

                                                                                                                frac14 207thorn 23 frac14 230 kN

                                                                                                                Figure Q813

                                                                                                                Bearing capacity 73

                                                                                                                Chapter 9

                                                                                                                Stability of slopes

                                                                                                                91

                                                                                                                Referring to Figure Q91

                                                                                                                W frac14 417 19 frac14 792 kN=m

                                                                                                                Q frac14 20 28 frac14 56 kN=m

                                                                                                                Arc lengthAB frac14

                                                                                                                180 73 90 frac14 115m

                                                                                                                Arc length BC frac14

                                                                                                                180 28 90 frac14 44m

                                                                                                                The factor of safety is given by

                                                                                                                F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                Depth of tension crack z0 frac14 2cu

                                                                                                                frac14 2 20

                                                                                                                19frac14 21m

                                                                                                                Arc length BD frac14

                                                                                                                180 13

                                                                                                                1

                                                                                                                2 90 frac14 21m

                                                                                                                F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                92

                                                                                                                u frac14 0

                                                                                                                Depth factor D frac14 11

                                                                                                                9frac14 122

                                                                                                                Using Equation 92 with F frac14 10

                                                                                                                Ns frac14 cu

                                                                                                                FHfrac14 30

                                                                                                                10 19 9frac14 0175

                                                                                                                Hence from Figure 93

                                                                                                                frac14 50

                                                                                                                For F frac14 12

                                                                                                                Ns frac14 30

                                                                                                                12 19 9frac14 0146

                                                                                                                frac14 27

                                                                                                                93

                                                                                                                Refer to Figure Q93

                                                                                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                74 m

                                                                                                                214 1deg

                                                                                                                213 1deg

                                                                                                                39 m

                                                                                                                WB

                                                                                                                D

                                                                                                                C

                                                                                                                28 m

                                                                                                                21 m

                                                                                                                A

                                                                                                                Q

                                                                                                                Soil (1)Soil (2)

                                                                                                                73deg

                                                                                                                Figure Q91

                                                                                                                Stability of slopes 75

                                                                                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                599 256 328 1372

                                                                                                                Figure Q93

                                                                                                                76 Stability of slopes

                                                                                                                XW cos frac14 b

                                                                                                                Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                W sin frac14 bX

                                                                                                                h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                Arc length La frac14

                                                                                                                180 57

                                                                                                                1

                                                                                                                2 326 frac14 327m

                                                                                                                The factor of safety is given by

                                                                                                                F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                W sin

                                                                                                                frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                frac14 091

                                                                                                                According to the limit state method

                                                                                                                0d frac14 tan1tan 32

                                                                                                                125

                                                                                                                frac14 265

                                                                                                                c0 frac14 8

                                                                                                                160frac14 5 kN=m2

                                                                                                                Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                Design disturbing moment frac14 1075 kN=m

                                                                                                                The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                94

                                                                                                                F frac14 1

                                                                                                                W sin

                                                                                                                Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                1thorn ethtan tan0=FTHORN

                                                                                                                c0 frac14 8 kN=m2

                                                                                                                0 frac14 32

                                                                                                                c0b frac14 8 2 frac14 16 kN=m

                                                                                                                W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                Try F frac14 100

                                                                                                                tan0

                                                                                                                Ffrac14 0625

                                                                                                                Stability of slopes 77

                                                                                                                Values of u are as obtained in Figure Q93

                                                                                                                SliceNo

                                                                                                                h(m)

                                                                                                                W frac14 bh(kNm)

                                                                                                                W sin(kNm)

                                                                                                                ub(kNm)

                                                                                                                c0bthorn (W ub) tan0(kNm)

                                                                                                                sec

                                                                                                                1thorn (tan tan0)FProduct(kNm)

                                                                                                                1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                23 33 30 1042 31

                                                                                                                3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                224 92 72 0931 67

                                                                                                                6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                12 58 112 90 0889 80

                                                                                                                8 60 252 1812

                                                                                                                80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                2154 88 116 0853 99

                                                                                                                14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                1074 1091

                                                                                                                F frac14 1091

                                                                                                                1074frac14 102 (assumed value 100)

                                                                                                                Thus

                                                                                                                F frac14 101

                                                                                                                95

                                                                                                                F frac14 1

                                                                                                                W sin

                                                                                                                XfWeth1 ruTHORN tan0g sec

                                                                                                                1thorn ethtan tan0THORN=F

                                                                                                                0 frac14 33

                                                                                                                ru frac14 020

                                                                                                                W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                Try F frac14 110

                                                                                                                tan 0

                                                                                                                Ffrac14 tan 33

                                                                                                                110frac14 0590

                                                                                                                78 Stability of slopes

                                                                                                                Referring to Figure Q95

                                                                                                                SliceNo

                                                                                                                h(m)

                                                                                                                W frac14 bh(kNm)

                                                                                                                W sin(kNm)

                                                                                                                W(1 ru) tan0(kNm)

                                                                                                                sec

                                                                                                                1thorn ( tan tan0)FProduct(kNm)

                                                                                                                1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                2120 234 0892 209

                                                                                                                4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                1185 1271

                                                                                                                Figure Q95

                                                                                                                Stability of slopes 79

                                                                                                                F frac14 1271

                                                                                                                1185frac14 107

                                                                                                                The trial value was 110 therefore take F to be 108

                                                                                                                96

                                                                                                                (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                F frac14 0

                                                                                                                sat

                                                                                                                tan0

                                                                                                                tan

                                                                                                                ptie 15 frac14 92

                                                                                                                19

                                                                                                                tan 36

                                                                                                                tan

                                                                                                                tan frac14 0234

                                                                                                                frac14 13

                                                                                                                Water table well below surface the factor of safety is given by Equation 911

                                                                                                                F frac14 tan0

                                                                                                                tan

                                                                                                                frac14 tan 36

                                                                                                                tan 13

                                                                                                                frac14 31

                                                                                                                (b) 0d frac14 tan1tan 36

                                                                                                                125

                                                                                                                frac14 30

                                                                                                                Depth of potential failure surface frac14 z

                                                                                                                Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                frac14 504z kN

                                                                                                                Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                frac14 19 z sin 13 cos 13

                                                                                                                frac14 416z kN

                                                                                                                Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                80 Stability of slopes

                                                                                                                • Book Cover
                                                                                                                • Title
                                                                                                                • Contents
                                                                                                                • Basic characteristics of soils
                                                                                                                • Seepage
                                                                                                                • Effective stress
                                                                                                                • Shear strength
                                                                                                                • Stresses and displacements
                                                                                                                • Lateral earth pressure
                                                                                                                • Consolidation theory
                                                                                                                • Bearing capacity
                                                                                                                • Stability of slopes

                                                                                                                  Figure Q71(a)

                                                                                                                  Figure Q71(b)

                                                                                                                  Final void ratio

                                                                                                                  e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                                                  e

                                                                                                                  Hfrac14 1thorn e0

                                                                                                                  H0frac14 1thorn e1 thorne

                                                                                                                  H0

                                                                                                                  ie

                                                                                                                  e

                                                                                                                  180frac14 1631thorne

                                                                                                                  1710

                                                                                                                  e frac14 2936

                                                                                                                  1530frac14 0192

                                                                                                                  Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                                                  mv frac14 1

                                                                                                                  1thorn e0 e0 e101 00

                                                                                                                  frac14 1

                                                                                                                  1823 0192

                                                                                                                  0107frac14 098m2=MN

                                                                                                                  k frac14 cvmvw frac14 265 098 98

                                                                                                                  60 1440 365 103frac14 81 1010 m=s

                                                                                                                  72

                                                                                                                  Using Equation 77 (one-dimensional method)

                                                                                                                  sc frac14 e0 e11thorn e0 H

                                                                                                                  Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                                                  Figure Q72

                                                                                                                  52 Consolidation theory

                                                                                                                  Settlement

                                                                                                                  Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                                                  1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                                                  318

                                                                                                                  Notes 5 92y 460thorn 84

                                                                                                                  Heave

                                                                                                                  Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                                                  1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                                                  38

                                                                                                                  73

                                                                                                                  U frac14 f ethTvTHORN frac14 f cvt

                                                                                                                  d2

                                                                                                                  Hence if cv is constant

                                                                                                                  t1

                                                                                                                  t2frac14 d

                                                                                                                  21

                                                                                                                  d22

                                                                                                                  where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                                                  d1 frac14 95mm and d2 frac14 2500mm

                                                                                                                  for U frac14 050 t2 frac14 t1 d22

                                                                                                                  d21

                                                                                                                  frac14 20

                                                                                                                  60 24 365 25002

                                                                                                                  952frac14 263 years

                                                                                                                  for U lt 060 Tv frac14

                                                                                                                  4U2 (Equation 724(a))

                                                                                                                  t030 frac14 t050 0302

                                                                                                                  0502

                                                                                                                  frac14 263 036 frac14 095 years

                                                                                                                  Consolidation theory 53

                                                                                                                  74

                                                                                                                  The layer is open

                                                                                                                  d frac14 8

                                                                                                                  2frac14 4m

                                                                                                                  Tv frac14 cvtd2frac14 24 3

                                                                                                                  42frac14 0450

                                                                                                                  ui frac14 frac14 84 kN=m2

                                                                                                                  The excess pore water pressure is given by Equation 721

                                                                                                                  ue frac14Xmfrac141mfrac140

                                                                                                                  2ui

                                                                                                                  Msin

                                                                                                                  Mz

                                                                                                                  d

                                                                                                                  expethM2TvTHORN

                                                                                                                  In this case z frac14 d

                                                                                                                  sinMz

                                                                                                                  d

                                                                                                                  frac14 sinM

                                                                                                                  where

                                                                                                                  M frac14

                                                                                                                  23

                                                                                                                  25

                                                                                                                  2

                                                                                                                  M sin M M2Tv exp (M2Tv)

                                                                                                                  2thorn1 1110 0329

                                                                                                                  3

                                                                                                                  21 9993 457 105

                                                                                                                  ue frac14 2 84 2

                                                                                                                  1 0329 ethother terms negligibleTHORN

                                                                                                                  frac14 352 kN=m2

                                                                                                                  75

                                                                                                                  The layer is open

                                                                                                                  d frac14 6

                                                                                                                  2frac14 3m

                                                                                                                  Tv frac14 cvtd2frac14 10 3

                                                                                                                  32frac14 0333

                                                                                                                  The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                                                  54 Consolidation theory

                                                                                                                  For an open layer

                                                                                                                  Tv frac14 4n

                                                                                                                  m2

                                                                                                                  n frac14 0333 62

                                                                                                                  4frac14 300

                                                                                                                  The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                                                  ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                                                  i j

                                                                                                                  0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                                                  0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                                                  The initial and 3-year isochrones are plotted in Figure Q75

                                                                                                                  Area under initial isochrone frac14 180 units

                                                                                                                  Area under 3-year isochrone frac14 63 units

                                                                                                                  The average degree of consolidation is given by Equation 725Thus

                                                                                                                  U frac14 1 63

                                                                                                                  180frac14 065

                                                                                                                  Figure Q75

                                                                                                                  Consolidation theory 55

                                                                                                                  76

                                                                                                                  At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                                  0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                                  The final consolidation settlement (one-dimensional method) is

                                                                                                                  sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                                  Corrected time t frac14 2 1

                                                                                                                  2

                                                                                                                  40

                                                                                                                  52

                                                                                                                  frac14 1615 years

                                                                                                                  Tv frac14 cvtd2frac14 44 1615

                                                                                                                  42frac14 0444

                                                                                                                  From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                                  77

                                                                                                                  The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                                  Figure Q77

                                                                                                                  56 Consolidation theory

                                                                                                                  Point m n Ir (kNm2) sc (mm)

                                                                                                                  13020frac14 15 20

                                                                                                                  20frac14 10 0194 (4) 113 124

                                                                                                                  260

                                                                                                                  20frac14 30

                                                                                                                  20

                                                                                                                  20frac14 10 0204 (2) 59 65

                                                                                                                  360

                                                                                                                  20frac14 30

                                                                                                                  40

                                                                                                                  20frac14 20 0238 (1) 35 38

                                                                                                                  430

                                                                                                                  20frac14 15

                                                                                                                  40

                                                                                                                  20frac14 20 0224 (2) 65 72

                                                                                                                  Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                                  78

                                                                                                                  Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                                  (a) Immediate settlement

                                                                                                                  H

                                                                                                                  Bfrac14 30

                                                                                                                  35frac14 086

                                                                                                                  D

                                                                                                                  Bfrac14 2

                                                                                                                  35frac14 006

                                                                                                                  Figure Q78

                                                                                                                  Consolidation theory 57

                                                                                                                  From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                                  si frac14 130131qB

                                                                                                                  Eufrac14 10 032 105 35

                                                                                                                  40frac14 30mm

                                                                                                                  (b) Consolidation settlement

                                                                                                                  Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                                  1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                                  3150

                                                                                                                  Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                                  Now

                                                                                                                  H

                                                                                                                  Bfrac14 30

                                                                                                                  35frac14 086 and A frac14 065

                                                                                                                  from Figure 712 13 frac14 079

                                                                                                                  sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                                  Total settlement

                                                                                                                  s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                                  79

                                                                                                                  Without sand drains

                                                                                                                  Uv frac14 025

                                                                                                                  Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                                  t frac14 Tvd2

                                                                                                                  cvfrac14 0049 82

                                                                                                                  cvWith sand drains

                                                                                                                  R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                                  n frac14 Rrfrac14 169

                                                                                                                  015frac14 113

                                                                                                                  Tr frac14 cht

                                                                                                                  4R2frac14 ch

                                                                                                                  4 1692 0049 82

                                                                                                                  cvethand ch frac14 cvTHORN

                                                                                                                  frac14 0275

                                                                                                                  Ur frac14 073 (from Figure 730)

                                                                                                                  58 Consolidation theory

                                                                                                                  Using Equation 740

                                                                                                                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                                  U frac14 080

                                                                                                                  710

                                                                                                                  Without sand drains

                                                                                                                  Uv frac14 090

                                                                                                                  Tv frac14 0848

                                                                                                                  t frac14 Tvd2

                                                                                                                  cvfrac14 0848 102

                                                                                                                  96frac14 88 years

                                                                                                                  With sand drains

                                                                                                                  R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                                  n frac14 Rrfrac14 226

                                                                                                                  015frac14 15

                                                                                                                  Tr

                                                                                                                  Tvfrac14 chcv

                                                                                                                  d2

                                                                                                                  4R2ethsame tTHORN

                                                                                                                  Tr

                                                                                                                  Tvfrac14 140

                                                                                                                  96 102

                                                                                                                  4 2262frac14 714 eth1THORN

                                                                                                                  Using Equation 740

                                                                                                                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                                  An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                                  Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                                  040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                                  Thus

                                                                                                                  Uv frac14 0295 and Ur frac14 086

                                                                                                                  t frac14 88 00683

                                                                                                                  0848frac14 07 years

                                                                                                                  Consolidation theory 59

                                                                                                                  Chapter 8

                                                                                                                  Bearing capacity

                                                                                                                  81

                                                                                                                  (a) The ultimate bearing capacity is given by Equation 83

                                                                                                                  qf frac14 cNc thorn DNq thorn 1

                                                                                                                  2BN

                                                                                                                  For u frac14 0

                                                                                                                  Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                  qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                                  The net ultimate bearing capacity is

                                                                                                                  qnf frac14 qf D frac14 540 kN=m2

                                                                                                                  The net foundation pressure is

                                                                                                                  qn frac14 q D frac14 425

                                                                                                                  2 eth21 1THORN frac14 192 kN=m2

                                                                                                                  The factor of safety (Equation 86) is

                                                                                                                  F frac14 qnfqnfrac14 540

                                                                                                                  192frac14 28

                                                                                                                  (b) For 0 frac14 28

                                                                                                                  Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                                  qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                                  2 112 2 13

                                                                                                                  frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                                  qnf frac14 574 112 frac14 563 kN=m2

                                                                                                                  F frac14 563

                                                                                                                  192frac14 29

                                                                                                                  (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                                  82

                                                                                                                  For 0 frac14 38

                                                                                                                  Nq frac14 49 N frac14 67

                                                                                                                  qnf frac14 DethNq 1THORN thorn 1

                                                                                                                  2BN ethfrom Equation 83THORN

                                                                                                                  frac14 eth18 075 48THORN thorn 1

                                                                                                                  2 18 15 67

                                                                                                                  frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                                  qn frac14 500

                                                                                                                  15 eth18 075THORN frac14 320 kN=m2

                                                                                                                  F frac14 qnfqnfrac14 1553

                                                                                                                  320frac14 48

                                                                                                                  0d frac14 tan1tan 38

                                                                                                                  125

                                                                                                                  frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                                  Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                                  2 18 15 25

                                                                                                                  frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                                  Design load (action) Vd frac14 500 kN=m

                                                                                                                  The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                                  83

                                                                                                                  D

                                                                                                                  Bfrac14 350

                                                                                                                  225frac14 155

                                                                                                                  From Figure 85 for a square foundation

                                                                                                                  Nc frac14 81

                                                                                                                  Bearing capacity 61

                                                                                                                  For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                                  Nc frac14 084thorn 016B

                                                                                                                  L

                                                                                                                  81 frac14 745

                                                                                                                  Using Equation 810

                                                                                                                  qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                                  For F frac14 3

                                                                                                                  qn frac14 1006

                                                                                                                  3frac14 335 kN=m2

                                                                                                                  q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                                  Design load frac14 405 450 225 frac14 4100 kN

                                                                                                                  Design undrained strength cud frac14 135

                                                                                                                  14frac14 96 kN=m2

                                                                                                                  Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                                  frac14 7241 kN

                                                                                                                  Design load Vd frac14 4100 kN

                                                                                                                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                  84

                                                                                                                  For 0 frac14 40

                                                                                                                  Nq frac14 64 N frac14 95

                                                                                                                  qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                                  (a) Water table 5m below ground level

                                                                                                                  qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                                  qn frac14 400 17 frac14 383 kN=m2

                                                                                                                  F frac14 2686

                                                                                                                  383frac14 70

                                                                                                                  (b) Water table 1m below ground level (ie at foundation level)

                                                                                                                  0 frac14 20 98 frac14 102 kN=m3

                                                                                                                  62 Bearing capacity

                                                                                                                  qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                                  F frac14 2040

                                                                                                                  383frac14 53

                                                                                                                  (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                                  eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                                  qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                                  F frac14 1296

                                                                                                                  392frac14 33

                                                                                                                  85

                                                                                                                  The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                                  Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                                  Design value of 0 frac14 tan1tan 39

                                                                                                                  125

                                                                                                                  frac14 33

                                                                                                                  For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                                  Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                  86

                                                                                                                  (a) Undrained shear for u frac14 0

                                                                                                                  Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                  qnf frac14 12cuNc

                                                                                                                  frac14 12 100 514 frac14 617 kN=m2

                                                                                                                  qn frac14 qnfFfrac14 617

                                                                                                                  3frac14 206 kN=m2

                                                                                                                  q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                                  Bearing capacity 63

                                                                                                                  Drained shear for 0 frac14 32

                                                                                                                  Nq frac14 23 N frac14 25

                                                                                                                  0 frac14 21 98 frac14 112 kN=m3

                                                                                                                  qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                  frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                  frac14 694 kN=m2

                                                                                                                  q frac14 694

                                                                                                                  3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                  Design load frac14 42 227 frac14 3632 kN

                                                                                                                  (b) Design undrained strength cud frac14 100

                                                                                                                  14frac14 71 kNm2

                                                                                                                  Design bearing resistance Rd frac14 12cudNe area

                                                                                                                  frac14 12 71 514 42

                                                                                                                  frac14 7007 kN

                                                                                                                  For drained shear 0d frac14 tan1tan 32

                                                                                                                  125

                                                                                                                  frac14 26

                                                                                                                  Nq frac14 12 N frac14 10

                                                                                                                  Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                  (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                  Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                  1 2 100 0175 0700qn 0182qn

                                                                                                                  2 6 033 0044 0176qn 0046qn

                                                                                                                  3 10 020 0017 0068qn 0018qn

                                                                                                                  0246qn

                                                                                                                  Diameter of equivalent circle B frac14 45m

                                                                                                                  H

                                                                                                                  Bfrac14 12

                                                                                                                  45frac14 27 and A frac14 042

                                                                                                                  13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                  64 Bearing capacity

                                                                                                                  For sc frac14 30mm

                                                                                                                  qn frac14 30

                                                                                                                  0147frac14 204 kN=m2

                                                                                                                  q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                  Design load frac14 42 225 frac14 3600 kN

                                                                                                                  The design load is 3600 kN settlement being the limiting criterion

                                                                                                                  87

                                                                                                                  D

                                                                                                                  Bfrac14 8

                                                                                                                  4frac14 20

                                                                                                                  From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                  F frac14 cuNc

                                                                                                                  Dfrac14 40 71

                                                                                                                  20 8frac14 18

                                                                                                                  88

                                                                                                                  Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                  Design value of 0 frac14 tan1tan 38

                                                                                                                  125

                                                                                                                  frac14 32

                                                                                                                  Figure Q86

                                                                                                                  Bearing capacity 65

                                                                                                                  For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                  Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                  The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                  Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                  For B frac14 250m qn frac14 3750

                                                                                                                  2502 17 frac14 583 kN=m2

                                                                                                                  From Figure 510 m frac14 n frac14 126

                                                                                                                  6frac14 021

                                                                                                                  Ir frac14 0019

                                                                                                                  Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                  Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                  The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                  89

                                                                                                                  Depth (m) N 0v (kNm2) CN N1

                                                                                                                  070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                  Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                  (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                  Cw frac14 05thorn 0530

                                                                                                                  47

                                                                                                                  frac14 082

                                                                                                                  66 Bearing capacity

                                                                                                                  Thus

                                                                                                                  qa frac14 150 082 frac14 120 kN=m2

                                                                                                                  (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                  Thus

                                                                                                                  qa frac14 90 15 frac14 135 kN=m2

                                                                                                                  (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                  Ic frac14 171

                                                                                                                  1014frac14 0068

                                                                                                                  From Equation 819(a) with s frac14 25mm

                                                                                                                  q frac14 25

                                                                                                                  3507 0068frac14 150 kN=m2

                                                                                                                  810

                                                                                                                  Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                  Izp frac14 05thorn 01130

                                                                                                                  16 27

                                                                                                                  05

                                                                                                                  frac14 067

                                                                                                                  Refer to Figure Q810

                                                                                                                  E frac14 25qc

                                                                                                                  Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                  Ez (mm3MN)

                                                                                                                  1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                  0203

                                                                                                                  C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                  130frac14 093

                                                                                                                  C2 frac14 1 ethsayTHORN

                                                                                                                  s frac14 C1C2qnX Iz

                                                                                                                  Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                  Bearing capacity 67

                                                                                                                  811

                                                                                                                  At pile base level

                                                                                                                  cu frac14 220 kN=m2

                                                                                                                  qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                  Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                  00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                  qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                  Then

                                                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                                                  frac14

                                                                                                                  4 32 1980

                                                                                                                  thorn eth 105 139 86THORN

                                                                                                                  frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                  0 01 02 03 04 05 06 07

                                                                                                                  0 2 4 6 8 10 12 14

                                                                                                                  1

                                                                                                                  2

                                                                                                                  3

                                                                                                                  4

                                                                                                                  5

                                                                                                                  6

                                                                                                                  7

                                                                                                                  8

                                                                                                                  (1)

                                                                                                                  (2)

                                                                                                                  (3)

                                                                                                                  (4)

                                                                                                                  (5)

                                                                                                                  qc

                                                                                                                  qc

                                                                                                                  Iz

                                                                                                                  Iz

                                                                                                                  (MNm2)

                                                                                                                  z (m)

                                                                                                                  Figure Q810

                                                                                                                  68 Bearing capacity

                                                                                                                  Allowable load

                                                                                                                  ethaTHORN Qf

                                                                                                                  2frac14 17 937

                                                                                                                  2frac14 8968 kN

                                                                                                                  ethbTHORN Abqb

                                                                                                                  3thorn Asqs frac14 13 996

                                                                                                                  3thorn 3941 frac14 8606 kN

                                                                                                                  ie allowable load frac14 8600 kN

                                                                                                                  Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                  According to the limit state method

                                                                                                                  Characteristic undrained strength at base level cuk frac14 220

                                                                                                                  150kN=m2

                                                                                                                  Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                  150frac14 1320 kN=m2

                                                                                                                  Characteristic shaft resistance qsk frac14 00150

                                                                                                                  frac14 86

                                                                                                                  150frac14 57 kN=m2

                                                                                                                  Characteristic base and shaft resistances

                                                                                                                  Rbk frac14

                                                                                                                  4 32 1320 frac14 9330 kN

                                                                                                                  Rsk frac14 105 139 86

                                                                                                                  150frac14 2629 kN

                                                                                                                  For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                  Design bearing resistance Rcd frac14 9330

                                                                                                                  160thorn 2629

                                                                                                                  130

                                                                                                                  frac14 5831thorn 2022

                                                                                                                  frac14 7850 kN

                                                                                                                  Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                  812

                                                                                                                  ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                  qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                  For a single pile

                                                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                                                  frac14

                                                                                                                  4 062 1305

                                                                                                                  thorn eth 06 15 42THORN

                                                                                                                  frac14 369thorn 1187 frac14 1556 kN

                                                                                                                  Bearing capacity 69

                                                                                                                  Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                  eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                  (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                  (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                  qbkfrac14 9cuk frac14 9 220

                                                                                                                  150frac14 1320 kN=m2

                                                                                                                  qskfrac14cuk frac14 040 105

                                                                                                                  150frac14 28 kN=m2

                                                                                                                  Rbkfrac14

                                                                                                                  4 0602 1320 frac14 373 kN

                                                                                                                  Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                  Rcdfrac14 373

                                                                                                                  160thorn 791

                                                                                                                  130frac14 233thorn 608 frac14 841 kN

                                                                                                                  Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                  Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                  (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                  q frac14 21 000

                                                                                                                  1762frac14 68 kN=m2

                                                                                                                  Immediate settlement

                                                                                                                  H

                                                                                                                  Bfrac14 15

                                                                                                                  176frac14 085

                                                                                                                  D

                                                                                                                  Bfrac14 13

                                                                                                                  176frac14 074

                                                                                                                  L

                                                                                                                  Bfrac14 1

                                                                                                                  Hence from Figure 515

                                                                                                                  130 frac14 078 and 131 frac14 041

                                                                                                                  70 Bearing capacity

                                                                                                                  Thus using Equation 528

                                                                                                                  si frac14 078 041 68 176

                                                                                                                  65frac14 6mm

                                                                                                                  Consolidation settlement

                                                                                                                  Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                  1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                  434 (sod)

                                                                                                                  Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                  sc frac14 056 434 frac14 24mm

                                                                                                                  The total settlement is (6thorn 24) frac14 30mm

                                                                                                                  813

                                                                                                                  At base level N frac14 26 Then using Equation 830

                                                                                                                  qb frac14 40NDb

                                                                                                                  Bfrac14 40 26 2

                                                                                                                  025frac14 8320 kN=m2

                                                                                                                  ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                  Figure Q812

                                                                                                                  Bearing capacity 71

                                                                                                                  Over the length embedded in sand

                                                                                                                  N frac14 21 ie18thorn 24

                                                                                                                  2

                                                                                                                  Using Equation 831

                                                                                                                  qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                  For a single pile

                                                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                                                  frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                  For the pile group assuming a group efficiency of 12

                                                                                                                  XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                  Then the load factor is

                                                                                                                  F frac14 6523

                                                                                                                  2000thorn 1000frac14 21

                                                                                                                  (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                  Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                  150frac14 5547 kNm2

                                                                                                                  Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                  150frac14 28 kNm2

                                                                                                                  Characteristic base and shaft resistances for a single pile

                                                                                                                  Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                  Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                  For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                  Design bearing resistance Rcd frac14 347

                                                                                                                  130thorn 56

                                                                                                                  130frac14 310 kN

                                                                                                                  For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                  Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                  (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                  From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                  72 Bearing capacity

                                                                                                                  N frac14 24thorn 26thorn 34

                                                                                                                  3frac14 28

                                                                                                                  Ic frac14 171

                                                                                                                  2814frac14 0016 ethEquation 818THORN

                                                                                                                  s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                  814

                                                                                                                  Using Equation 841

                                                                                                                  Tf frac14 DLcu thorn

                                                                                                                  4ethD2 d2THORNcuNc

                                                                                                                  frac14 eth 02 5 06 110THORN thorn

                                                                                                                  4eth022 012THORN110 9

                                                                                                                  frac14 207thorn 23 frac14 230 kN

                                                                                                                  Figure Q813

                                                                                                                  Bearing capacity 73

                                                                                                                  Chapter 9

                                                                                                                  Stability of slopes

                                                                                                                  91

                                                                                                                  Referring to Figure Q91

                                                                                                                  W frac14 417 19 frac14 792 kN=m

                                                                                                                  Q frac14 20 28 frac14 56 kN=m

                                                                                                                  Arc lengthAB frac14

                                                                                                                  180 73 90 frac14 115m

                                                                                                                  Arc length BC frac14

                                                                                                                  180 28 90 frac14 44m

                                                                                                                  The factor of safety is given by

                                                                                                                  F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                  90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                  Depth of tension crack z0 frac14 2cu

                                                                                                                  frac14 2 20

                                                                                                                  19frac14 21m

                                                                                                                  Arc length BD frac14

                                                                                                                  180 13

                                                                                                                  1

                                                                                                                  2 90 frac14 21m

                                                                                                                  F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                  The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                  Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                  14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                  Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                  The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                  92

                                                                                                                  u frac14 0

                                                                                                                  Depth factor D frac14 11

                                                                                                                  9frac14 122

                                                                                                                  Using Equation 92 with F frac14 10

                                                                                                                  Ns frac14 cu

                                                                                                                  FHfrac14 30

                                                                                                                  10 19 9frac14 0175

                                                                                                                  Hence from Figure 93

                                                                                                                  frac14 50

                                                                                                                  For F frac14 12

                                                                                                                  Ns frac14 30

                                                                                                                  12 19 9frac14 0146

                                                                                                                  frac14 27

                                                                                                                  93

                                                                                                                  Refer to Figure Q93

                                                                                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                  1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                  74 m

                                                                                                                  214 1deg

                                                                                                                  213 1deg

                                                                                                                  39 m

                                                                                                                  WB

                                                                                                                  D

                                                                                                                  C

                                                                                                                  28 m

                                                                                                                  21 m

                                                                                                                  A

                                                                                                                  Q

                                                                                                                  Soil (1)Soil (2)

                                                                                                                  73deg

                                                                                                                  Figure Q91

                                                                                                                  Stability of slopes 75

                                                                                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                  9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                  599 256 328 1372

                                                                                                                  Figure Q93

                                                                                                                  76 Stability of slopes

                                                                                                                  XW cos frac14 b

                                                                                                                  Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                  W sin frac14 bX

                                                                                                                  h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                  Arc length La frac14

                                                                                                                  180 57

                                                                                                                  1

                                                                                                                  2 326 frac14 327m

                                                                                                                  The factor of safety is given by

                                                                                                                  F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                  W sin

                                                                                                                  frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                  frac14 091

                                                                                                                  According to the limit state method

                                                                                                                  0d frac14 tan1tan 32

                                                                                                                  125

                                                                                                                  frac14 265

                                                                                                                  c0 frac14 8

                                                                                                                  160frac14 5 kN=m2

                                                                                                                  Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                  Design disturbing moment frac14 1075 kN=m

                                                                                                                  The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                  94

                                                                                                                  F frac14 1

                                                                                                                  W sin

                                                                                                                  Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                  1thorn ethtan tan0=FTHORN

                                                                                                                  c0 frac14 8 kN=m2

                                                                                                                  0 frac14 32

                                                                                                                  c0b frac14 8 2 frac14 16 kN=m

                                                                                                                  W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                  Try F frac14 100

                                                                                                                  tan0

                                                                                                                  Ffrac14 0625

                                                                                                                  Stability of slopes 77

                                                                                                                  Values of u are as obtained in Figure Q93

                                                                                                                  SliceNo

                                                                                                                  h(m)

                                                                                                                  W frac14 bh(kNm)

                                                                                                                  W sin(kNm)

                                                                                                                  ub(kNm)

                                                                                                                  c0bthorn (W ub) tan0(kNm)

                                                                                                                  sec

                                                                                                                  1thorn (tan tan0)FProduct(kNm)

                                                                                                                  1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                  23 33 30 1042 31

                                                                                                                  3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                  224 92 72 0931 67

                                                                                                                  6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                  12 58 112 90 0889 80

                                                                                                                  8 60 252 1812

                                                                                                                  80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                  10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                  2154 88 116 0853 99

                                                                                                                  14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                  1074 1091

                                                                                                                  F frac14 1091

                                                                                                                  1074frac14 102 (assumed value 100)

                                                                                                                  Thus

                                                                                                                  F frac14 101

                                                                                                                  95

                                                                                                                  F frac14 1

                                                                                                                  W sin

                                                                                                                  XfWeth1 ruTHORN tan0g sec

                                                                                                                  1thorn ethtan tan0THORN=F

                                                                                                                  0 frac14 33

                                                                                                                  ru frac14 020

                                                                                                                  W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                  eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                  Try F frac14 110

                                                                                                                  tan 0

                                                                                                                  Ffrac14 tan 33

                                                                                                                  110frac14 0590

                                                                                                                  78 Stability of slopes

                                                                                                                  Referring to Figure Q95

                                                                                                                  SliceNo

                                                                                                                  h(m)

                                                                                                                  W frac14 bh(kNm)

                                                                                                                  W sin(kNm)

                                                                                                                  W(1 ru) tan0(kNm)

                                                                                                                  sec

                                                                                                                  1thorn ( tan tan0)FProduct(kNm)

                                                                                                                  1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                  2120 234 0892 209

                                                                                                                  4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                  1185 1271

                                                                                                                  Figure Q95

                                                                                                                  Stability of slopes 79

                                                                                                                  F frac14 1271

                                                                                                                  1185frac14 107

                                                                                                                  The trial value was 110 therefore take F to be 108

                                                                                                                  96

                                                                                                                  (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                  F frac14 0

                                                                                                                  sat

                                                                                                                  tan0

                                                                                                                  tan

                                                                                                                  ptie 15 frac14 92

                                                                                                                  19

                                                                                                                  tan 36

                                                                                                                  tan

                                                                                                                  tan frac14 0234

                                                                                                                  frac14 13

                                                                                                                  Water table well below surface the factor of safety is given by Equation 911

                                                                                                                  F frac14 tan0

                                                                                                                  tan

                                                                                                                  frac14 tan 36

                                                                                                                  tan 13

                                                                                                                  frac14 31

                                                                                                                  (b) 0d frac14 tan1tan 36

                                                                                                                  125

                                                                                                                  frac14 30

                                                                                                                  Depth of potential failure surface frac14 z

                                                                                                                  Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                  frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                  frac14 504z kN

                                                                                                                  Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                  frac14 19 z sin 13 cos 13

                                                                                                                  frac14 416z kN

                                                                                                                  Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                  80 Stability of slopes

                                                                                                                  • Book Cover
                                                                                                                  • Title
                                                                                                                  • Contents
                                                                                                                  • Basic characteristics of soils
                                                                                                                  • Seepage
                                                                                                                  • Effective stress
                                                                                                                  • Shear strength
                                                                                                                  • Stresses and displacements
                                                                                                                  • Lateral earth pressure
                                                                                                                  • Consolidation theory
                                                                                                                  • Bearing capacity
                                                                                                                  • Stability of slopes

                                                                                                                    Final void ratio

                                                                                                                    e1 frac14 w1Gs frac14 0232 272 frac14 0631

                                                                                                                    e

                                                                                                                    Hfrac14 1thorn e0

                                                                                                                    H0frac14 1thorn e1 thorne

                                                                                                                    H0

                                                                                                                    ie

                                                                                                                    e

                                                                                                                    180frac14 1631thorne

                                                                                                                    1710

                                                                                                                    e frac14 2936

                                                                                                                    1530frac14 0192

                                                                                                                    Initial void ratio e0 frac14 0631thorn 0192 frac14 0823Then

                                                                                                                    mv frac14 1

                                                                                                                    1thorn e0 e0 e101 00

                                                                                                                    frac14 1

                                                                                                                    1823 0192

                                                                                                                    0107frac14 098m2=MN

                                                                                                                    k frac14 cvmvw frac14 265 098 98

                                                                                                                    60 1440 365 103frac14 81 1010 m=s

                                                                                                                    72

                                                                                                                    Using Equation 77 (one-dimensional method)

                                                                                                                    sc frac14 e0 e11thorn e0 H

                                                                                                                    Appropriate values of e are obtained from Figure Q72 The clay will be divided intofour sublayers hence H frac14 2000mm

                                                                                                                    Figure Q72

                                                                                                                    52 Consolidation theory

                                                                                                                    Settlement

                                                                                                                    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                                                    1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                                                    318

                                                                                                                    Notes 5 92y 460thorn 84

                                                                                                                    Heave

                                                                                                                    Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                                                    1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                                                    38

                                                                                                                    73

                                                                                                                    U frac14 f ethTvTHORN frac14 f cvt

                                                                                                                    d2

                                                                                                                    Hence if cv is constant

                                                                                                                    t1

                                                                                                                    t2frac14 d

                                                                                                                    21

                                                                                                                    d22

                                                                                                                    where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                                                    d1 frac14 95mm and d2 frac14 2500mm

                                                                                                                    for U frac14 050 t2 frac14 t1 d22

                                                                                                                    d21

                                                                                                                    frac14 20

                                                                                                                    60 24 365 25002

                                                                                                                    952frac14 263 years

                                                                                                                    for U lt 060 Tv frac14

                                                                                                                    4U2 (Equation 724(a))

                                                                                                                    t030 frac14 t050 0302

                                                                                                                    0502

                                                                                                                    frac14 263 036 frac14 095 years

                                                                                                                    Consolidation theory 53

                                                                                                                    74

                                                                                                                    The layer is open

                                                                                                                    d frac14 8

                                                                                                                    2frac14 4m

                                                                                                                    Tv frac14 cvtd2frac14 24 3

                                                                                                                    42frac14 0450

                                                                                                                    ui frac14 frac14 84 kN=m2

                                                                                                                    The excess pore water pressure is given by Equation 721

                                                                                                                    ue frac14Xmfrac141mfrac140

                                                                                                                    2ui

                                                                                                                    Msin

                                                                                                                    Mz

                                                                                                                    d

                                                                                                                    expethM2TvTHORN

                                                                                                                    In this case z frac14 d

                                                                                                                    sinMz

                                                                                                                    d

                                                                                                                    frac14 sinM

                                                                                                                    where

                                                                                                                    M frac14

                                                                                                                    23

                                                                                                                    25

                                                                                                                    2

                                                                                                                    M sin M M2Tv exp (M2Tv)

                                                                                                                    2thorn1 1110 0329

                                                                                                                    3

                                                                                                                    21 9993 457 105

                                                                                                                    ue frac14 2 84 2

                                                                                                                    1 0329 ethother terms negligibleTHORN

                                                                                                                    frac14 352 kN=m2

                                                                                                                    75

                                                                                                                    The layer is open

                                                                                                                    d frac14 6

                                                                                                                    2frac14 3m

                                                                                                                    Tv frac14 cvtd2frac14 10 3

                                                                                                                    32frac14 0333

                                                                                                                    The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                                                    54 Consolidation theory

                                                                                                                    For an open layer

                                                                                                                    Tv frac14 4n

                                                                                                                    m2

                                                                                                                    n frac14 0333 62

                                                                                                                    4frac14 300

                                                                                                                    The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                                                    ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                                                    i j

                                                                                                                    0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                                                    0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                                                    The initial and 3-year isochrones are plotted in Figure Q75

                                                                                                                    Area under initial isochrone frac14 180 units

                                                                                                                    Area under 3-year isochrone frac14 63 units

                                                                                                                    The average degree of consolidation is given by Equation 725Thus

                                                                                                                    U frac14 1 63

                                                                                                                    180frac14 065

                                                                                                                    Figure Q75

                                                                                                                    Consolidation theory 55

                                                                                                                    76

                                                                                                                    At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                                    0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                                    The final consolidation settlement (one-dimensional method) is

                                                                                                                    sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                                    Corrected time t frac14 2 1

                                                                                                                    2

                                                                                                                    40

                                                                                                                    52

                                                                                                                    frac14 1615 years

                                                                                                                    Tv frac14 cvtd2frac14 44 1615

                                                                                                                    42frac14 0444

                                                                                                                    From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                                    77

                                                                                                                    The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                                    Figure Q77

                                                                                                                    56 Consolidation theory

                                                                                                                    Point m n Ir (kNm2) sc (mm)

                                                                                                                    13020frac14 15 20

                                                                                                                    20frac14 10 0194 (4) 113 124

                                                                                                                    260

                                                                                                                    20frac14 30

                                                                                                                    20

                                                                                                                    20frac14 10 0204 (2) 59 65

                                                                                                                    360

                                                                                                                    20frac14 30

                                                                                                                    40

                                                                                                                    20frac14 20 0238 (1) 35 38

                                                                                                                    430

                                                                                                                    20frac14 15

                                                                                                                    40

                                                                                                                    20frac14 20 0224 (2) 65 72

                                                                                                                    Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                                    78

                                                                                                                    Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                                    (a) Immediate settlement

                                                                                                                    H

                                                                                                                    Bfrac14 30

                                                                                                                    35frac14 086

                                                                                                                    D

                                                                                                                    Bfrac14 2

                                                                                                                    35frac14 006

                                                                                                                    Figure Q78

                                                                                                                    Consolidation theory 57

                                                                                                                    From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                                    si frac14 130131qB

                                                                                                                    Eufrac14 10 032 105 35

                                                                                                                    40frac14 30mm

                                                                                                                    (b) Consolidation settlement

                                                                                                                    Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                                    1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                                    3150

                                                                                                                    Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                                    Now

                                                                                                                    H

                                                                                                                    Bfrac14 30

                                                                                                                    35frac14 086 and A frac14 065

                                                                                                                    from Figure 712 13 frac14 079

                                                                                                                    sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                                    Total settlement

                                                                                                                    s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                                    79

                                                                                                                    Without sand drains

                                                                                                                    Uv frac14 025

                                                                                                                    Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                                    t frac14 Tvd2

                                                                                                                    cvfrac14 0049 82

                                                                                                                    cvWith sand drains

                                                                                                                    R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                                    n frac14 Rrfrac14 169

                                                                                                                    015frac14 113

                                                                                                                    Tr frac14 cht

                                                                                                                    4R2frac14 ch

                                                                                                                    4 1692 0049 82

                                                                                                                    cvethand ch frac14 cvTHORN

                                                                                                                    frac14 0275

                                                                                                                    Ur frac14 073 (from Figure 730)

                                                                                                                    58 Consolidation theory

                                                                                                                    Using Equation 740

                                                                                                                    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                                    U frac14 080

                                                                                                                    710

                                                                                                                    Without sand drains

                                                                                                                    Uv frac14 090

                                                                                                                    Tv frac14 0848

                                                                                                                    t frac14 Tvd2

                                                                                                                    cvfrac14 0848 102

                                                                                                                    96frac14 88 years

                                                                                                                    With sand drains

                                                                                                                    R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                                    n frac14 Rrfrac14 226

                                                                                                                    015frac14 15

                                                                                                                    Tr

                                                                                                                    Tvfrac14 chcv

                                                                                                                    d2

                                                                                                                    4R2ethsame tTHORN

                                                                                                                    Tr

                                                                                                                    Tvfrac14 140

                                                                                                                    96 102

                                                                                                                    4 2262frac14 714 eth1THORN

                                                                                                                    Using Equation 740

                                                                                                                    eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                                    An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                                    Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                                    040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                                    Thus

                                                                                                                    Uv frac14 0295 and Ur frac14 086

                                                                                                                    t frac14 88 00683

                                                                                                                    0848frac14 07 years

                                                                                                                    Consolidation theory 59

                                                                                                                    Chapter 8

                                                                                                                    Bearing capacity

                                                                                                                    81

                                                                                                                    (a) The ultimate bearing capacity is given by Equation 83

                                                                                                                    qf frac14 cNc thorn DNq thorn 1

                                                                                                                    2BN

                                                                                                                    For u frac14 0

                                                                                                                    Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                    qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                                    The net ultimate bearing capacity is

                                                                                                                    qnf frac14 qf D frac14 540 kN=m2

                                                                                                                    The net foundation pressure is

                                                                                                                    qn frac14 q D frac14 425

                                                                                                                    2 eth21 1THORN frac14 192 kN=m2

                                                                                                                    The factor of safety (Equation 86) is

                                                                                                                    F frac14 qnfqnfrac14 540

                                                                                                                    192frac14 28

                                                                                                                    (b) For 0 frac14 28

                                                                                                                    Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                                    qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                                    2 112 2 13

                                                                                                                    frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                                    qnf frac14 574 112 frac14 563 kN=m2

                                                                                                                    F frac14 563

                                                                                                                    192frac14 29

                                                                                                                    (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                                    82

                                                                                                                    For 0 frac14 38

                                                                                                                    Nq frac14 49 N frac14 67

                                                                                                                    qnf frac14 DethNq 1THORN thorn 1

                                                                                                                    2BN ethfrom Equation 83THORN

                                                                                                                    frac14 eth18 075 48THORN thorn 1

                                                                                                                    2 18 15 67

                                                                                                                    frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                                    qn frac14 500

                                                                                                                    15 eth18 075THORN frac14 320 kN=m2

                                                                                                                    F frac14 qnfqnfrac14 1553

                                                                                                                    320frac14 48

                                                                                                                    0d frac14 tan1tan 38

                                                                                                                    125

                                                                                                                    frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                                    Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                                    2 18 15 25

                                                                                                                    frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                                    Design load (action) Vd frac14 500 kN=m

                                                                                                                    The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                                    83

                                                                                                                    D

                                                                                                                    Bfrac14 350

                                                                                                                    225frac14 155

                                                                                                                    From Figure 85 for a square foundation

                                                                                                                    Nc frac14 81

                                                                                                                    Bearing capacity 61

                                                                                                                    For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                                    Nc frac14 084thorn 016B

                                                                                                                    L

                                                                                                                    81 frac14 745

                                                                                                                    Using Equation 810

                                                                                                                    qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                                    For F frac14 3

                                                                                                                    qn frac14 1006

                                                                                                                    3frac14 335 kN=m2

                                                                                                                    q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                                    Design load frac14 405 450 225 frac14 4100 kN

                                                                                                                    Design undrained strength cud frac14 135

                                                                                                                    14frac14 96 kN=m2

                                                                                                                    Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                                    frac14 7241 kN

                                                                                                                    Design load Vd frac14 4100 kN

                                                                                                                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                    84

                                                                                                                    For 0 frac14 40

                                                                                                                    Nq frac14 64 N frac14 95

                                                                                                                    qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                                    (a) Water table 5m below ground level

                                                                                                                    qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                                    qn frac14 400 17 frac14 383 kN=m2

                                                                                                                    F frac14 2686

                                                                                                                    383frac14 70

                                                                                                                    (b) Water table 1m below ground level (ie at foundation level)

                                                                                                                    0 frac14 20 98 frac14 102 kN=m3

                                                                                                                    62 Bearing capacity

                                                                                                                    qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                                    F frac14 2040

                                                                                                                    383frac14 53

                                                                                                                    (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                                    eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                                    qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                                    F frac14 1296

                                                                                                                    392frac14 33

                                                                                                                    85

                                                                                                                    The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                                    Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                                    Design value of 0 frac14 tan1tan 39

                                                                                                                    125

                                                                                                                    frac14 33

                                                                                                                    For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                                    Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                    86

                                                                                                                    (a) Undrained shear for u frac14 0

                                                                                                                    Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                    qnf frac14 12cuNc

                                                                                                                    frac14 12 100 514 frac14 617 kN=m2

                                                                                                                    qn frac14 qnfFfrac14 617

                                                                                                                    3frac14 206 kN=m2

                                                                                                                    q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                                    Bearing capacity 63

                                                                                                                    Drained shear for 0 frac14 32

                                                                                                                    Nq frac14 23 N frac14 25

                                                                                                                    0 frac14 21 98 frac14 112 kN=m3

                                                                                                                    qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                    frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                    frac14 694 kN=m2

                                                                                                                    q frac14 694

                                                                                                                    3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                    Design load frac14 42 227 frac14 3632 kN

                                                                                                                    (b) Design undrained strength cud frac14 100

                                                                                                                    14frac14 71 kNm2

                                                                                                                    Design bearing resistance Rd frac14 12cudNe area

                                                                                                                    frac14 12 71 514 42

                                                                                                                    frac14 7007 kN

                                                                                                                    For drained shear 0d frac14 tan1tan 32

                                                                                                                    125

                                                                                                                    frac14 26

                                                                                                                    Nq frac14 12 N frac14 10

                                                                                                                    Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                    (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                    Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                    1 2 100 0175 0700qn 0182qn

                                                                                                                    2 6 033 0044 0176qn 0046qn

                                                                                                                    3 10 020 0017 0068qn 0018qn

                                                                                                                    0246qn

                                                                                                                    Diameter of equivalent circle B frac14 45m

                                                                                                                    H

                                                                                                                    Bfrac14 12

                                                                                                                    45frac14 27 and A frac14 042

                                                                                                                    13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                    64 Bearing capacity

                                                                                                                    For sc frac14 30mm

                                                                                                                    qn frac14 30

                                                                                                                    0147frac14 204 kN=m2

                                                                                                                    q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                    Design load frac14 42 225 frac14 3600 kN

                                                                                                                    The design load is 3600 kN settlement being the limiting criterion

                                                                                                                    87

                                                                                                                    D

                                                                                                                    Bfrac14 8

                                                                                                                    4frac14 20

                                                                                                                    From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                    F frac14 cuNc

                                                                                                                    Dfrac14 40 71

                                                                                                                    20 8frac14 18

                                                                                                                    88

                                                                                                                    Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                    Design value of 0 frac14 tan1tan 38

                                                                                                                    125

                                                                                                                    frac14 32

                                                                                                                    Figure Q86

                                                                                                                    Bearing capacity 65

                                                                                                                    For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                    Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                    The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                    Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                    For B frac14 250m qn frac14 3750

                                                                                                                    2502 17 frac14 583 kN=m2

                                                                                                                    From Figure 510 m frac14 n frac14 126

                                                                                                                    6frac14 021

                                                                                                                    Ir frac14 0019

                                                                                                                    Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                    Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                    The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                    89

                                                                                                                    Depth (m) N 0v (kNm2) CN N1

                                                                                                                    070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                    Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                    (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                    Cw frac14 05thorn 0530

                                                                                                                    47

                                                                                                                    frac14 082

                                                                                                                    66 Bearing capacity

                                                                                                                    Thus

                                                                                                                    qa frac14 150 082 frac14 120 kN=m2

                                                                                                                    (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                    Thus

                                                                                                                    qa frac14 90 15 frac14 135 kN=m2

                                                                                                                    (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                    Ic frac14 171

                                                                                                                    1014frac14 0068

                                                                                                                    From Equation 819(a) with s frac14 25mm

                                                                                                                    q frac14 25

                                                                                                                    3507 0068frac14 150 kN=m2

                                                                                                                    810

                                                                                                                    Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                    Izp frac14 05thorn 01130

                                                                                                                    16 27

                                                                                                                    05

                                                                                                                    frac14 067

                                                                                                                    Refer to Figure Q810

                                                                                                                    E frac14 25qc

                                                                                                                    Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                    Ez (mm3MN)

                                                                                                                    1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                    0203

                                                                                                                    C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                    130frac14 093

                                                                                                                    C2 frac14 1 ethsayTHORN

                                                                                                                    s frac14 C1C2qnX Iz

                                                                                                                    Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                    Bearing capacity 67

                                                                                                                    811

                                                                                                                    At pile base level

                                                                                                                    cu frac14 220 kN=m2

                                                                                                                    qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                    Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                    00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                    qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                    Then

                                                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                                                    frac14

                                                                                                                    4 32 1980

                                                                                                                    thorn eth 105 139 86THORN

                                                                                                                    frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                    0 01 02 03 04 05 06 07

                                                                                                                    0 2 4 6 8 10 12 14

                                                                                                                    1

                                                                                                                    2

                                                                                                                    3

                                                                                                                    4

                                                                                                                    5

                                                                                                                    6

                                                                                                                    7

                                                                                                                    8

                                                                                                                    (1)

                                                                                                                    (2)

                                                                                                                    (3)

                                                                                                                    (4)

                                                                                                                    (5)

                                                                                                                    qc

                                                                                                                    qc

                                                                                                                    Iz

                                                                                                                    Iz

                                                                                                                    (MNm2)

                                                                                                                    z (m)

                                                                                                                    Figure Q810

                                                                                                                    68 Bearing capacity

                                                                                                                    Allowable load

                                                                                                                    ethaTHORN Qf

                                                                                                                    2frac14 17 937

                                                                                                                    2frac14 8968 kN

                                                                                                                    ethbTHORN Abqb

                                                                                                                    3thorn Asqs frac14 13 996

                                                                                                                    3thorn 3941 frac14 8606 kN

                                                                                                                    ie allowable load frac14 8600 kN

                                                                                                                    Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                    According to the limit state method

                                                                                                                    Characteristic undrained strength at base level cuk frac14 220

                                                                                                                    150kN=m2

                                                                                                                    Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                    150frac14 1320 kN=m2

                                                                                                                    Characteristic shaft resistance qsk frac14 00150

                                                                                                                    frac14 86

                                                                                                                    150frac14 57 kN=m2

                                                                                                                    Characteristic base and shaft resistances

                                                                                                                    Rbk frac14

                                                                                                                    4 32 1320 frac14 9330 kN

                                                                                                                    Rsk frac14 105 139 86

                                                                                                                    150frac14 2629 kN

                                                                                                                    For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                    Design bearing resistance Rcd frac14 9330

                                                                                                                    160thorn 2629

                                                                                                                    130

                                                                                                                    frac14 5831thorn 2022

                                                                                                                    frac14 7850 kN

                                                                                                                    Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                    812

                                                                                                                    ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                    qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                    For a single pile

                                                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                                                    frac14

                                                                                                                    4 062 1305

                                                                                                                    thorn eth 06 15 42THORN

                                                                                                                    frac14 369thorn 1187 frac14 1556 kN

                                                                                                                    Bearing capacity 69

                                                                                                                    Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                    eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                    (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                    (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                    qbkfrac14 9cuk frac14 9 220

                                                                                                                    150frac14 1320 kN=m2

                                                                                                                    qskfrac14cuk frac14 040 105

                                                                                                                    150frac14 28 kN=m2

                                                                                                                    Rbkfrac14

                                                                                                                    4 0602 1320 frac14 373 kN

                                                                                                                    Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                    Rcdfrac14 373

                                                                                                                    160thorn 791

                                                                                                                    130frac14 233thorn 608 frac14 841 kN

                                                                                                                    Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                    Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                    (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                    q frac14 21 000

                                                                                                                    1762frac14 68 kN=m2

                                                                                                                    Immediate settlement

                                                                                                                    H

                                                                                                                    Bfrac14 15

                                                                                                                    176frac14 085

                                                                                                                    D

                                                                                                                    Bfrac14 13

                                                                                                                    176frac14 074

                                                                                                                    L

                                                                                                                    Bfrac14 1

                                                                                                                    Hence from Figure 515

                                                                                                                    130 frac14 078 and 131 frac14 041

                                                                                                                    70 Bearing capacity

                                                                                                                    Thus using Equation 528

                                                                                                                    si frac14 078 041 68 176

                                                                                                                    65frac14 6mm

                                                                                                                    Consolidation settlement

                                                                                                                    Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                    1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                    434 (sod)

                                                                                                                    Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                    sc frac14 056 434 frac14 24mm

                                                                                                                    The total settlement is (6thorn 24) frac14 30mm

                                                                                                                    813

                                                                                                                    At base level N frac14 26 Then using Equation 830

                                                                                                                    qb frac14 40NDb

                                                                                                                    Bfrac14 40 26 2

                                                                                                                    025frac14 8320 kN=m2

                                                                                                                    ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                    Figure Q812

                                                                                                                    Bearing capacity 71

                                                                                                                    Over the length embedded in sand

                                                                                                                    N frac14 21 ie18thorn 24

                                                                                                                    2

                                                                                                                    Using Equation 831

                                                                                                                    qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                    For a single pile

                                                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                                                    frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                    For the pile group assuming a group efficiency of 12

                                                                                                                    XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                    Then the load factor is

                                                                                                                    F frac14 6523

                                                                                                                    2000thorn 1000frac14 21

                                                                                                                    (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                    Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                    150frac14 5547 kNm2

                                                                                                                    Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                    150frac14 28 kNm2

                                                                                                                    Characteristic base and shaft resistances for a single pile

                                                                                                                    Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                    Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                    For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                    Design bearing resistance Rcd frac14 347

                                                                                                                    130thorn 56

                                                                                                                    130frac14 310 kN

                                                                                                                    For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                    Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                    (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                    From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                    72 Bearing capacity

                                                                                                                    N frac14 24thorn 26thorn 34

                                                                                                                    3frac14 28

                                                                                                                    Ic frac14 171

                                                                                                                    2814frac14 0016 ethEquation 818THORN

                                                                                                                    s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                    814

                                                                                                                    Using Equation 841

                                                                                                                    Tf frac14 DLcu thorn

                                                                                                                    4ethD2 d2THORNcuNc

                                                                                                                    frac14 eth 02 5 06 110THORN thorn

                                                                                                                    4eth022 012THORN110 9

                                                                                                                    frac14 207thorn 23 frac14 230 kN

                                                                                                                    Figure Q813

                                                                                                                    Bearing capacity 73

                                                                                                                    Chapter 9

                                                                                                                    Stability of slopes

                                                                                                                    91

                                                                                                                    Referring to Figure Q91

                                                                                                                    W frac14 417 19 frac14 792 kN=m

                                                                                                                    Q frac14 20 28 frac14 56 kN=m

                                                                                                                    Arc lengthAB frac14

                                                                                                                    180 73 90 frac14 115m

                                                                                                                    Arc length BC frac14

                                                                                                                    180 28 90 frac14 44m

                                                                                                                    The factor of safety is given by

                                                                                                                    F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                    90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                    Depth of tension crack z0 frac14 2cu

                                                                                                                    frac14 2 20

                                                                                                                    19frac14 21m

                                                                                                                    Arc length BD frac14

                                                                                                                    180 13

                                                                                                                    1

                                                                                                                    2 90 frac14 21m

                                                                                                                    F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                    The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                    Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                    14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                    Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                    The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                    92

                                                                                                                    u frac14 0

                                                                                                                    Depth factor D frac14 11

                                                                                                                    9frac14 122

                                                                                                                    Using Equation 92 with F frac14 10

                                                                                                                    Ns frac14 cu

                                                                                                                    FHfrac14 30

                                                                                                                    10 19 9frac14 0175

                                                                                                                    Hence from Figure 93

                                                                                                                    frac14 50

                                                                                                                    For F frac14 12

                                                                                                                    Ns frac14 30

                                                                                                                    12 19 9frac14 0146

                                                                                                                    frac14 27

                                                                                                                    93

                                                                                                                    Refer to Figure Q93

                                                                                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                    1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                    74 m

                                                                                                                    214 1deg

                                                                                                                    213 1deg

                                                                                                                    39 m

                                                                                                                    WB

                                                                                                                    D

                                                                                                                    C

                                                                                                                    28 m

                                                                                                                    21 m

                                                                                                                    A

                                                                                                                    Q

                                                                                                                    Soil (1)Soil (2)

                                                                                                                    73deg

                                                                                                                    Figure Q91

                                                                                                                    Stability of slopes 75

                                                                                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                    9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                    599 256 328 1372

                                                                                                                    Figure Q93

                                                                                                                    76 Stability of slopes

                                                                                                                    XW cos frac14 b

                                                                                                                    Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                    W sin frac14 bX

                                                                                                                    h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                    Arc length La frac14

                                                                                                                    180 57

                                                                                                                    1

                                                                                                                    2 326 frac14 327m

                                                                                                                    The factor of safety is given by

                                                                                                                    F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                    W sin

                                                                                                                    frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                    frac14 091

                                                                                                                    According to the limit state method

                                                                                                                    0d frac14 tan1tan 32

                                                                                                                    125

                                                                                                                    frac14 265

                                                                                                                    c0 frac14 8

                                                                                                                    160frac14 5 kN=m2

                                                                                                                    Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                    Design disturbing moment frac14 1075 kN=m

                                                                                                                    The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                    94

                                                                                                                    F frac14 1

                                                                                                                    W sin

                                                                                                                    Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                    1thorn ethtan tan0=FTHORN

                                                                                                                    c0 frac14 8 kN=m2

                                                                                                                    0 frac14 32

                                                                                                                    c0b frac14 8 2 frac14 16 kN=m

                                                                                                                    W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                    Try F frac14 100

                                                                                                                    tan0

                                                                                                                    Ffrac14 0625

                                                                                                                    Stability of slopes 77

                                                                                                                    Values of u are as obtained in Figure Q93

                                                                                                                    SliceNo

                                                                                                                    h(m)

                                                                                                                    W frac14 bh(kNm)

                                                                                                                    W sin(kNm)

                                                                                                                    ub(kNm)

                                                                                                                    c0bthorn (W ub) tan0(kNm)

                                                                                                                    sec

                                                                                                                    1thorn (tan tan0)FProduct(kNm)

                                                                                                                    1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                    23 33 30 1042 31

                                                                                                                    3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                    224 92 72 0931 67

                                                                                                                    6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                    12 58 112 90 0889 80

                                                                                                                    8 60 252 1812

                                                                                                                    80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                    10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                    2154 88 116 0853 99

                                                                                                                    14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                    1074 1091

                                                                                                                    F frac14 1091

                                                                                                                    1074frac14 102 (assumed value 100)

                                                                                                                    Thus

                                                                                                                    F frac14 101

                                                                                                                    95

                                                                                                                    F frac14 1

                                                                                                                    W sin

                                                                                                                    XfWeth1 ruTHORN tan0g sec

                                                                                                                    1thorn ethtan tan0THORN=F

                                                                                                                    0 frac14 33

                                                                                                                    ru frac14 020

                                                                                                                    W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                    eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                    Try F frac14 110

                                                                                                                    tan 0

                                                                                                                    Ffrac14 tan 33

                                                                                                                    110frac14 0590

                                                                                                                    78 Stability of slopes

                                                                                                                    Referring to Figure Q95

                                                                                                                    SliceNo

                                                                                                                    h(m)

                                                                                                                    W frac14 bh(kNm)

                                                                                                                    W sin(kNm)

                                                                                                                    W(1 ru) tan0(kNm)

                                                                                                                    sec

                                                                                                                    1thorn ( tan tan0)FProduct(kNm)

                                                                                                                    1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                    2120 234 0892 209

                                                                                                                    4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                    1185 1271

                                                                                                                    Figure Q95

                                                                                                                    Stability of slopes 79

                                                                                                                    F frac14 1271

                                                                                                                    1185frac14 107

                                                                                                                    The trial value was 110 therefore take F to be 108

                                                                                                                    96

                                                                                                                    (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                    F frac14 0

                                                                                                                    sat

                                                                                                                    tan0

                                                                                                                    tan

                                                                                                                    ptie 15 frac14 92

                                                                                                                    19

                                                                                                                    tan 36

                                                                                                                    tan

                                                                                                                    tan frac14 0234

                                                                                                                    frac14 13

                                                                                                                    Water table well below surface the factor of safety is given by Equation 911

                                                                                                                    F frac14 tan0

                                                                                                                    tan

                                                                                                                    frac14 tan 36

                                                                                                                    tan 13

                                                                                                                    frac14 31

                                                                                                                    (b) 0d frac14 tan1tan 36

                                                                                                                    125

                                                                                                                    frac14 30

                                                                                                                    Depth of potential failure surface frac14 z

                                                                                                                    Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                    frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                    frac14 504z kN

                                                                                                                    Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                    frac14 19 z sin 13 cos 13

                                                                                                                    frac14 416z kN

                                                                                                                    Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                    80 Stability of slopes

                                                                                                                    • Book Cover
                                                                                                                    • Title
                                                                                                                    • Contents
                                                                                                                    • Basic characteristics of soils
                                                                                                                    • Seepage
                                                                                                                    • Effective stress
                                                                                                                    • Shear strength
                                                                                                                    • Stresses and displacements
                                                                                                                    • Lateral earth pressure
                                                                                                                    • Consolidation theory
                                                                                                                    • Bearing capacity
                                                                                                                    • Stability of slopes

                                                                                                                      Settlement

                                                                                                                      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 ndash e1 sc (mm)

                                                                                                                      1 460 1300y 1236 1123 0113 1012 644 1484 1200 1108 0092 843 828 1668 1172 1095 0077 714 1012 1852 1150 1083 0067 62

                                                                                                                      318

                                                                                                                      Notes 5 92y 460thorn 84

                                                                                                                      Heave

                                                                                                                      Layer 00 (kNm2) 01 (kNm2) e0 e1 e0 e1 sc (mm)

                                                                                                                      1 1300 460 1123 1136 0013 122 1484 644 1108 1119 0011 103 1668 828 1095 1104 0009 94 1852 1012 1083 1091 0008 7

                                                                                                                      38

                                                                                                                      73

                                                                                                                      U frac14 f ethTvTHORN frac14 f cvt

                                                                                                                      d2

                                                                                                                      Hence if cv is constant

                                                                                                                      t1

                                                                                                                      t2frac14 d

                                                                                                                      21

                                                                                                                      d22

                                                                                                                      where lsquo1rsquo refers to the oedometer specimen and lsquo2rsquo the clay layerFor open layers

                                                                                                                      d1 frac14 95mm and d2 frac14 2500mm

                                                                                                                      for U frac14 050 t2 frac14 t1 d22

                                                                                                                      d21

                                                                                                                      frac14 20

                                                                                                                      60 24 365 25002

                                                                                                                      952frac14 263 years

                                                                                                                      for U lt 060 Tv frac14

                                                                                                                      4U2 (Equation 724(a))

                                                                                                                      t030 frac14 t050 0302

                                                                                                                      0502

                                                                                                                      frac14 263 036 frac14 095 years

                                                                                                                      Consolidation theory 53

                                                                                                                      74

                                                                                                                      The layer is open

                                                                                                                      d frac14 8

                                                                                                                      2frac14 4m

                                                                                                                      Tv frac14 cvtd2frac14 24 3

                                                                                                                      42frac14 0450

                                                                                                                      ui frac14 frac14 84 kN=m2

                                                                                                                      The excess pore water pressure is given by Equation 721

                                                                                                                      ue frac14Xmfrac141mfrac140

                                                                                                                      2ui

                                                                                                                      Msin

                                                                                                                      Mz

                                                                                                                      d

                                                                                                                      expethM2TvTHORN

                                                                                                                      In this case z frac14 d

                                                                                                                      sinMz

                                                                                                                      d

                                                                                                                      frac14 sinM

                                                                                                                      where

                                                                                                                      M frac14

                                                                                                                      23

                                                                                                                      25

                                                                                                                      2

                                                                                                                      M sin M M2Tv exp (M2Tv)

                                                                                                                      2thorn1 1110 0329

                                                                                                                      3

                                                                                                                      21 9993 457 105

                                                                                                                      ue frac14 2 84 2

                                                                                                                      1 0329 ethother terms negligibleTHORN

                                                                                                                      frac14 352 kN=m2

                                                                                                                      75

                                                                                                                      The layer is open

                                                                                                                      d frac14 6

                                                                                                                      2frac14 3m

                                                                                                                      Tv frac14 cvtd2frac14 10 3

                                                                                                                      32frac14 0333

                                                                                                                      The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                                                      54 Consolidation theory

                                                                                                                      For an open layer

                                                                                                                      Tv frac14 4n

                                                                                                                      m2

                                                                                                                      n frac14 0333 62

                                                                                                                      4frac14 300

                                                                                                                      The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                                                      ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                                                      i j

                                                                                                                      0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                                                      0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                                                      The initial and 3-year isochrones are plotted in Figure Q75

                                                                                                                      Area under initial isochrone frac14 180 units

                                                                                                                      Area under 3-year isochrone frac14 63 units

                                                                                                                      The average degree of consolidation is given by Equation 725Thus

                                                                                                                      U frac14 1 63

                                                                                                                      180frac14 065

                                                                                                                      Figure Q75

                                                                                                                      Consolidation theory 55

                                                                                                                      76

                                                                                                                      At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                                      0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                                      The final consolidation settlement (one-dimensional method) is

                                                                                                                      sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                                      Corrected time t frac14 2 1

                                                                                                                      2

                                                                                                                      40

                                                                                                                      52

                                                                                                                      frac14 1615 years

                                                                                                                      Tv frac14 cvtd2frac14 44 1615

                                                                                                                      42frac14 0444

                                                                                                                      From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                                      77

                                                                                                                      The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                                      Figure Q77

                                                                                                                      56 Consolidation theory

                                                                                                                      Point m n Ir (kNm2) sc (mm)

                                                                                                                      13020frac14 15 20

                                                                                                                      20frac14 10 0194 (4) 113 124

                                                                                                                      260

                                                                                                                      20frac14 30

                                                                                                                      20

                                                                                                                      20frac14 10 0204 (2) 59 65

                                                                                                                      360

                                                                                                                      20frac14 30

                                                                                                                      40

                                                                                                                      20frac14 20 0238 (1) 35 38

                                                                                                                      430

                                                                                                                      20frac14 15

                                                                                                                      40

                                                                                                                      20frac14 20 0224 (2) 65 72

                                                                                                                      Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                                      78

                                                                                                                      Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                                      (a) Immediate settlement

                                                                                                                      H

                                                                                                                      Bfrac14 30

                                                                                                                      35frac14 086

                                                                                                                      D

                                                                                                                      Bfrac14 2

                                                                                                                      35frac14 006

                                                                                                                      Figure Q78

                                                                                                                      Consolidation theory 57

                                                                                                                      From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                                      si frac14 130131qB

                                                                                                                      Eufrac14 10 032 105 35

                                                                                                                      40frac14 30mm

                                                                                                                      (b) Consolidation settlement

                                                                                                                      Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                                      1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                                      3150

                                                                                                                      Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                                      Now

                                                                                                                      H

                                                                                                                      Bfrac14 30

                                                                                                                      35frac14 086 and A frac14 065

                                                                                                                      from Figure 712 13 frac14 079

                                                                                                                      sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                                      Total settlement

                                                                                                                      s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                                      79

                                                                                                                      Without sand drains

                                                                                                                      Uv frac14 025

                                                                                                                      Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                                      t frac14 Tvd2

                                                                                                                      cvfrac14 0049 82

                                                                                                                      cvWith sand drains

                                                                                                                      R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                                      n frac14 Rrfrac14 169

                                                                                                                      015frac14 113

                                                                                                                      Tr frac14 cht

                                                                                                                      4R2frac14 ch

                                                                                                                      4 1692 0049 82

                                                                                                                      cvethand ch frac14 cvTHORN

                                                                                                                      frac14 0275

                                                                                                                      Ur frac14 073 (from Figure 730)

                                                                                                                      58 Consolidation theory

                                                                                                                      Using Equation 740

                                                                                                                      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                                      U frac14 080

                                                                                                                      710

                                                                                                                      Without sand drains

                                                                                                                      Uv frac14 090

                                                                                                                      Tv frac14 0848

                                                                                                                      t frac14 Tvd2

                                                                                                                      cvfrac14 0848 102

                                                                                                                      96frac14 88 years

                                                                                                                      With sand drains

                                                                                                                      R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                                      n frac14 Rrfrac14 226

                                                                                                                      015frac14 15

                                                                                                                      Tr

                                                                                                                      Tvfrac14 chcv

                                                                                                                      d2

                                                                                                                      4R2ethsame tTHORN

                                                                                                                      Tr

                                                                                                                      Tvfrac14 140

                                                                                                                      96 102

                                                                                                                      4 2262frac14 714 eth1THORN

                                                                                                                      Using Equation 740

                                                                                                                      eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                                      An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                                      Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                                      040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                                      Thus

                                                                                                                      Uv frac14 0295 and Ur frac14 086

                                                                                                                      t frac14 88 00683

                                                                                                                      0848frac14 07 years

                                                                                                                      Consolidation theory 59

                                                                                                                      Chapter 8

                                                                                                                      Bearing capacity

                                                                                                                      81

                                                                                                                      (a) The ultimate bearing capacity is given by Equation 83

                                                                                                                      qf frac14 cNc thorn DNq thorn 1

                                                                                                                      2BN

                                                                                                                      For u frac14 0

                                                                                                                      Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                      qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                                      The net ultimate bearing capacity is

                                                                                                                      qnf frac14 qf D frac14 540 kN=m2

                                                                                                                      The net foundation pressure is

                                                                                                                      qn frac14 q D frac14 425

                                                                                                                      2 eth21 1THORN frac14 192 kN=m2

                                                                                                                      The factor of safety (Equation 86) is

                                                                                                                      F frac14 qnfqnfrac14 540

                                                                                                                      192frac14 28

                                                                                                                      (b) For 0 frac14 28

                                                                                                                      Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                                      qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                                      2 112 2 13

                                                                                                                      frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                                      qnf frac14 574 112 frac14 563 kN=m2

                                                                                                                      F frac14 563

                                                                                                                      192frac14 29

                                                                                                                      (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                                      82

                                                                                                                      For 0 frac14 38

                                                                                                                      Nq frac14 49 N frac14 67

                                                                                                                      qnf frac14 DethNq 1THORN thorn 1

                                                                                                                      2BN ethfrom Equation 83THORN

                                                                                                                      frac14 eth18 075 48THORN thorn 1

                                                                                                                      2 18 15 67

                                                                                                                      frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                                      qn frac14 500

                                                                                                                      15 eth18 075THORN frac14 320 kN=m2

                                                                                                                      F frac14 qnfqnfrac14 1553

                                                                                                                      320frac14 48

                                                                                                                      0d frac14 tan1tan 38

                                                                                                                      125

                                                                                                                      frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                                      Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                                      2 18 15 25

                                                                                                                      frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                                      Design load (action) Vd frac14 500 kN=m

                                                                                                                      The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                                      83

                                                                                                                      D

                                                                                                                      Bfrac14 350

                                                                                                                      225frac14 155

                                                                                                                      From Figure 85 for a square foundation

                                                                                                                      Nc frac14 81

                                                                                                                      Bearing capacity 61

                                                                                                                      For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                                      Nc frac14 084thorn 016B

                                                                                                                      L

                                                                                                                      81 frac14 745

                                                                                                                      Using Equation 810

                                                                                                                      qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                                      For F frac14 3

                                                                                                                      qn frac14 1006

                                                                                                                      3frac14 335 kN=m2

                                                                                                                      q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                                      Design load frac14 405 450 225 frac14 4100 kN

                                                                                                                      Design undrained strength cud frac14 135

                                                                                                                      14frac14 96 kN=m2

                                                                                                                      Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                                      frac14 7241 kN

                                                                                                                      Design load Vd frac14 4100 kN

                                                                                                                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                      84

                                                                                                                      For 0 frac14 40

                                                                                                                      Nq frac14 64 N frac14 95

                                                                                                                      qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                                      (a) Water table 5m below ground level

                                                                                                                      qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                                      qn frac14 400 17 frac14 383 kN=m2

                                                                                                                      F frac14 2686

                                                                                                                      383frac14 70

                                                                                                                      (b) Water table 1m below ground level (ie at foundation level)

                                                                                                                      0 frac14 20 98 frac14 102 kN=m3

                                                                                                                      62 Bearing capacity

                                                                                                                      qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                                      F frac14 2040

                                                                                                                      383frac14 53

                                                                                                                      (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                                      eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                                      qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                                      F frac14 1296

                                                                                                                      392frac14 33

                                                                                                                      85

                                                                                                                      The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                                      Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                                      Design value of 0 frac14 tan1tan 39

                                                                                                                      125

                                                                                                                      frac14 33

                                                                                                                      For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                                      Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                      86

                                                                                                                      (a) Undrained shear for u frac14 0

                                                                                                                      Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                      qnf frac14 12cuNc

                                                                                                                      frac14 12 100 514 frac14 617 kN=m2

                                                                                                                      qn frac14 qnfFfrac14 617

                                                                                                                      3frac14 206 kN=m2

                                                                                                                      q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                                      Bearing capacity 63

                                                                                                                      Drained shear for 0 frac14 32

                                                                                                                      Nq frac14 23 N frac14 25

                                                                                                                      0 frac14 21 98 frac14 112 kN=m3

                                                                                                                      qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                      frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                      frac14 694 kN=m2

                                                                                                                      q frac14 694

                                                                                                                      3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                      Design load frac14 42 227 frac14 3632 kN

                                                                                                                      (b) Design undrained strength cud frac14 100

                                                                                                                      14frac14 71 kNm2

                                                                                                                      Design bearing resistance Rd frac14 12cudNe area

                                                                                                                      frac14 12 71 514 42

                                                                                                                      frac14 7007 kN

                                                                                                                      For drained shear 0d frac14 tan1tan 32

                                                                                                                      125

                                                                                                                      frac14 26

                                                                                                                      Nq frac14 12 N frac14 10

                                                                                                                      Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                      (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                      Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                      1 2 100 0175 0700qn 0182qn

                                                                                                                      2 6 033 0044 0176qn 0046qn

                                                                                                                      3 10 020 0017 0068qn 0018qn

                                                                                                                      0246qn

                                                                                                                      Diameter of equivalent circle B frac14 45m

                                                                                                                      H

                                                                                                                      Bfrac14 12

                                                                                                                      45frac14 27 and A frac14 042

                                                                                                                      13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                      64 Bearing capacity

                                                                                                                      For sc frac14 30mm

                                                                                                                      qn frac14 30

                                                                                                                      0147frac14 204 kN=m2

                                                                                                                      q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                      Design load frac14 42 225 frac14 3600 kN

                                                                                                                      The design load is 3600 kN settlement being the limiting criterion

                                                                                                                      87

                                                                                                                      D

                                                                                                                      Bfrac14 8

                                                                                                                      4frac14 20

                                                                                                                      From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                      F frac14 cuNc

                                                                                                                      Dfrac14 40 71

                                                                                                                      20 8frac14 18

                                                                                                                      88

                                                                                                                      Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                      Design value of 0 frac14 tan1tan 38

                                                                                                                      125

                                                                                                                      frac14 32

                                                                                                                      Figure Q86

                                                                                                                      Bearing capacity 65

                                                                                                                      For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                      Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                      The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                      Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                      For B frac14 250m qn frac14 3750

                                                                                                                      2502 17 frac14 583 kN=m2

                                                                                                                      From Figure 510 m frac14 n frac14 126

                                                                                                                      6frac14 021

                                                                                                                      Ir frac14 0019

                                                                                                                      Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                      Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                      The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                      89

                                                                                                                      Depth (m) N 0v (kNm2) CN N1

                                                                                                                      070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                      Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                      (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                      Cw frac14 05thorn 0530

                                                                                                                      47

                                                                                                                      frac14 082

                                                                                                                      66 Bearing capacity

                                                                                                                      Thus

                                                                                                                      qa frac14 150 082 frac14 120 kN=m2

                                                                                                                      (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                      Thus

                                                                                                                      qa frac14 90 15 frac14 135 kN=m2

                                                                                                                      (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                      Ic frac14 171

                                                                                                                      1014frac14 0068

                                                                                                                      From Equation 819(a) with s frac14 25mm

                                                                                                                      q frac14 25

                                                                                                                      3507 0068frac14 150 kN=m2

                                                                                                                      810

                                                                                                                      Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                      Izp frac14 05thorn 01130

                                                                                                                      16 27

                                                                                                                      05

                                                                                                                      frac14 067

                                                                                                                      Refer to Figure Q810

                                                                                                                      E frac14 25qc

                                                                                                                      Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                      Ez (mm3MN)

                                                                                                                      1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                      0203

                                                                                                                      C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                      130frac14 093

                                                                                                                      C2 frac14 1 ethsayTHORN

                                                                                                                      s frac14 C1C2qnX Iz

                                                                                                                      Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                      Bearing capacity 67

                                                                                                                      811

                                                                                                                      At pile base level

                                                                                                                      cu frac14 220 kN=m2

                                                                                                                      qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                      Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                      00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                      qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                      Then

                                                                                                                      Qf frac14 Abqb thorn Asqs

                                                                                                                      frac14

                                                                                                                      4 32 1980

                                                                                                                      thorn eth 105 139 86THORN

                                                                                                                      frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                      0 01 02 03 04 05 06 07

                                                                                                                      0 2 4 6 8 10 12 14

                                                                                                                      1

                                                                                                                      2

                                                                                                                      3

                                                                                                                      4

                                                                                                                      5

                                                                                                                      6

                                                                                                                      7

                                                                                                                      8

                                                                                                                      (1)

                                                                                                                      (2)

                                                                                                                      (3)

                                                                                                                      (4)

                                                                                                                      (5)

                                                                                                                      qc

                                                                                                                      qc

                                                                                                                      Iz

                                                                                                                      Iz

                                                                                                                      (MNm2)

                                                                                                                      z (m)

                                                                                                                      Figure Q810

                                                                                                                      68 Bearing capacity

                                                                                                                      Allowable load

                                                                                                                      ethaTHORN Qf

                                                                                                                      2frac14 17 937

                                                                                                                      2frac14 8968 kN

                                                                                                                      ethbTHORN Abqb

                                                                                                                      3thorn Asqs frac14 13 996

                                                                                                                      3thorn 3941 frac14 8606 kN

                                                                                                                      ie allowable load frac14 8600 kN

                                                                                                                      Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                      According to the limit state method

                                                                                                                      Characteristic undrained strength at base level cuk frac14 220

                                                                                                                      150kN=m2

                                                                                                                      Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                      150frac14 1320 kN=m2

                                                                                                                      Characteristic shaft resistance qsk frac14 00150

                                                                                                                      frac14 86

                                                                                                                      150frac14 57 kN=m2

                                                                                                                      Characteristic base and shaft resistances

                                                                                                                      Rbk frac14

                                                                                                                      4 32 1320 frac14 9330 kN

                                                                                                                      Rsk frac14 105 139 86

                                                                                                                      150frac14 2629 kN

                                                                                                                      For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                      Design bearing resistance Rcd frac14 9330

                                                                                                                      160thorn 2629

                                                                                                                      130

                                                                                                                      frac14 5831thorn 2022

                                                                                                                      frac14 7850 kN

                                                                                                                      Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                      812

                                                                                                                      ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                      qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                      For a single pile

                                                                                                                      Qf frac14 Abqb thorn Asqs

                                                                                                                      frac14

                                                                                                                      4 062 1305

                                                                                                                      thorn eth 06 15 42THORN

                                                                                                                      frac14 369thorn 1187 frac14 1556 kN

                                                                                                                      Bearing capacity 69

                                                                                                                      Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                      eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                      (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                      (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                      qbkfrac14 9cuk frac14 9 220

                                                                                                                      150frac14 1320 kN=m2

                                                                                                                      qskfrac14cuk frac14 040 105

                                                                                                                      150frac14 28 kN=m2

                                                                                                                      Rbkfrac14

                                                                                                                      4 0602 1320 frac14 373 kN

                                                                                                                      Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                      Rcdfrac14 373

                                                                                                                      160thorn 791

                                                                                                                      130frac14 233thorn 608 frac14 841 kN

                                                                                                                      Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                      Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                      (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                      q frac14 21 000

                                                                                                                      1762frac14 68 kN=m2

                                                                                                                      Immediate settlement

                                                                                                                      H

                                                                                                                      Bfrac14 15

                                                                                                                      176frac14 085

                                                                                                                      D

                                                                                                                      Bfrac14 13

                                                                                                                      176frac14 074

                                                                                                                      L

                                                                                                                      Bfrac14 1

                                                                                                                      Hence from Figure 515

                                                                                                                      130 frac14 078 and 131 frac14 041

                                                                                                                      70 Bearing capacity

                                                                                                                      Thus using Equation 528

                                                                                                                      si frac14 078 041 68 176

                                                                                                                      65frac14 6mm

                                                                                                                      Consolidation settlement

                                                                                                                      Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                      1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                      434 (sod)

                                                                                                                      Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                      sc frac14 056 434 frac14 24mm

                                                                                                                      The total settlement is (6thorn 24) frac14 30mm

                                                                                                                      813

                                                                                                                      At base level N frac14 26 Then using Equation 830

                                                                                                                      qb frac14 40NDb

                                                                                                                      Bfrac14 40 26 2

                                                                                                                      025frac14 8320 kN=m2

                                                                                                                      ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                      Figure Q812

                                                                                                                      Bearing capacity 71

                                                                                                                      Over the length embedded in sand

                                                                                                                      N frac14 21 ie18thorn 24

                                                                                                                      2

                                                                                                                      Using Equation 831

                                                                                                                      qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                      For a single pile

                                                                                                                      Qf frac14 Abqb thorn Asqs

                                                                                                                      frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                      For the pile group assuming a group efficiency of 12

                                                                                                                      XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                      Then the load factor is

                                                                                                                      F frac14 6523

                                                                                                                      2000thorn 1000frac14 21

                                                                                                                      (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                      Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                      150frac14 5547 kNm2

                                                                                                                      Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                      150frac14 28 kNm2

                                                                                                                      Characteristic base and shaft resistances for a single pile

                                                                                                                      Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                      Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                      For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                      Design bearing resistance Rcd frac14 347

                                                                                                                      130thorn 56

                                                                                                                      130frac14 310 kN

                                                                                                                      For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                      Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                      (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                      From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                      72 Bearing capacity

                                                                                                                      N frac14 24thorn 26thorn 34

                                                                                                                      3frac14 28

                                                                                                                      Ic frac14 171

                                                                                                                      2814frac14 0016 ethEquation 818THORN

                                                                                                                      s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                      814

                                                                                                                      Using Equation 841

                                                                                                                      Tf frac14 DLcu thorn

                                                                                                                      4ethD2 d2THORNcuNc

                                                                                                                      frac14 eth 02 5 06 110THORN thorn

                                                                                                                      4eth022 012THORN110 9

                                                                                                                      frac14 207thorn 23 frac14 230 kN

                                                                                                                      Figure Q813

                                                                                                                      Bearing capacity 73

                                                                                                                      Chapter 9

                                                                                                                      Stability of slopes

                                                                                                                      91

                                                                                                                      Referring to Figure Q91

                                                                                                                      W frac14 417 19 frac14 792 kN=m

                                                                                                                      Q frac14 20 28 frac14 56 kN=m

                                                                                                                      Arc lengthAB frac14

                                                                                                                      180 73 90 frac14 115m

                                                                                                                      Arc length BC frac14

                                                                                                                      180 28 90 frac14 44m

                                                                                                                      The factor of safety is given by

                                                                                                                      F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                      90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                      Depth of tension crack z0 frac14 2cu

                                                                                                                      frac14 2 20

                                                                                                                      19frac14 21m

                                                                                                                      Arc length BD frac14

                                                                                                                      180 13

                                                                                                                      1

                                                                                                                      2 90 frac14 21m

                                                                                                                      F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                      The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                      Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                      14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                      Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                      The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                      92

                                                                                                                      u frac14 0

                                                                                                                      Depth factor D frac14 11

                                                                                                                      9frac14 122

                                                                                                                      Using Equation 92 with F frac14 10

                                                                                                                      Ns frac14 cu

                                                                                                                      FHfrac14 30

                                                                                                                      10 19 9frac14 0175

                                                                                                                      Hence from Figure 93

                                                                                                                      frac14 50

                                                                                                                      For F frac14 12

                                                                                                                      Ns frac14 30

                                                                                                                      12 19 9frac14 0146

                                                                                                                      frac14 27

                                                                                                                      93

                                                                                                                      Refer to Figure Q93

                                                                                                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                      1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                      74 m

                                                                                                                      214 1deg

                                                                                                                      213 1deg

                                                                                                                      39 m

                                                                                                                      WB

                                                                                                                      D

                                                                                                                      C

                                                                                                                      28 m

                                                                                                                      21 m

                                                                                                                      A

                                                                                                                      Q

                                                                                                                      Soil (1)Soil (2)

                                                                                                                      73deg

                                                                                                                      Figure Q91

                                                                                                                      Stability of slopes 75

                                                                                                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                      9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                      599 256 328 1372

                                                                                                                      Figure Q93

                                                                                                                      76 Stability of slopes

                                                                                                                      XW cos frac14 b

                                                                                                                      Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                      W sin frac14 bX

                                                                                                                      h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                      Arc length La frac14

                                                                                                                      180 57

                                                                                                                      1

                                                                                                                      2 326 frac14 327m

                                                                                                                      The factor of safety is given by

                                                                                                                      F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                      W sin

                                                                                                                      frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                      frac14 091

                                                                                                                      According to the limit state method

                                                                                                                      0d frac14 tan1tan 32

                                                                                                                      125

                                                                                                                      frac14 265

                                                                                                                      c0 frac14 8

                                                                                                                      160frac14 5 kN=m2

                                                                                                                      Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                      Design disturbing moment frac14 1075 kN=m

                                                                                                                      The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                      94

                                                                                                                      F frac14 1

                                                                                                                      W sin

                                                                                                                      Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                      1thorn ethtan tan0=FTHORN

                                                                                                                      c0 frac14 8 kN=m2

                                                                                                                      0 frac14 32

                                                                                                                      c0b frac14 8 2 frac14 16 kN=m

                                                                                                                      W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                      Try F frac14 100

                                                                                                                      tan0

                                                                                                                      Ffrac14 0625

                                                                                                                      Stability of slopes 77

                                                                                                                      Values of u are as obtained in Figure Q93

                                                                                                                      SliceNo

                                                                                                                      h(m)

                                                                                                                      W frac14 bh(kNm)

                                                                                                                      W sin(kNm)

                                                                                                                      ub(kNm)

                                                                                                                      c0bthorn (W ub) tan0(kNm)

                                                                                                                      sec

                                                                                                                      1thorn (tan tan0)FProduct(kNm)

                                                                                                                      1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                      23 33 30 1042 31

                                                                                                                      3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                      224 92 72 0931 67

                                                                                                                      6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                      12 58 112 90 0889 80

                                                                                                                      8 60 252 1812

                                                                                                                      80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                      10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                      2154 88 116 0853 99

                                                                                                                      14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                      1074 1091

                                                                                                                      F frac14 1091

                                                                                                                      1074frac14 102 (assumed value 100)

                                                                                                                      Thus

                                                                                                                      F frac14 101

                                                                                                                      95

                                                                                                                      F frac14 1

                                                                                                                      W sin

                                                                                                                      XfWeth1 ruTHORN tan0g sec

                                                                                                                      1thorn ethtan tan0THORN=F

                                                                                                                      0 frac14 33

                                                                                                                      ru frac14 020

                                                                                                                      W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                      eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                      Try F frac14 110

                                                                                                                      tan 0

                                                                                                                      Ffrac14 tan 33

                                                                                                                      110frac14 0590

                                                                                                                      78 Stability of slopes

                                                                                                                      Referring to Figure Q95

                                                                                                                      SliceNo

                                                                                                                      h(m)

                                                                                                                      W frac14 bh(kNm)

                                                                                                                      W sin(kNm)

                                                                                                                      W(1 ru) tan0(kNm)

                                                                                                                      sec

                                                                                                                      1thorn ( tan tan0)FProduct(kNm)

                                                                                                                      1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                      2120 234 0892 209

                                                                                                                      4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                      1185 1271

                                                                                                                      Figure Q95

                                                                                                                      Stability of slopes 79

                                                                                                                      F frac14 1271

                                                                                                                      1185frac14 107

                                                                                                                      The trial value was 110 therefore take F to be 108

                                                                                                                      96

                                                                                                                      (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                      F frac14 0

                                                                                                                      sat

                                                                                                                      tan0

                                                                                                                      tan

                                                                                                                      ptie 15 frac14 92

                                                                                                                      19

                                                                                                                      tan 36

                                                                                                                      tan

                                                                                                                      tan frac14 0234

                                                                                                                      frac14 13

                                                                                                                      Water table well below surface the factor of safety is given by Equation 911

                                                                                                                      F frac14 tan0

                                                                                                                      tan

                                                                                                                      frac14 tan 36

                                                                                                                      tan 13

                                                                                                                      frac14 31

                                                                                                                      (b) 0d frac14 tan1tan 36

                                                                                                                      125

                                                                                                                      frac14 30

                                                                                                                      Depth of potential failure surface frac14 z

                                                                                                                      Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                      frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                      frac14 504z kN

                                                                                                                      Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                      frac14 19 z sin 13 cos 13

                                                                                                                      frac14 416z kN

                                                                                                                      Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                      80 Stability of slopes

                                                                                                                      • Book Cover
                                                                                                                      • Title
                                                                                                                      • Contents
                                                                                                                      • Basic characteristics of soils
                                                                                                                      • Seepage
                                                                                                                      • Effective stress
                                                                                                                      • Shear strength
                                                                                                                      • Stresses and displacements
                                                                                                                      • Lateral earth pressure
                                                                                                                      • Consolidation theory
                                                                                                                      • Bearing capacity
                                                                                                                      • Stability of slopes

                                                                                                                        74

                                                                                                                        The layer is open

                                                                                                                        d frac14 8

                                                                                                                        2frac14 4m

                                                                                                                        Tv frac14 cvtd2frac14 24 3

                                                                                                                        42frac14 0450

                                                                                                                        ui frac14 frac14 84 kN=m2

                                                                                                                        The excess pore water pressure is given by Equation 721

                                                                                                                        ue frac14Xmfrac141mfrac140

                                                                                                                        2ui

                                                                                                                        Msin

                                                                                                                        Mz

                                                                                                                        d

                                                                                                                        expethM2TvTHORN

                                                                                                                        In this case z frac14 d

                                                                                                                        sinMz

                                                                                                                        d

                                                                                                                        frac14 sinM

                                                                                                                        where

                                                                                                                        M frac14

                                                                                                                        23

                                                                                                                        25

                                                                                                                        2

                                                                                                                        M sin M M2Tv exp (M2Tv)

                                                                                                                        2thorn1 1110 0329

                                                                                                                        3

                                                                                                                        21 9993 457 105

                                                                                                                        ue frac14 2 84 2

                                                                                                                        1 0329 ethother terms negligibleTHORN

                                                                                                                        frac14 352 kN=m2

                                                                                                                        75

                                                                                                                        The layer is open

                                                                                                                        d frac14 6

                                                                                                                        2frac14 3m

                                                                                                                        Tv frac14 cvtd2frac14 10 3

                                                                                                                        32frac14 0333

                                                                                                                        The layer thickness will be divided into six equal parts ie m frac14 6

                                                                                                                        54 Consolidation theory

                                                                                                                        For an open layer

                                                                                                                        Tv frac14 4n

                                                                                                                        m2

                                                                                                                        n frac14 0333 62

                                                                                                                        4frac14 300

                                                                                                                        The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                                                        ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                                                        i j

                                                                                                                        0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                                                        0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                                                        The initial and 3-year isochrones are plotted in Figure Q75

                                                                                                                        Area under initial isochrone frac14 180 units

                                                                                                                        Area under 3-year isochrone frac14 63 units

                                                                                                                        The average degree of consolidation is given by Equation 725Thus

                                                                                                                        U frac14 1 63

                                                                                                                        180frac14 065

                                                                                                                        Figure Q75

                                                                                                                        Consolidation theory 55

                                                                                                                        76

                                                                                                                        At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                                        0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                                        The final consolidation settlement (one-dimensional method) is

                                                                                                                        sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                                        Corrected time t frac14 2 1

                                                                                                                        2

                                                                                                                        40

                                                                                                                        52

                                                                                                                        frac14 1615 years

                                                                                                                        Tv frac14 cvtd2frac14 44 1615

                                                                                                                        42frac14 0444

                                                                                                                        From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                                        77

                                                                                                                        The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                                        Figure Q77

                                                                                                                        56 Consolidation theory

                                                                                                                        Point m n Ir (kNm2) sc (mm)

                                                                                                                        13020frac14 15 20

                                                                                                                        20frac14 10 0194 (4) 113 124

                                                                                                                        260

                                                                                                                        20frac14 30

                                                                                                                        20

                                                                                                                        20frac14 10 0204 (2) 59 65

                                                                                                                        360

                                                                                                                        20frac14 30

                                                                                                                        40

                                                                                                                        20frac14 20 0238 (1) 35 38

                                                                                                                        430

                                                                                                                        20frac14 15

                                                                                                                        40

                                                                                                                        20frac14 20 0224 (2) 65 72

                                                                                                                        Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                                        78

                                                                                                                        Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                                        (a) Immediate settlement

                                                                                                                        H

                                                                                                                        Bfrac14 30

                                                                                                                        35frac14 086

                                                                                                                        D

                                                                                                                        Bfrac14 2

                                                                                                                        35frac14 006

                                                                                                                        Figure Q78

                                                                                                                        Consolidation theory 57

                                                                                                                        From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                                        si frac14 130131qB

                                                                                                                        Eufrac14 10 032 105 35

                                                                                                                        40frac14 30mm

                                                                                                                        (b) Consolidation settlement

                                                                                                                        Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                                        1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                                        3150

                                                                                                                        Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                                        Now

                                                                                                                        H

                                                                                                                        Bfrac14 30

                                                                                                                        35frac14 086 and A frac14 065

                                                                                                                        from Figure 712 13 frac14 079

                                                                                                                        sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                                        Total settlement

                                                                                                                        s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                                        79

                                                                                                                        Without sand drains

                                                                                                                        Uv frac14 025

                                                                                                                        Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                                        t frac14 Tvd2

                                                                                                                        cvfrac14 0049 82

                                                                                                                        cvWith sand drains

                                                                                                                        R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                                        n frac14 Rrfrac14 169

                                                                                                                        015frac14 113

                                                                                                                        Tr frac14 cht

                                                                                                                        4R2frac14 ch

                                                                                                                        4 1692 0049 82

                                                                                                                        cvethand ch frac14 cvTHORN

                                                                                                                        frac14 0275

                                                                                                                        Ur frac14 073 (from Figure 730)

                                                                                                                        58 Consolidation theory

                                                                                                                        Using Equation 740

                                                                                                                        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                                        U frac14 080

                                                                                                                        710

                                                                                                                        Without sand drains

                                                                                                                        Uv frac14 090

                                                                                                                        Tv frac14 0848

                                                                                                                        t frac14 Tvd2

                                                                                                                        cvfrac14 0848 102

                                                                                                                        96frac14 88 years

                                                                                                                        With sand drains

                                                                                                                        R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                                        n frac14 Rrfrac14 226

                                                                                                                        015frac14 15

                                                                                                                        Tr

                                                                                                                        Tvfrac14 chcv

                                                                                                                        d2

                                                                                                                        4R2ethsame tTHORN

                                                                                                                        Tr

                                                                                                                        Tvfrac14 140

                                                                                                                        96 102

                                                                                                                        4 2262frac14 714 eth1THORN

                                                                                                                        Using Equation 740

                                                                                                                        eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                                        An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                                        Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                                        040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                                        Thus

                                                                                                                        Uv frac14 0295 and Ur frac14 086

                                                                                                                        t frac14 88 00683

                                                                                                                        0848frac14 07 years

                                                                                                                        Consolidation theory 59

                                                                                                                        Chapter 8

                                                                                                                        Bearing capacity

                                                                                                                        81

                                                                                                                        (a) The ultimate bearing capacity is given by Equation 83

                                                                                                                        qf frac14 cNc thorn DNq thorn 1

                                                                                                                        2BN

                                                                                                                        For u frac14 0

                                                                                                                        Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                        qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                                        The net ultimate bearing capacity is

                                                                                                                        qnf frac14 qf D frac14 540 kN=m2

                                                                                                                        The net foundation pressure is

                                                                                                                        qn frac14 q D frac14 425

                                                                                                                        2 eth21 1THORN frac14 192 kN=m2

                                                                                                                        The factor of safety (Equation 86) is

                                                                                                                        F frac14 qnfqnfrac14 540

                                                                                                                        192frac14 28

                                                                                                                        (b) For 0 frac14 28

                                                                                                                        Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                                        qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                                        2 112 2 13

                                                                                                                        frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                                        qnf frac14 574 112 frac14 563 kN=m2

                                                                                                                        F frac14 563

                                                                                                                        192frac14 29

                                                                                                                        (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                                        82

                                                                                                                        For 0 frac14 38

                                                                                                                        Nq frac14 49 N frac14 67

                                                                                                                        qnf frac14 DethNq 1THORN thorn 1

                                                                                                                        2BN ethfrom Equation 83THORN

                                                                                                                        frac14 eth18 075 48THORN thorn 1

                                                                                                                        2 18 15 67

                                                                                                                        frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                                        qn frac14 500

                                                                                                                        15 eth18 075THORN frac14 320 kN=m2

                                                                                                                        F frac14 qnfqnfrac14 1553

                                                                                                                        320frac14 48

                                                                                                                        0d frac14 tan1tan 38

                                                                                                                        125

                                                                                                                        frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                                        Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                                        2 18 15 25

                                                                                                                        frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                                        Design load (action) Vd frac14 500 kN=m

                                                                                                                        The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                                        83

                                                                                                                        D

                                                                                                                        Bfrac14 350

                                                                                                                        225frac14 155

                                                                                                                        From Figure 85 for a square foundation

                                                                                                                        Nc frac14 81

                                                                                                                        Bearing capacity 61

                                                                                                                        For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                                        Nc frac14 084thorn 016B

                                                                                                                        L

                                                                                                                        81 frac14 745

                                                                                                                        Using Equation 810

                                                                                                                        qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                                        For F frac14 3

                                                                                                                        qn frac14 1006

                                                                                                                        3frac14 335 kN=m2

                                                                                                                        q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                                        Design load frac14 405 450 225 frac14 4100 kN

                                                                                                                        Design undrained strength cud frac14 135

                                                                                                                        14frac14 96 kN=m2

                                                                                                                        Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                                        frac14 7241 kN

                                                                                                                        Design load Vd frac14 4100 kN

                                                                                                                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                        84

                                                                                                                        For 0 frac14 40

                                                                                                                        Nq frac14 64 N frac14 95

                                                                                                                        qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                                        (a) Water table 5m below ground level

                                                                                                                        qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                                        qn frac14 400 17 frac14 383 kN=m2

                                                                                                                        F frac14 2686

                                                                                                                        383frac14 70

                                                                                                                        (b) Water table 1m below ground level (ie at foundation level)

                                                                                                                        0 frac14 20 98 frac14 102 kN=m3

                                                                                                                        62 Bearing capacity

                                                                                                                        qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                                        F frac14 2040

                                                                                                                        383frac14 53

                                                                                                                        (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                                        eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                                        qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                                        F frac14 1296

                                                                                                                        392frac14 33

                                                                                                                        85

                                                                                                                        The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                                        Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                                        Design value of 0 frac14 tan1tan 39

                                                                                                                        125

                                                                                                                        frac14 33

                                                                                                                        For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                                        Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                        86

                                                                                                                        (a) Undrained shear for u frac14 0

                                                                                                                        Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                        qnf frac14 12cuNc

                                                                                                                        frac14 12 100 514 frac14 617 kN=m2

                                                                                                                        qn frac14 qnfFfrac14 617

                                                                                                                        3frac14 206 kN=m2

                                                                                                                        q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                                        Bearing capacity 63

                                                                                                                        Drained shear for 0 frac14 32

                                                                                                                        Nq frac14 23 N frac14 25

                                                                                                                        0 frac14 21 98 frac14 112 kN=m3

                                                                                                                        qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                        frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                        frac14 694 kN=m2

                                                                                                                        q frac14 694

                                                                                                                        3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                        Design load frac14 42 227 frac14 3632 kN

                                                                                                                        (b) Design undrained strength cud frac14 100

                                                                                                                        14frac14 71 kNm2

                                                                                                                        Design bearing resistance Rd frac14 12cudNe area

                                                                                                                        frac14 12 71 514 42

                                                                                                                        frac14 7007 kN

                                                                                                                        For drained shear 0d frac14 tan1tan 32

                                                                                                                        125

                                                                                                                        frac14 26

                                                                                                                        Nq frac14 12 N frac14 10

                                                                                                                        Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                        (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                        Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                        1 2 100 0175 0700qn 0182qn

                                                                                                                        2 6 033 0044 0176qn 0046qn

                                                                                                                        3 10 020 0017 0068qn 0018qn

                                                                                                                        0246qn

                                                                                                                        Diameter of equivalent circle B frac14 45m

                                                                                                                        H

                                                                                                                        Bfrac14 12

                                                                                                                        45frac14 27 and A frac14 042

                                                                                                                        13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                        64 Bearing capacity

                                                                                                                        For sc frac14 30mm

                                                                                                                        qn frac14 30

                                                                                                                        0147frac14 204 kN=m2

                                                                                                                        q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                        Design load frac14 42 225 frac14 3600 kN

                                                                                                                        The design load is 3600 kN settlement being the limiting criterion

                                                                                                                        87

                                                                                                                        D

                                                                                                                        Bfrac14 8

                                                                                                                        4frac14 20

                                                                                                                        From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                        F frac14 cuNc

                                                                                                                        Dfrac14 40 71

                                                                                                                        20 8frac14 18

                                                                                                                        88

                                                                                                                        Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                        Design value of 0 frac14 tan1tan 38

                                                                                                                        125

                                                                                                                        frac14 32

                                                                                                                        Figure Q86

                                                                                                                        Bearing capacity 65

                                                                                                                        For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                        Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                        The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                        Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                        For B frac14 250m qn frac14 3750

                                                                                                                        2502 17 frac14 583 kN=m2

                                                                                                                        From Figure 510 m frac14 n frac14 126

                                                                                                                        6frac14 021

                                                                                                                        Ir frac14 0019

                                                                                                                        Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                        Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                        The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                        89

                                                                                                                        Depth (m) N 0v (kNm2) CN N1

                                                                                                                        070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                        Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                        (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                        Cw frac14 05thorn 0530

                                                                                                                        47

                                                                                                                        frac14 082

                                                                                                                        66 Bearing capacity

                                                                                                                        Thus

                                                                                                                        qa frac14 150 082 frac14 120 kN=m2

                                                                                                                        (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                        Thus

                                                                                                                        qa frac14 90 15 frac14 135 kN=m2

                                                                                                                        (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                        Ic frac14 171

                                                                                                                        1014frac14 0068

                                                                                                                        From Equation 819(a) with s frac14 25mm

                                                                                                                        q frac14 25

                                                                                                                        3507 0068frac14 150 kN=m2

                                                                                                                        810

                                                                                                                        Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                        Izp frac14 05thorn 01130

                                                                                                                        16 27

                                                                                                                        05

                                                                                                                        frac14 067

                                                                                                                        Refer to Figure Q810

                                                                                                                        E frac14 25qc

                                                                                                                        Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                        Ez (mm3MN)

                                                                                                                        1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                        0203

                                                                                                                        C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                        130frac14 093

                                                                                                                        C2 frac14 1 ethsayTHORN

                                                                                                                        s frac14 C1C2qnX Iz

                                                                                                                        Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                        Bearing capacity 67

                                                                                                                        811

                                                                                                                        At pile base level

                                                                                                                        cu frac14 220 kN=m2

                                                                                                                        qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                        Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                        00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                        qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                        Then

                                                                                                                        Qf frac14 Abqb thorn Asqs

                                                                                                                        frac14

                                                                                                                        4 32 1980

                                                                                                                        thorn eth 105 139 86THORN

                                                                                                                        frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                        0 01 02 03 04 05 06 07

                                                                                                                        0 2 4 6 8 10 12 14

                                                                                                                        1

                                                                                                                        2

                                                                                                                        3

                                                                                                                        4

                                                                                                                        5

                                                                                                                        6

                                                                                                                        7

                                                                                                                        8

                                                                                                                        (1)

                                                                                                                        (2)

                                                                                                                        (3)

                                                                                                                        (4)

                                                                                                                        (5)

                                                                                                                        qc

                                                                                                                        qc

                                                                                                                        Iz

                                                                                                                        Iz

                                                                                                                        (MNm2)

                                                                                                                        z (m)

                                                                                                                        Figure Q810

                                                                                                                        68 Bearing capacity

                                                                                                                        Allowable load

                                                                                                                        ethaTHORN Qf

                                                                                                                        2frac14 17 937

                                                                                                                        2frac14 8968 kN

                                                                                                                        ethbTHORN Abqb

                                                                                                                        3thorn Asqs frac14 13 996

                                                                                                                        3thorn 3941 frac14 8606 kN

                                                                                                                        ie allowable load frac14 8600 kN

                                                                                                                        Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                        According to the limit state method

                                                                                                                        Characteristic undrained strength at base level cuk frac14 220

                                                                                                                        150kN=m2

                                                                                                                        Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                        150frac14 1320 kN=m2

                                                                                                                        Characteristic shaft resistance qsk frac14 00150

                                                                                                                        frac14 86

                                                                                                                        150frac14 57 kN=m2

                                                                                                                        Characteristic base and shaft resistances

                                                                                                                        Rbk frac14

                                                                                                                        4 32 1320 frac14 9330 kN

                                                                                                                        Rsk frac14 105 139 86

                                                                                                                        150frac14 2629 kN

                                                                                                                        For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                        Design bearing resistance Rcd frac14 9330

                                                                                                                        160thorn 2629

                                                                                                                        130

                                                                                                                        frac14 5831thorn 2022

                                                                                                                        frac14 7850 kN

                                                                                                                        Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                        812

                                                                                                                        ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                        qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                        For a single pile

                                                                                                                        Qf frac14 Abqb thorn Asqs

                                                                                                                        frac14

                                                                                                                        4 062 1305

                                                                                                                        thorn eth 06 15 42THORN

                                                                                                                        frac14 369thorn 1187 frac14 1556 kN

                                                                                                                        Bearing capacity 69

                                                                                                                        Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                        eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                        (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                        (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                        qbkfrac14 9cuk frac14 9 220

                                                                                                                        150frac14 1320 kN=m2

                                                                                                                        qskfrac14cuk frac14 040 105

                                                                                                                        150frac14 28 kN=m2

                                                                                                                        Rbkfrac14

                                                                                                                        4 0602 1320 frac14 373 kN

                                                                                                                        Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                        Rcdfrac14 373

                                                                                                                        160thorn 791

                                                                                                                        130frac14 233thorn 608 frac14 841 kN

                                                                                                                        Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                        Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                        (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                        q frac14 21 000

                                                                                                                        1762frac14 68 kN=m2

                                                                                                                        Immediate settlement

                                                                                                                        H

                                                                                                                        Bfrac14 15

                                                                                                                        176frac14 085

                                                                                                                        D

                                                                                                                        Bfrac14 13

                                                                                                                        176frac14 074

                                                                                                                        L

                                                                                                                        Bfrac14 1

                                                                                                                        Hence from Figure 515

                                                                                                                        130 frac14 078 and 131 frac14 041

                                                                                                                        70 Bearing capacity

                                                                                                                        Thus using Equation 528

                                                                                                                        si frac14 078 041 68 176

                                                                                                                        65frac14 6mm

                                                                                                                        Consolidation settlement

                                                                                                                        Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                        1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                        434 (sod)

                                                                                                                        Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                        sc frac14 056 434 frac14 24mm

                                                                                                                        The total settlement is (6thorn 24) frac14 30mm

                                                                                                                        813

                                                                                                                        At base level N frac14 26 Then using Equation 830

                                                                                                                        qb frac14 40NDb

                                                                                                                        Bfrac14 40 26 2

                                                                                                                        025frac14 8320 kN=m2

                                                                                                                        ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                        Figure Q812

                                                                                                                        Bearing capacity 71

                                                                                                                        Over the length embedded in sand

                                                                                                                        N frac14 21 ie18thorn 24

                                                                                                                        2

                                                                                                                        Using Equation 831

                                                                                                                        qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                        For a single pile

                                                                                                                        Qf frac14 Abqb thorn Asqs

                                                                                                                        frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                        For the pile group assuming a group efficiency of 12

                                                                                                                        XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                        Then the load factor is

                                                                                                                        F frac14 6523

                                                                                                                        2000thorn 1000frac14 21

                                                                                                                        (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                        Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                        150frac14 5547 kNm2

                                                                                                                        Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                        150frac14 28 kNm2

                                                                                                                        Characteristic base and shaft resistances for a single pile

                                                                                                                        Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                        Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                        For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                        Design bearing resistance Rcd frac14 347

                                                                                                                        130thorn 56

                                                                                                                        130frac14 310 kN

                                                                                                                        For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                        Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                        (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                        From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                        72 Bearing capacity

                                                                                                                        N frac14 24thorn 26thorn 34

                                                                                                                        3frac14 28

                                                                                                                        Ic frac14 171

                                                                                                                        2814frac14 0016 ethEquation 818THORN

                                                                                                                        s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                        814

                                                                                                                        Using Equation 841

                                                                                                                        Tf frac14 DLcu thorn

                                                                                                                        4ethD2 d2THORNcuNc

                                                                                                                        frac14 eth 02 5 06 110THORN thorn

                                                                                                                        4eth022 012THORN110 9

                                                                                                                        frac14 207thorn 23 frac14 230 kN

                                                                                                                        Figure Q813

                                                                                                                        Bearing capacity 73

                                                                                                                        Chapter 9

                                                                                                                        Stability of slopes

                                                                                                                        91

                                                                                                                        Referring to Figure Q91

                                                                                                                        W frac14 417 19 frac14 792 kN=m

                                                                                                                        Q frac14 20 28 frac14 56 kN=m

                                                                                                                        Arc lengthAB frac14

                                                                                                                        180 73 90 frac14 115m

                                                                                                                        Arc length BC frac14

                                                                                                                        180 28 90 frac14 44m

                                                                                                                        The factor of safety is given by

                                                                                                                        F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                        90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                        Depth of tension crack z0 frac14 2cu

                                                                                                                        frac14 2 20

                                                                                                                        19frac14 21m

                                                                                                                        Arc length BD frac14

                                                                                                                        180 13

                                                                                                                        1

                                                                                                                        2 90 frac14 21m

                                                                                                                        F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                        The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                        Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                        14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                        Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                        The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                        92

                                                                                                                        u frac14 0

                                                                                                                        Depth factor D frac14 11

                                                                                                                        9frac14 122

                                                                                                                        Using Equation 92 with F frac14 10

                                                                                                                        Ns frac14 cu

                                                                                                                        FHfrac14 30

                                                                                                                        10 19 9frac14 0175

                                                                                                                        Hence from Figure 93

                                                                                                                        frac14 50

                                                                                                                        For F frac14 12

                                                                                                                        Ns frac14 30

                                                                                                                        12 19 9frac14 0146

                                                                                                                        frac14 27

                                                                                                                        93

                                                                                                                        Refer to Figure Q93

                                                                                                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                        1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                        74 m

                                                                                                                        214 1deg

                                                                                                                        213 1deg

                                                                                                                        39 m

                                                                                                                        WB

                                                                                                                        D

                                                                                                                        C

                                                                                                                        28 m

                                                                                                                        21 m

                                                                                                                        A

                                                                                                                        Q

                                                                                                                        Soil (1)Soil (2)

                                                                                                                        73deg

                                                                                                                        Figure Q91

                                                                                                                        Stability of slopes 75

                                                                                                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                        9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                        599 256 328 1372

                                                                                                                        Figure Q93

                                                                                                                        76 Stability of slopes

                                                                                                                        XW cos frac14 b

                                                                                                                        Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                        W sin frac14 bX

                                                                                                                        h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                        Arc length La frac14

                                                                                                                        180 57

                                                                                                                        1

                                                                                                                        2 326 frac14 327m

                                                                                                                        The factor of safety is given by

                                                                                                                        F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                        W sin

                                                                                                                        frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                        frac14 091

                                                                                                                        According to the limit state method

                                                                                                                        0d frac14 tan1tan 32

                                                                                                                        125

                                                                                                                        frac14 265

                                                                                                                        c0 frac14 8

                                                                                                                        160frac14 5 kN=m2

                                                                                                                        Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                        Design disturbing moment frac14 1075 kN=m

                                                                                                                        The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                        94

                                                                                                                        F frac14 1

                                                                                                                        W sin

                                                                                                                        Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                        1thorn ethtan tan0=FTHORN

                                                                                                                        c0 frac14 8 kN=m2

                                                                                                                        0 frac14 32

                                                                                                                        c0b frac14 8 2 frac14 16 kN=m

                                                                                                                        W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                        Try F frac14 100

                                                                                                                        tan0

                                                                                                                        Ffrac14 0625

                                                                                                                        Stability of slopes 77

                                                                                                                        Values of u are as obtained in Figure Q93

                                                                                                                        SliceNo

                                                                                                                        h(m)

                                                                                                                        W frac14 bh(kNm)

                                                                                                                        W sin(kNm)

                                                                                                                        ub(kNm)

                                                                                                                        c0bthorn (W ub) tan0(kNm)

                                                                                                                        sec

                                                                                                                        1thorn (tan tan0)FProduct(kNm)

                                                                                                                        1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                        23 33 30 1042 31

                                                                                                                        3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                        224 92 72 0931 67

                                                                                                                        6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                        12 58 112 90 0889 80

                                                                                                                        8 60 252 1812

                                                                                                                        80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                        10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                        2154 88 116 0853 99

                                                                                                                        14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                        1074 1091

                                                                                                                        F frac14 1091

                                                                                                                        1074frac14 102 (assumed value 100)

                                                                                                                        Thus

                                                                                                                        F frac14 101

                                                                                                                        95

                                                                                                                        F frac14 1

                                                                                                                        W sin

                                                                                                                        XfWeth1 ruTHORN tan0g sec

                                                                                                                        1thorn ethtan tan0THORN=F

                                                                                                                        0 frac14 33

                                                                                                                        ru frac14 020

                                                                                                                        W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                        eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                        Try F frac14 110

                                                                                                                        tan 0

                                                                                                                        Ffrac14 tan 33

                                                                                                                        110frac14 0590

                                                                                                                        78 Stability of slopes

                                                                                                                        Referring to Figure Q95

                                                                                                                        SliceNo

                                                                                                                        h(m)

                                                                                                                        W frac14 bh(kNm)

                                                                                                                        W sin(kNm)

                                                                                                                        W(1 ru) tan0(kNm)

                                                                                                                        sec

                                                                                                                        1thorn ( tan tan0)FProduct(kNm)

                                                                                                                        1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                        2120 234 0892 209

                                                                                                                        4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                        1185 1271

                                                                                                                        Figure Q95

                                                                                                                        Stability of slopes 79

                                                                                                                        F frac14 1271

                                                                                                                        1185frac14 107

                                                                                                                        The trial value was 110 therefore take F to be 108

                                                                                                                        96

                                                                                                                        (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                        F frac14 0

                                                                                                                        sat

                                                                                                                        tan0

                                                                                                                        tan

                                                                                                                        ptie 15 frac14 92

                                                                                                                        19

                                                                                                                        tan 36

                                                                                                                        tan

                                                                                                                        tan frac14 0234

                                                                                                                        frac14 13

                                                                                                                        Water table well below surface the factor of safety is given by Equation 911

                                                                                                                        F frac14 tan0

                                                                                                                        tan

                                                                                                                        frac14 tan 36

                                                                                                                        tan 13

                                                                                                                        frac14 31

                                                                                                                        (b) 0d frac14 tan1tan 36

                                                                                                                        125

                                                                                                                        frac14 30

                                                                                                                        Depth of potential failure surface frac14 z

                                                                                                                        Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                        frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                        frac14 504z kN

                                                                                                                        Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                        frac14 19 z sin 13 cos 13

                                                                                                                        frac14 416z kN

                                                                                                                        Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                        80 Stability of slopes

                                                                                                                        • Book Cover
                                                                                                                        • Title
                                                                                                                        • Contents
                                                                                                                        • Basic characteristics of soils
                                                                                                                        • Seepage
                                                                                                                        • Effective stress
                                                                                                                        • Shear strength
                                                                                                                        • Stresses and displacements
                                                                                                                        • Lateral earth pressure
                                                                                                                        • Consolidation theory
                                                                                                                        • Bearing capacity
                                                                                                                        • Stability of slopes

                                                                                                                          For an open layer

                                                                                                                          Tv frac14 4n

                                                                                                                          m2

                                                                                                                          n frac14 0333 62

                                                                                                                          4frac14 300

                                                                                                                          The value of n will be taken as 12 (ie t frac14 312 frac14 14 year) making frac14 025 Thecomputation is set out below all pressures having been multiplied by 10

                                                                                                                          ui jthorn1 frac14 ui j thorn 025ethui1 j thorn uithorn1 j 2ui jTHORN

                                                                                                                          i j

                                                                                                                          0 1 2 3 4 5 6 7 8 9 10 11 12

                                                                                                                          0 0 0 0 0 0 0 0 0 0 0 0 0 01 500 350 275 228 195 171 151 136 123 112 102 94 872 400 400 362 325 292 264 240 219 201 185 171 158 1463 300 300 300 292 277 261 245 230 215 201 187 175 1634 200 200 200 200 198 193 189 180 171 163 154 145 1375 100 100 100 100 100 995 98 96 93 89 85 81 776 0 0 0 0 0 0 0 0 0 0 0 0 0

                                                                                                                          The initial and 3-year isochrones are plotted in Figure Q75

                                                                                                                          Area under initial isochrone frac14 180 units

                                                                                                                          Area under 3-year isochrone frac14 63 units

                                                                                                                          The average degree of consolidation is given by Equation 725Thus

                                                                                                                          U frac14 1 63

                                                                                                                          180frac14 065

                                                                                                                          Figure Q75

                                                                                                                          Consolidation theory 55

                                                                                                                          76

                                                                                                                          At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                                          0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                                          The final consolidation settlement (one-dimensional method) is

                                                                                                                          sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                                          Corrected time t frac14 2 1

                                                                                                                          2

                                                                                                                          40

                                                                                                                          52

                                                                                                                          frac14 1615 years

                                                                                                                          Tv frac14 cvtd2frac14 44 1615

                                                                                                                          42frac14 0444

                                                                                                                          From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                                          77

                                                                                                                          The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                                          Figure Q77

                                                                                                                          56 Consolidation theory

                                                                                                                          Point m n Ir (kNm2) sc (mm)

                                                                                                                          13020frac14 15 20

                                                                                                                          20frac14 10 0194 (4) 113 124

                                                                                                                          260

                                                                                                                          20frac14 30

                                                                                                                          20

                                                                                                                          20frac14 10 0204 (2) 59 65

                                                                                                                          360

                                                                                                                          20frac14 30

                                                                                                                          40

                                                                                                                          20frac14 20 0238 (1) 35 38

                                                                                                                          430

                                                                                                                          20frac14 15

                                                                                                                          40

                                                                                                                          20frac14 20 0224 (2) 65 72

                                                                                                                          Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                                          78

                                                                                                                          Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                                          (a) Immediate settlement

                                                                                                                          H

                                                                                                                          Bfrac14 30

                                                                                                                          35frac14 086

                                                                                                                          D

                                                                                                                          Bfrac14 2

                                                                                                                          35frac14 006

                                                                                                                          Figure Q78

                                                                                                                          Consolidation theory 57

                                                                                                                          From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                                          si frac14 130131qB

                                                                                                                          Eufrac14 10 032 105 35

                                                                                                                          40frac14 30mm

                                                                                                                          (b) Consolidation settlement

                                                                                                                          Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                                          1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                                          3150

                                                                                                                          Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                                          Now

                                                                                                                          H

                                                                                                                          Bfrac14 30

                                                                                                                          35frac14 086 and A frac14 065

                                                                                                                          from Figure 712 13 frac14 079

                                                                                                                          sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                                          Total settlement

                                                                                                                          s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                                          79

                                                                                                                          Without sand drains

                                                                                                                          Uv frac14 025

                                                                                                                          Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                                          t frac14 Tvd2

                                                                                                                          cvfrac14 0049 82

                                                                                                                          cvWith sand drains

                                                                                                                          R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                                          n frac14 Rrfrac14 169

                                                                                                                          015frac14 113

                                                                                                                          Tr frac14 cht

                                                                                                                          4R2frac14 ch

                                                                                                                          4 1692 0049 82

                                                                                                                          cvethand ch frac14 cvTHORN

                                                                                                                          frac14 0275

                                                                                                                          Ur frac14 073 (from Figure 730)

                                                                                                                          58 Consolidation theory

                                                                                                                          Using Equation 740

                                                                                                                          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                                          U frac14 080

                                                                                                                          710

                                                                                                                          Without sand drains

                                                                                                                          Uv frac14 090

                                                                                                                          Tv frac14 0848

                                                                                                                          t frac14 Tvd2

                                                                                                                          cvfrac14 0848 102

                                                                                                                          96frac14 88 years

                                                                                                                          With sand drains

                                                                                                                          R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                                          n frac14 Rrfrac14 226

                                                                                                                          015frac14 15

                                                                                                                          Tr

                                                                                                                          Tvfrac14 chcv

                                                                                                                          d2

                                                                                                                          4R2ethsame tTHORN

                                                                                                                          Tr

                                                                                                                          Tvfrac14 140

                                                                                                                          96 102

                                                                                                                          4 2262frac14 714 eth1THORN

                                                                                                                          Using Equation 740

                                                                                                                          eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                                          An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                                          Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                                          040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                                          Thus

                                                                                                                          Uv frac14 0295 and Ur frac14 086

                                                                                                                          t frac14 88 00683

                                                                                                                          0848frac14 07 years

                                                                                                                          Consolidation theory 59

                                                                                                                          Chapter 8

                                                                                                                          Bearing capacity

                                                                                                                          81

                                                                                                                          (a) The ultimate bearing capacity is given by Equation 83

                                                                                                                          qf frac14 cNc thorn DNq thorn 1

                                                                                                                          2BN

                                                                                                                          For u frac14 0

                                                                                                                          Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                          qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                                          The net ultimate bearing capacity is

                                                                                                                          qnf frac14 qf D frac14 540 kN=m2

                                                                                                                          The net foundation pressure is

                                                                                                                          qn frac14 q D frac14 425

                                                                                                                          2 eth21 1THORN frac14 192 kN=m2

                                                                                                                          The factor of safety (Equation 86) is

                                                                                                                          F frac14 qnfqnfrac14 540

                                                                                                                          192frac14 28

                                                                                                                          (b) For 0 frac14 28

                                                                                                                          Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                                          qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                                          2 112 2 13

                                                                                                                          frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                                          qnf frac14 574 112 frac14 563 kN=m2

                                                                                                                          F frac14 563

                                                                                                                          192frac14 29

                                                                                                                          (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                                          82

                                                                                                                          For 0 frac14 38

                                                                                                                          Nq frac14 49 N frac14 67

                                                                                                                          qnf frac14 DethNq 1THORN thorn 1

                                                                                                                          2BN ethfrom Equation 83THORN

                                                                                                                          frac14 eth18 075 48THORN thorn 1

                                                                                                                          2 18 15 67

                                                                                                                          frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                                          qn frac14 500

                                                                                                                          15 eth18 075THORN frac14 320 kN=m2

                                                                                                                          F frac14 qnfqnfrac14 1553

                                                                                                                          320frac14 48

                                                                                                                          0d frac14 tan1tan 38

                                                                                                                          125

                                                                                                                          frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                                          Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                                          2 18 15 25

                                                                                                                          frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                                          Design load (action) Vd frac14 500 kN=m

                                                                                                                          The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                                          83

                                                                                                                          D

                                                                                                                          Bfrac14 350

                                                                                                                          225frac14 155

                                                                                                                          From Figure 85 for a square foundation

                                                                                                                          Nc frac14 81

                                                                                                                          Bearing capacity 61

                                                                                                                          For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                                          Nc frac14 084thorn 016B

                                                                                                                          L

                                                                                                                          81 frac14 745

                                                                                                                          Using Equation 810

                                                                                                                          qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                                          For F frac14 3

                                                                                                                          qn frac14 1006

                                                                                                                          3frac14 335 kN=m2

                                                                                                                          q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                                          Design load frac14 405 450 225 frac14 4100 kN

                                                                                                                          Design undrained strength cud frac14 135

                                                                                                                          14frac14 96 kN=m2

                                                                                                                          Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                                          frac14 7241 kN

                                                                                                                          Design load Vd frac14 4100 kN

                                                                                                                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                          84

                                                                                                                          For 0 frac14 40

                                                                                                                          Nq frac14 64 N frac14 95

                                                                                                                          qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                                          (a) Water table 5m below ground level

                                                                                                                          qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                                          qn frac14 400 17 frac14 383 kN=m2

                                                                                                                          F frac14 2686

                                                                                                                          383frac14 70

                                                                                                                          (b) Water table 1m below ground level (ie at foundation level)

                                                                                                                          0 frac14 20 98 frac14 102 kN=m3

                                                                                                                          62 Bearing capacity

                                                                                                                          qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                                          F frac14 2040

                                                                                                                          383frac14 53

                                                                                                                          (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                                          eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                                          qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                                          F frac14 1296

                                                                                                                          392frac14 33

                                                                                                                          85

                                                                                                                          The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                                          Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                                          Design value of 0 frac14 tan1tan 39

                                                                                                                          125

                                                                                                                          frac14 33

                                                                                                                          For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                                          Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                          86

                                                                                                                          (a) Undrained shear for u frac14 0

                                                                                                                          Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                          qnf frac14 12cuNc

                                                                                                                          frac14 12 100 514 frac14 617 kN=m2

                                                                                                                          qn frac14 qnfFfrac14 617

                                                                                                                          3frac14 206 kN=m2

                                                                                                                          q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                                          Bearing capacity 63

                                                                                                                          Drained shear for 0 frac14 32

                                                                                                                          Nq frac14 23 N frac14 25

                                                                                                                          0 frac14 21 98 frac14 112 kN=m3

                                                                                                                          qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                          frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                          frac14 694 kN=m2

                                                                                                                          q frac14 694

                                                                                                                          3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                          Design load frac14 42 227 frac14 3632 kN

                                                                                                                          (b) Design undrained strength cud frac14 100

                                                                                                                          14frac14 71 kNm2

                                                                                                                          Design bearing resistance Rd frac14 12cudNe area

                                                                                                                          frac14 12 71 514 42

                                                                                                                          frac14 7007 kN

                                                                                                                          For drained shear 0d frac14 tan1tan 32

                                                                                                                          125

                                                                                                                          frac14 26

                                                                                                                          Nq frac14 12 N frac14 10

                                                                                                                          Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                          (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                          Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                          1 2 100 0175 0700qn 0182qn

                                                                                                                          2 6 033 0044 0176qn 0046qn

                                                                                                                          3 10 020 0017 0068qn 0018qn

                                                                                                                          0246qn

                                                                                                                          Diameter of equivalent circle B frac14 45m

                                                                                                                          H

                                                                                                                          Bfrac14 12

                                                                                                                          45frac14 27 and A frac14 042

                                                                                                                          13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                          64 Bearing capacity

                                                                                                                          For sc frac14 30mm

                                                                                                                          qn frac14 30

                                                                                                                          0147frac14 204 kN=m2

                                                                                                                          q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                          Design load frac14 42 225 frac14 3600 kN

                                                                                                                          The design load is 3600 kN settlement being the limiting criterion

                                                                                                                          87

                                                                                                                          D

                                                                                                                          Bfrac14 8

                                                                                                                          4frac14 20

                                                                                                                          From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                          F frac14 cuNc

                                                                                                                          Dfrac14 40 71

                                                                                                                          20 8frac14 18

                                                                                                                          88

                                                                                                                          Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                          Design value of 0 frac14 tan1tan 38

                                                                                                                          125

                                                                                                                          frac14 32

                                                                                                                          Figure Q86

                                                                                                                          Bearing capacity 65

                                                                                                                          For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                          Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                          The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                          Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                          For B frac14 250m qn frac14 3750

                                                                                                                          2502 17 frac14 583 kN=m2

                                                                                                                          From Figure 510 m frac14 n frac14 126

                                                                                                                          6frac14 021

                                                                                                                          Ir frac14 0019

                                                                                                                          Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                          Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                          The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                          89

                                                                                                                          Depth (m) N 0v (kNm2) CN N1

                                                                                                                          070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                          Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                          (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                          Cw frac14 05thorn 0530

                                                                                                                          47

                                                                                                                          frac14 082

                                                                                                                          66 Bearing capacity

                                                                                                                          Thus

                                                                                                                          qa frac14 150 082 frac14 120 kN=m2

                                                                                                                          (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                          Thus

                                                                                                                          qa frac14 90 15 frac14 135 kN=m2

                                                                                                                          (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                          Ic frac14 171

                                                                                                                          1014frac14 0068

                                                                                                                          From Equation 819(a) with s frac14 25mm

                                                                                                                          q frac14 25

                                                                                                                          3507 0068frac14 150 kN=m2

                                                                                                                          810

                                                                                                                          Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                          Izp frac14 05thorn 01130

                                                                                                                          16 27

                                                                                                                          05

                                                                                                                          frac14 067

                                                                                                                          Refer to Figure Q810

                                                                                                                          E frac14 25qc

                                                                                                                          Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                          Ez (mm3MN)

                                                                                                                          1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                          0203

                                                                                                                          C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                          130frac14 093

                                                                                                                          C2 frac14 1 ethsayTHORN

                                                                                                                          s frac14 C1C2qnX Iz

                                                                                                                          Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                          Bearing capacity 67

                                                                                                                          811

                                                                                                                          At pile base level

                                                                                                                          cu frac14 220 kN=m2

                                                                                                                          qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                          Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                          00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                          qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                          Then

                                                                                                                          Qf frac14 Abqb thorn Asqs

                                                                                                                          frac14

                                                                                                                          4 32 1980

                                                                                                                          thorn eth 105 139 86THORN

                                                                                                                          frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                          0 01 02 03 04 05 06 07

                                                                                                                          0 2 4 6 8 10 12 14

                                                                                                                          1

                                                                                                                          2

                                                                                                                          3

                                                                                                                          4

                                                                                                                          5

                                                                                                                          6

                                                                                                                          7

                                                                                                                          8

                                                                                                                          (1)

                                                                                                                          (2)

                                                                                                                          (3)

                                                                                                                          (4)

                                                                                                                          (5)

                                                                                                                          qc

                                                                                                                          qc

                                                                                                                          Iz

                                                                                                                          Iz

                                                                                                                          (MNm2)

                                                                                                                          z (m)

                                                                                                                          Figure Q810

                                                                                                                          68 Bearing capacity

                                                                                                                          Allowable load

                                                                                                                          ethaTHORN Qf

                                                                                                                          2frac14 17 937

                                                                                                                          2frac14 8968 kN

                                                                                                                          ethbTHORN Abqb

                                                                                                                          3thorn Asqs frac14 13 996

                                                                                                                          3thorn 3941 frac14 8606 kN

                                                                                                                          ie allowable load frac14 8600 kN

                                                                                                                          Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                          According to the limit state method

                                                                                                                          Characteristic undrained strength at base level cuk frac14 220

                                                                                                                          150kN=m2

                                                                                                                          Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                          150frac14 1320 kN=m2

                                                                                                                          Characteristic shaft resistance qsk frac14 00150

                                                                                                                          frac14 86

                                                                                                                          150frac14 57 kN=m2

                                                                                                                          Characteristic base and shaft resistances

                                                                                                                          Rbk frac14

                                                                                                                          4 32 1320 frac14 9330 kN

                                                                                                                          Rsk frac14 105 139 86

                                                                                                                          150frac14 2629 kN

                                                                                                                          For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                          Design bearing resistance Rcd frac14 9330

                                                                                                                          160thorn 2629

                                                                                                                          130

                                                                                                                          frac14 5831thorn 2022

                                                                                                                          frac14 7850 kN

                                                                                                                          Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                          812

                                                                                                                          ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                          qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                          For a single pile

                                                                                                                          Qf frac14 Abqb thorn Asqs

                                                                                                                          frac14

                                                                                                                          4 062 1305

                                                                                                                          thorn eth 06 15 42THORN

                                                                                                                          frac14 369thorn 1187 frac14 1556 kN

                                                                                                                          Bearing capacity 69

                                                                                                                          Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                          eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                          (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                          (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                          qbkfrac14 9cuk frac14 9 220

                                                                                                                          150frac14 1320 kN=m2

                                                                                                                          qskfrac14cuk frac14 040 105

                                                                                                                          150frac14 28 kN=m2

                                                                                                                          Rbkfrac14

                                                                                                                          4 0602 1320 frac14 373 kN

                                                                                                                          Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                          Rcdfrac14 373

                                                                                                                          160thorn 791

                                                                                                                          130frac14 233thorn 608 frac14 841 kN

                                                                                                                          Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                          Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                          (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                          q frac14 21 000

                                                                                                                          1762frac14 68 kN=m2

                                                                                                                          Immediate settlement

                                                                                                                          H

                                                                                                                          Bfrac14 15

                                                                                                                          176frac14 085

                                                                                                                          D

                                                                                                                          Bfrac14 13

                                                                                                                          176frac14 074

                                                                                                                          L

                                                                                                                          Bfrac14 1

                                                                                                                          Hence from Figure 515

                                                                                                                          130 frac14 078 and 131 frac14 041

                                                                                                                          70 Bearing capacity

                                                                                                                          Thus using Equation 528

                                                                                                                          si frac14 078 041 68 176

                                                                                                                          65frac14 6mm

                                                                                                                          Consolidation settlement

                                                                                                                          Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                          1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                          434 (sod)

                                                                                                                          Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                          sc frac14 056 434 frac14 24mm

                                                                                                                          The total settlement is (6thorn 24) frac14 30mm

                                                                                                                          813

                                                                                                                          At base level N frac14 26 Then using Equation 830

                                                                                                                          qb frac14 40NDb

                                                                                                                          Bfrac14 40 26 2

                                                                                                                          025frac14 8320 kN=m2

                                                                                                                          ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                          Figure Q812

                                                                                                                          Bearing capacity 71

                                                                                                                          Over the length embedded in sand

                                                                                                                          N frac14 21 ie18thorn 24

                                                                                                                          2

                                                                                                                          Using Equation 831

                                                                                                                          qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                          For a single pile

                                                                                                                          Qf frac14 Abqb thorn Asqs

                                                                                                                          frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                          For the pile group assuming a group efficiency of 12

                                                                                                                          XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                          Then the load factor is

                                                                                                                          F frac14 6523

                                                                                                                          2000thorn 1000frac14 21

                                                                                                                          (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                          Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                          150frac14 5547 kNm2

                                                                                                                          Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                          150frac14 28 kNm2

                                                                                                                          Characteristic base and shaft resistances for a single pile

                                                                                                                          Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                          Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                          For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                          Design bearing resistance Rcd frac14 347

                                                                                                                          130thorn 56

                                                                                                                          130frac14 310 kN

                                                                                                                          For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                          Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                          (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                          From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                          72 Bearing capacity

                                                                                                                          N frac14 24thorn 26thorn 34

                                                                                                                          3frac14 28

                                                                                                                          Ic frac14 171

                                                                                                                          2814frac14 0016 ethEquation 818THORN

                                                                                                                          s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                          814

                                                                                                                          Using Equation 841

                                                                                                                          Tf frac14 DLcu thorn

                                                                                                                          4ethD2 d2THORNcuNc

                                                                                                                          frac14 eth 02 5 06 110THORN thorn

                                                                                                                          4eth022 012THORN110 9

                                                                                                                          frac14 207thorn 23 frac14 230 kN

                                                                                                                          Figure Q813

                                                                                                                          Bearing capacity 73

                                                                                                                          Chapter 9

                                                                                                                          Stability of slopes

                                                                                                                          91

                                                                                                                          Referring to Figure Q91

                                                                                                                          W frac14 417 19 frac14 792 kN=m

                                                                                                                          Q frac14 20 28 frac14 56 kN=m

                                                                                                                          Arc lengthAB frac14

                                                                                                                          180 73 90 frac14 115m

                                                                                                                          Arc length BC frac14

                                                                                                                          180 28 90 frac14 44m

                                                                                                                          The factor of safety is given by

                                                                                                                          F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                          90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                          Depth of tension crack z0 frac14 2cu

                                                                                                                          frac14 2 20

                                                                                                                          19frac14 21m

                                                                                                                          Arc length BD frac14

                                                                                                                          180 13

                                                                                                                          1

                                                                                                                          2 90 frac14 21m

                                                                                                                          F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                          The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                          Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                          14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                          Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                          The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                          92

                                                                                                                          u frac14 0

                                                                                                                          Depth factor D frac14 11

                                                                                                                          9frac14 122

                                                                                                                          Using Equation 92 with F frac14 10

                                                                                                                          Ns frac14 cu

                                                                                                                          FHfrac14 30

                                                                                                                          10 19 9frac14 0175

                                                                                                                          Hence from Figure 93

                                                                                                                          frac14 50

                                                                                                                          For F frac14 12

                                                                                                                          Ns frac14 30

                                                                                                                          12 19 9frac14 0146

                                                                                                                          frac14 27

                                                                                                                          93

                                                                                                                          Refer to Figure Q93

                                                                                                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                          1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                          74 m

                                                                                                                          214 1deg

                                                                                                                          213 1deg

                                                                                                                          39 m

                                                                                                                          WB

                                                                                                                          D

                                                                                                                          C

                                                                                                                          28 m

                                                                                                                          21 m

                                                                                                                          A

                                                                                                                          Q

                                                                                                                          Soil (1)Soil (2)

                                                                                                                          73deg

                                                                                                                          Figure Q91

                                                                                                                          Stability of slopes 75

                                                                                                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                          9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                          599 256 328 1372

                                                                                                                          Figure Q93

                                                                                                                          76 Stability of slopes

                                                                                                                          XW cos frac14 b

                                                                                                                          Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                          W sin frac14 bX

                                                                                                                          h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                          Arc length La frac14

                                                                                                                          180 57

                                                                                                                          1

                                                                                                                          2 326 frac14 327m

                                                                                                                          The factor of safety is given by

                                                                                                                          F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                          W sin

                                                                                                                          frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                          frac14 091

                                                                                                                          According to the limit state method

                                                                                                                          0d frac14 tan1tan 32

                                                                                                                          125

                                                                                                                          frac14 265

                                                                                                                          c0 frac14 8

                                                                                                                          160frac14 5 kN=m2

                                                                                                                          Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                          Design disturbing moment frac14 1075 kN=m

                                                                                                                          The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                          94

                                                                                                                          F frac14 1

                                                                                                                          W sin

                                                                                                                          Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                          1thorn ethtan tan0=FTHORN

                                                                                                                          c0 frac14 8 kN=m2

                                                                                                                          0 frac14 32

                                                                                                                          c0b frac14 8 2 frac14 16 kN=m

                                                                                                                          W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                          Try F frac14 100

                                                                                                                          tan0

                                                                                                                          Ffrac14 0625

                                                                                                                          Stability of slopes 77

                                                                                                                          Values of u are as obtained in Figure Q93

                                                                                                                          SliceNo

                                                                                                                          h(m)

                                                                                                                          W frac14 bh(kNm)

                                                                                                                          W sin(kNm)

                                                                                                                          ub(kNm)

                                                                                                                          c0bthorn (W ub) tan0(kNm)

                                                                                                                          sec

                                                                                                                          1thorn (tan tan0)FProduct(kNm)

                                                                                                                          1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                          23 33 30 1042 31

                                                                                                                          3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                          224 92 72 0931 67

                                                                                                                          6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                          12 58 112 90 0889 80

                                                                                                                          8 60 252 1812

                                                                                                                          80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                          10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                          2154 88 116 0853 99

                                                                                                                          14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                          1074 1091

                                                                                                                          F frac14 1091

                                                                                                                          1074frac14 102 (assumed value 100)

                                                                                                                          Thus

                                                                                                                          F frac14 101

                                                                                                                          95

                                                                                                                          F frac14 1

                                                                                                                          W sin

                                                                                                                          XfWeth1 ruTHORN tan0g sec

                                                                                                                          1thorn ethtan tan0THORN=F

                                                                                                                          0 frac14 33

                                                                                                                          ru frac14 020

                                                                                                                          W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                          eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                          Try F frac14 110

                                                                                                                          tan 0

                                                                                                                          Ffrac14 tan 33

                                                                                                                          110frac14 0590

                                                                                                                          78 Stability of slopes

                                                                                                                          Referring to Figure Q95

                                                                                                                          SliceNo

                                                                                                                          h(m)

                                                                                                                          W frac14 bh(kNm)

                                                                                                                          W sin(kNm)

                                                                                                                          W(1 ru) tan0(kNm)

                                                                                                                          sec

                                                                                                                          1thorn ( tan tan0)FProduct(kNm)

                                                                                                                          1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                          2120 234 0892 209

                                                                                                                          4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                          1185 1271

                                                                                                                          Figure Q95

                                                                                                                          Stability of slopes 79

                                                                                                                          F frac14 1271

                                                                                                                          1185frac14 107

                                                                                                                          The trial value was 110 therefore take F to be 108

                                                                                                                          96

                                                                                                                          (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                          F frac14 0

                                                                                                                          sat

                                                                                                                          tan0

                                                                                                                          tan

                                                                                                                          ptie 15 frac14 92

                                                                                                                          19

                                                                                                                          tan 36

                                                                                                                          tan

                                                                                                                          tan frac14 0234

                                                                                                                          frac14 13

                                                                                                                          Water table well below surface the factor of safety is given by Equation 911

                                                                                                                          F frac14 tan0

                                                                                                                          tan

                                                                                                                          frac14 tan 36

                                                                                                                          tan 13

                                                                                                                          frac14 31

                                                                                                                          (b) 0d frac14 tan1tan 36

                                                                                                                          125

                                                                                                                          frac14 30

                                                                                                                          Depth of potential failure surface frac14 z

                                                                                                                          Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                          frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                          frac14 504z kN

                                                                                                                          Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                          frac14 19 z sin 13 cos 13

                                                                                                                          frac14 416z kN

                                                                                                                          Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                          80 Stability of slopes

                                                                                                                          • Book Cover
                                                                                                                          • Title
                                                                                                                          • Contents
                                                                                                                          • Basic characteristics of soils
                                                                                                                          • Seepage
                                                                                                                          • Effective stress
                                                                                                                          • Shear strength
                                                                                                                          • Stresses and displacements
                                                                                                                          • Lateral earth pressure
                                                                                                                          • Consolidation theory
                                                                                                                          • Bearing capacity
                                                                                                                          • Stability of slopes

                                                                                                                            76

                                                                                                                            At the top of the clay layer the decrease in pore water pressure is 4w At the bottom of theclay layer the pore water pressure remains constant Hence at the centre of the clay layer

                                                                                                                            0 frac14 2w frac14 2 98 frac14 196 kN=m2

                                                                                                                            The final consolidation settlement (one-dimensional method) is

                                                                                                                            sc frac14 mv0H frac14 083 196 8 frac14 130mm

                                                                                                                            Corrected time t frac14 2 1

                                                                                                                            2

                                                                                                                            40

                                                                                                                            52

                                                                                                                            frac14 1615 years

                                                                                                                            Tv frac14 cvtd2frac14 44 1615

                                                                                                                            42frac14 0444

                                                                                                                            From Figure 718 (curve 1) U frac14 073Settlement after 2 years frac14 Usc frac14 073 130 frac14 95mm

                                                                                                                            77

                                                                                                                            The clay layer is thin relative to the dimensions of the raft and therefore theone-dimensional method is appropriate The clay layer can be considered as a whole(see Figure Q77)

                                                                                                                            Figure Q77

                                                                                                                            56 Consolidation theory

                                                                                                                            Point m n Ir (kNm2) sc (mm)

                                                                                                                            13020frac14 15 20

                                                                                                                            20frac14 10 0194 (4) 113 124

                                                                                                                            260

                                                                                                                            20frac14 30

                                                                                                                            20

                                                                                                                            20frac14 10 0204 (2) 59 65

                                                                                                                            360

                                                                                                                            20frac14 30

                                                                                                                            40

                                                                                                                            20frac14 20 0238 (1) 35 38

                                                                                                                            430

                                                                                                                            20frac14 15

                                                                                                                            40

                                                                                                                            20frac14 20 0224 (2) 65 72

                                                                                                                            Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                                            78

                                                                                                                            Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                                            (a) Immediate settlement

                                                                                                                            H

                                                                                                                            Bfrac14 30

                                                                                                                            35frac14 086

                                                                                                                            D

                                                                                                                            Bfrac14 2

                                                                                                                            35frac14 006

                                                                                                                            Figure Q78

                                                                                                                            Consolidation theory 57

                                                                                                                            From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                                            si frac14 130131qB

                                                                                                                            Eufrac14 10 032 105 35

                                                                                                                            40frac14 30mm

                                                                                                                            (b) Consolidation settlement

                                                                                                                            Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                                            1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                                            3150

                                                                                                                            Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                                            Now

                                                                                                                            H

                                                                                                                            Bfrac14 30

                                                                                                                            35frac14 086 and A frac14 065

                                                                                                                            from Figure 712 13 frac14 079

                                                                                                                            sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                                            Total settlement

                                                                                                                            s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                                            79

                                                                                                                            Without sand drains

                                                                                                                            Uv frac14 025

                                                                                                                            Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                                            t frac14 Tvd2

                                                                                                                            cvfrac14 0049 82

                                                                                                                            cvWith sand drains

                                                                                                                            R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                                            n frac14 Rrfrac14 169

                                                                                                                            015frac14 113

                                                                                                                            Tr frac14 cht

                                                                                                                            4R2frac14 ch

                                                                                                                            4 1692 0049 82

                                                                                                                            cvethand ch frac14 cvTHORN

                                                                                                                            frac14 0275

                                                                                                                            Ur frac14 073 (from Figure 730)

                                                                                                                            58 Consolidation theory

                                                                                                                            Using Equation 740

                                                                                                                            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                                            U frac14 080

                                                                                                                            710

                                                                                                                            Without sand drains

                                                                                                                            Uv frac14 090

                                                                                                                            Tv frac14 0848

                                                                                                                            t frac14 Tvd2

                                                                                                                            cvfrac14 0848 102

                                                                                                                            96frac14 88 years

                                                                                                                            With sand drains

                                                                                                                            R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                                            n frac14 Rrfrac14 226

                                                                                                                            015frac14 15

                                                                                                                            Tr

                                                                                                                            Tvfrac14 chcv

                                                                                                                            d2

                                                                                                                            4R2ethsame tTHORN

                                                                                                                            Tr

                                                                                                                            Tvfrac14 140

                                                                                                                            96 102

                                                                                                                            4 2262frac14 714 eth1THORN

                                                                                                                            Using Equation 740

                                                                                                                            eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                                            An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                                            Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                                            040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                                            Thus

                                                                                                                            Uv frac14 0295 and Ur frac14 086

                                                                                                                            t frac14 88 00683

                                                                                                                            0848frac14 07 years

                                                                                                                            Consolidation theory 59

                                                                                                                            Chapter 8

                                                                                                                            Bearing capacity

                                                                                                                            81

                                                                                                                            (a) The ultimate bearing capacity is given by Equation 83

                                                                                                                            qf frac14 cNc thorn DNq thorn 1

                                                                                                                            2BN

                                                                                                                            For u frac14 0

                                                                                                                            Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                            qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                                            The net ultimate bearing capacity is

                                                                                                                            qnf frac14 qf D frac14 540 kN=m2

                                                                                                                            The net foundation pressure is

                                                                                                                            qn frac14 q D frac14 425

                                                                                                                            2 eth21 1THORN frac14 192 kN=m2

                                                                                                                            The factor of safety (Equation 86) is

                                                                                                                            F frac14 qnfqnfrac14 540

                                                                                                                            192frac14 28

                                                                                                                            (b) For 0 frac14 28

                                                                                                                            Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                                            qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                                            2 112 2 13

                                                                                                                            frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                                            qnf frac14 574 112 frac14 563 kN=m2

                                                                                                                            F frac14 563

                                                                                                                            192frac14 29

                                                                                                                            (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                                            82

                                                                                                                            For 0 frac14 38

                                                                                                                            Nq frac14 49 N frac14 67

                                                                                                                            qnf frac14 DethNq 1THORN thorn 1

                                                                                                                            2BN ethfrom Equation 83THORN

                                                                                                                            frac14 eth18 075 48THORN thorn 1

                                                                                                                            2 18 15 67

                                                                                                                            frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                                            qn frac14 500

                                                                                                                            15 eth18 075THORN frac14 320 kN=m2

                                                                                                                            F frac14 qnfqnfrac14 1553

                                                                                                                            320frac14 48

                                                                                                                            0d frac14 tan1tan 38

                                                                                                                            125

                                                                                                                            frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                                            Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                                            2 18 15 25

                                                                                                                            frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                                            Design load (action) Vd frac14 500 kN=m

                                                                                                                            The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                                            83

                                                                                                                            D

                                                                                                                            Bfrac14 350

                                                                                                                            225frac14 155

                                                                                                                            From Figure 85 for a square foundation

                                                                                                                            Nc frac14 81

                                                                                                                            Bearing capacity 61

                                                                                                                            For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                                            Nc frac14 084thorn 016B

                                                                                                                            L

                                                                                                                            81 frac14 745

                                                                                                                            Using Equation 810

                                                                                                                            qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                                            For F frac14 3

                                                                                                                            qn frac14 1006

                                                                                                                            3frac14 335 kN=m2

                                                                                                                            q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                                            Design load frac14 405 450 225 frac14 4100 kN

                                                                                                                            Design undrained strength cud frac14 135

                                                                                                                            14frac14 96 kN=m2

                                                                                                                            Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                                            frac14 7241 kN

                                                                                                                            Design load Vd frac14 4100 kN

                                                                                                                            Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                            84

                                                                                                                            For 0 frac14 40

                                                                                                                            Nq frac14 64 N frac14 95

                                                                                                                            qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                                            (a) Water table 5m below ground level

                                                                                                                            qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                                            qn frac14 400 17 frac14 383 kN=m2

                                                                                                                            F frac14 2686

                                                                                                                            383frac14 70

                                                                                                                            (b) Water table 1m below ground level (ie at foundation level)

                                                                                                                            0 frac14 20 98 frac14 102 kN=m3

                                                                                                                            62 Bearing capacity

                                                                                                                            qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                                            F frac14 2040

                                                                                                                            383frac14 53

                                                                                                                            (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                                            eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                                            qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                                            F frac14 1296

                                                                                                                            392frac14 33

                                                                                                                            85

                                                                                                                            The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                                            Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                                            Design value of 0 frac14 tan1tan 39

                                                                                                                            125

                                                                                                                            frac14 33

                                                                                                                            For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                                            Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                                            Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                            86

                                                                                                                            (a) Undrained shear for u frac14 0

                                                                                                                            Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                            qnf frac14 12cuNc

                                                                                                                            frac14 12 100 514 frac14 617 kN=m2

                                                                                                                            qn frac14 qnfFfrac14 617

                                                                                                                            3frac14 206 kN=m2

                                                                                                                            q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                                            Bearing capacity 63

                                                                                                                            Drained shear for 0 frac14 32

                                                                                                                            Nq frac14 23 N frac14 25

                                                                                                                            0 frac14 21 98 frac14 112 kN=m3

                                                                                                                            qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                            frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                            frac14 694 kN=m2

                                                                                                                            q frac14 694

                                                                                                                            3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                            Design load frac14 42 227 frac14 3632 kN

                                                                                                                            (b) Design undrained strength cud frac14 100

                                                                                                                            14frac14 71 kNm2

                                                                                                                            Design bearing resistance Rd frac14 12cudNe area

                                                                                                                            frac14 12 71 514 42

                                                                                                                            frac14 7007 kN

                                                                                                                            For drained shear 0d frac14 tan1tan 32

                                                                                                                            125

                                                                                                                            frac14 26

                                                                                                                            Nq frac14 12 N frac14 10

                                                                                                                            Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                            (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                            Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                            1 2 100 0175 0700qn 0182qn

                                                                                                                            2 6 033 0044 0176qn 0046qn

                                                                                                                            3 10 020 0017 0068qn 0018qn

                                                                                                                            0246qn

                                                                                                                            Diameter of equivalent circle B frac14 45m

                                                                                                                            H

                                                                                                                            Bfrac14 12

                                                                                                                            45frac14 27 and A frac14 042

                                                                                                                            13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                            64 Bearing capacity

                                                                                                                            For sc frac14 30mm

                                                                                                                            qn frac14 30

                                                                                                                            0147frac14 204 kN=m2

                                                                                                                            q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                            Design load frac14 42 225 frac14 3600 kN

                                                                                                                            The design load is 3600 kN settlement being the limiting criterion

                                                                                                                            87

                                                                                                                            D

                                                                                                                            Bfrac14 8

                                                                                                                            4frac14 20

                                                                                                                            From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                            F frac14 cuNc

                                                                                                                            Dfrac14 40 71

                                                                                                                            20 8frac14 18

                                                                                                                            88

                                                                                                                            Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                            Design value of 0 frac14 tan1tan 38

                                                                                                                            125

                                                                                                                            frac14 32

                                                                                                                            Figure Q86

                                                                                                                            Bearing capacity 65

                                                                                                                            For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                            Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                            The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                            Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                            For B frac14 250m qn frac14 3750

                                                                                                                            2502 17 frac14 583 kN=m2

                                                                                                                            From Figure 510 m frac14 n frac14 126

                                                                                                                            6frac14 021

                                                                                                                            Ir frac14 0019

                                                                                                                            Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                            Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                            The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                            89

                                                                                                                            Depth (m) N 0v (kNm2) CN N1

                                                                                                                            070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                            Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                            (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                            Cw frac14 05thorn 0530

                                                                                                                            47

                                                                                                                            frac14 082

                                                                                                                            66 Bearing capacity

                                                                                                                            Thus

                                                                                                                            qa frac14 150 082 frac14 120 kN=m2

                                                                                                                            (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                            Thus

                                                                                                                            qa frac14 90 15 frac14 135 kN=m2

                                                                                                                            (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                            Ic frac14 171

                                                                                                                            1014frac14 0068

                                                                                                                            From Equation 819(a) with s frac14 25mm

                                                                                                                            q frac14 25

                                                                                                                            3507 0068frac14 150 kN=m2

                                                                                                                            810

                                                                                                                            Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                            Izp frac14 05thorn 01130

                                                                                                                            16 27

                                                                                                                            05

                                                                                                                            frac14 067

                                                                                                                            Refer to Figure Q810

                                                                                                                            E frac14 25qc

                                                                                                                            Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                            Ez (mm3MN)

                                                                                                                            1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                            0203

                                                                                                                            C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                            130frac14 093

                                                                                                                            C2 frac14 1 ethsayTHORN

                                                                                                                            s frac14 C1C2qnX Iz

                                                                                                                            Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                            Bearing capacity 67

                                                                                                                            811

                                                                                                                            At pile base level

                                                                                                                            cu frac14 220 kN=m2

                                                                                                                            qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                            Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                            00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                            qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                            Then

                                                                                                                            Qf frac14 Abqb thorn Asqs

                                                                                                                            frac14

                                                                                                                            4 32 1980

                                                                                                                            thorn eth 105 139 86THORN

                                                                                                                            frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                            0 01 02 03 04 05 06 07

                                                                                                                            0 2 4 6 8 10 12 14

                                                                                                                            1

                                                                                                                            2

                                                                                                                            3

                                                                                                                            4

                                                                                                                            5

                                                                                                                            6

                                                                                                                            7

                                                                                                                            8

                                                                                                                            (1)

                                                                                                                            (2)

                                                                                                                            (3)

                                                                                                                            (4)

                                                                                                                            (5)

                                                                                                                            qc

                                                                                                                            qc

                                                                                                                            Iz

                                                                                                                            Iz

                                                                                                                            (MNm2)

                                                                                                                            z (m)

                                                                                                                            Figure Q810

                                                                                                                            68 Bearing capacity

                                                                                                                            Allowable load

                                                                                                                            ethaTHORN Qf

                                                                                                                            2frac14 17 937

                                                                                                                            2frac14 8968 kN

                                                                                                                            ethbTHORN Abqb

                                                                                                                            3thorn Asqs frac14 13 996

                                                                                                                            3thorn 3941 frac14 8606 kN

                                                                                                                            ie allowable load frac14 8600 kN

                                                                                                                            Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                            According to the limit state method

                                                                                                                            Characteristic undrained strength at base level cuk frac14 220

                                                                                                                            150kN=m2

                                                                                                                            Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                            150frac14 1320 kN=m2

                                                                                                                            Characteristic shaft resistance qsk frac14 00150

                                                                                                                            frac14 86

                                                                                                                            150frac14 57 kN=m2

                                                                                                                            Characteristic base and shaft resistances

                                                                                                                            Rbk frac14

                                                                                                                            4 32 1320 frac14 9330 kN

                                                                                                                            Rsk frac14 105 139 86

                                                                                                                            150frac14 2629 kN

                                                                                                                            For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                            Design bearing resistance Rcd frac14 9330

                                                                                                                            160thorn 2629

                                                                                                                            130

                                                                                                                            frac14 5831thorn 2022

                                                                                                                            frac14 7850 kN

                                                                                                                            Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                            812

                                                                                                                            ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                            qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                            For a single pile

                                                                                                                            Qf frac14 Abqb thorn Asqs

                                                                                                                            frac14

                                                                                                                            4 062 1305

                                                                                                                            thorn eth 06 15 42THORN

                                                                                                                            frac14 369thorn 1187 frac14 1556 kN

                                                                                                                            Bearing capacity 69

                                                                                                                            Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                            eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                            (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                            (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                            qbkfrac14 9cuk frac14 9 220

                                                                                                                            150frac14 1320 kN=m2

                                                                                                                            qskfrac14cuk frac14 040 105

                                                                                                                            150frac14 28 kN=m2

                                                                                                                            Rbkfrac14

                                                                                                                            4 0602 1320 frac14 373 kN

                                                                                                                            Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                            Rcdfrac14 373

                                                                                                                            160thorn 791

                                                                                                                            130frac14 233thorn 608 frac14 841 kN

                                                                                                                            Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                            Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                            (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                            q frac14 21 000

                                                                                                                            1762frac14 68 kN=m2

                                                                                                                            Immediate settlement

                                                                                                                            H

                                                                                                                            Bfrac14 15

                                                                                                                            176frac14 085

                                                                                                                            D

                                                                                                                            Bfrac14 13

                                                                                                                            176frac14 074

                                                                                                                            L

                                                                                                                            Bfrac14 1

                                                                                                                            Hence from Figure 515

                                                                                                                            130 frac14 078 and 131 frac14 041

                                                                                                                            70 Bearing capacity

                                                                                                                            Thus using Equation 528

                                                                                                                            si frac14 078 041 68 176

                                                                                                                            65frac14 6mm

                                                                                                                            Consolidation settlement

                                                                                                                            Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                            1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                            434 (sod)

                                                                                                                            Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                            sc frac14 056 434 frac14 24mm

                                                                                                                            The total settlement is (6thorn 24) frac14 30mm

                                                                                                                            813

                                                                                                                            At base level N frac14 26 Then using Equation 830

                                                                                                                            qb frac14 40NDb

                                                                                                                            Bfrac14 40 26 2

                                                                                                                            025frac14 8320 kN=m2

                                                                                                                            ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                            Figure Q812

                                                                                                                            Bearing capacity 71

                                                                                                                            Over the length embedded in sand

                                                                                                                            N frac14 21 ie18thorn 24

                                                                                                                            2

                                                                                                                            Using Equation 831

                                                                                                                            qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                            For a single pile

                                                                                                                            Qf frac14 Abqb thorn Asqs

                                                                                                                            frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                            For the pile group assuming a group efficiency of 12

                                                                                                                            XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                            Then the load factor is

                                                                                                                            F frac14 6523

                                                                                                                            2000thorn 1000frac14 21

                                                                                                                            (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                            Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                            150frac14 5547 kNm2

                                                                                                                            Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                            150frac14 28 kNm2

                                                                                                                            Characteristic base and shaft resistances for a single pile

                                                                                                                            Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                            Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                            For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                            Design bearing resistance Rcd frac14 347

                                                                                                                            130thorn 56

                                                                                                                            130frac14 310 kN

                                                                                                                            For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                            Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                            (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                            From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                            72 Bearing capacity

                                                                                                                            N frac14 24thorn 26thorn 34

                                                                                                                            3frac14 28

                                                                                                                            Ic frac14 171

                                                                                                                            2814frac14 0016 ethEquation 818THORN

                                                                                                                            s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                            814

                                                                                                                            Using Equation 841

                                                                                                                            Tf frac14 DLcu thorn

                                                                                                                            4ethD2 d2THORNcuNc

                                                                                                                            frac14 eth 02 5 06 110THORN thorn

                                                                                                                            4eth022 012THORN110 9

                                                                                                                            frac14 207thorn 23 frac14 230 kN

                                                                                                                            Figure Q813

                                                                                                                            Bearing capacity 73

                                                                                                                            Chapter 9

                                                                                                                            Stability of slopes

                                                                                                                            91

                                                                                                                            Referring to Figure Q91

                                                                                                                            W frac14 417 19 frac14 792 kN=m

                                                                                                                            Q frac14 20 28 frac14 56 kN=m

                                                                                                                            Arc lengthAB frac14

                                                                                                                            180 73 90 frac14 115m

                                                                                                                            Arc length BC frac14

                                                                                                                            180 28 90 frac14 44m

                                                                                                                            The factor of safety is given by

                                                                                                                            F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                            90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                            Depth of tension crack z0 frac14 2cu

                                                                                                                            frac14 2 20

                                                                                                                            19frac14 21m

                                                                                                                            Arc length BD frac14

                                                                                                                            180 13

                                                                                                                            1

                                                                                                                            2 90 frac14 21m

                                                                                                                            F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                            The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                            Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                            14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                            Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                            The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                            92

                                                                                                                            u frac14 0

                                                                                                                            Depth factor D frac14 11

                                                                                                                            9frac14 122

                                                                                                                            Using Equation 92 with F frac14 10

                                                                                                                            Ns frac14 cu

                                                                                                                            FHfrac14 30

                                                                                                                            10 19 9frac14 0175

                                                                                                                            Hence from Figure 93

                                                                                                                            frac14 50

                                                                                                                            For F frac14 12

                                                                                                                            Ns frac14 30

                                                                                                                            12 19 9frac14 0146

                                                                                                                            frac14 27

                                                                                                                            93

                                                                                                                            Refer to Figure Q93

                                                                                                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                            1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                            74 m

                                                                                                                            214 1deg

                                                                                                                            213 1deg

                                                                                                                            39 m

                                                                                                                            WB

                                                                                                                            D

                                                                                                                            C

                                                                                                                            28 m

                                                                                                                            21 m

                                                                                                                            A

                                                                                                                            Q

                                                                                                                            Soil (1)Soil (2)

                                                                                                                            73deg

                                                                                                                            Figure Q91

                                                                                                                            Stability of slopes 75

                                                                                                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                            9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                            599 256 328 1372

                                                                                                                            Figure Q93

                                                                                                                            76 Stability of slopes

                                                                                                                            XW cos frac14 b

                                                                                                                            Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                            W sin frac14 bX

                                                                                                                            h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                            Arc length La frac14

                                                                                                                            180 57

                                                                                                                            1

                                                                                                                            2 326 frac14 327m

                                                                                                                            The factor of safety is given by

                                                                                                                            F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                            W sin

                                                                                                                            frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                            frac14 091

                                                                                                                            According to the limit state method

                                                                                                                            0d frac14 tan1tan 32

                                                                                                                            125

                                                                                                                            frac14 265

                                                                                                                            c0 frac14 8

                                                                                                                            160frac14 5 kN=m2

                                                                                                                            Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                            Design disturbing moment frac14 1075 kN=m

                                                                                                                            The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                            94

                                                                                                                            F frac14 1

                                                                                                                            W sin

                                                                                                                            Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                            1thorn ethtan tan0=FTHORN

                                                                                                                            c0 frac14 8 kN=m2

                                                                                                                            0 frac14 32

                                                                                                                            c0b frac14 8 2 frac14 16 kN=m

                                                                                                                            W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                            Try F frac14 100

                                                                                                                            tan0

                                                                                                                            Ffrac14 0625

                                                                                                                            Stability of slopes 77

                                                                                                                            Values of u are as obtained in Figure Q93

                                                                                                                            SliceNo

                                                                                                                            h(m)

                                                                                                                            W frac14 bh(kNm)

                                                                                                                            W sin(kNm)

                                                                                                                            ub(kNm)

                                                                                                                            c0bthorn (W ub) tan0(kNm)

                                                                                                                            sec

                                                                                                                            1thorn (tan tan0)FProduct(kNm)

                                                                                                                            1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                            23 33 30 1042 31

                                                                                                                            3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                            224 92 72 0931 67

                                                                                                                            6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                            12 58 112 90 0889 80

                                                                                                                            8 60 252 1812

                                                                                                                            80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                            10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                            2154 88 116 0853 99

                                                                                                                            14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                            1074 1091

                                                                                                                            F frac14 1091

                                                                                                                            1074frac14 102 (assumed value 100)

                                                                                                                            Thus

                                                                                                                            F frac14 101

                                                                                                                            95

                                                                                                                            F frac14 1

                                                                                                                            W sin

                                                                                                                            XfWeth1 ruTHORN tan0g sec

                                                                                                                            1thorn ethtan tan0THORN=F

                                                                                                                            0 frac14 33

                                                                                                                            ru frac14 020

                                                                                                                            W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                            eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                            Try F frac14 110

                                                                                                                            tan 0

                                                                                                                            Ffrac14 tan 33

                                                                                                                            110frac14 0590

                                                                                                                            78 Stability of slopes

                                                                                                                            Referring to Figure Q95

                                                                                                                            SliceNo

                                                                                                                            h(m)

                                                                                                                            W frac14 bh(kNm)

                                                                                                                            W sin(kNm)

                                                                                                                            W(1 ru) tan0(kNm)

                                                                                                                            sec

                                                                                                                            1thorn ( tan tan0)FProduct(kNm)

                                                                                                                            1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                            2120 234 0892 209

                                                                                                                            4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                            1185 1271

                                                                                                                            Figure Q95

                                                                                                                            Stability of slopes 79

                                                                                                                            F frac14 1271

                                                                                                                            1185frac14 107

                                                                                                                            The trial value was 110 therefore take F to be 108

                                                                                                                            96

                                                                                                                            (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                            F frac14 0

                                                                                                                            sat

                                                                                                                            tan0

                                                                                                                            tan

                                                                                                                            ptie 15 frac14 92

                                                                                                                            19

                                                                                                                            tan 36

                                                                                                                            tan

                                                                                                                            tan frac14 0234

                                                                                                                            frac14 13

                                                                                                                            Water table well below surface the factor of safety is given by Equation 911

                                                                                                                            F frac14 tan0

                                                                                                                            tan

                                                                                                                            frac14 tan 36

                                                                                                                            tan 13

                                                                                                                            frac14 31

                                                                                                                            (b) 0d frac14 tan1tan 36

                                                                                                                            125

                                                                                                                            frac14 30

                                                                                                                            Depth of potential failure surface frac14 z

                                                                                                                            Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                            frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                            frac14 504z kN

                                                                                                                            Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                            frac14 19 z sin 13 cos 13

                                                                                                                            frac14 416z kN

                                                                                                                            Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                            80 Stability of slopes

                                                                                                                            • Book Cover
                                                                                                                            • Title
                                                                                                                            • Contents
                                                                                                                            • Basic characteristics of soils
                                                                                                                            • Seepage
                                                                                                                            • Effective stress
                                                                                                                            • Shear strength
                                                                                                                            • Stresses and displacements
                                                                                                                            • Lateral earth pressure
                                                                                                                            • Consolidation theory
                                                                                                                            • Bearing capacity
                                                                                                                            • Stability of slopes

                                                                                                                              Point m n Ir (kNm2) sc (mm)

                                                                                                                              13020frac14 15 20

                                                                                                                              20frac14 10 0194 (4) 113 124

                                                                                                                              260

                                                                                                                              20frac14 30

                                                                                                                              20

                                                                                                                              20frac14 10 0204 (2) 59 65

                                                                                                                              360

                                                                                                                              20frac14 30

                                                                                                                              40

                                                                                                                              20frac14 20 0238 (1) 35 38

                                                                                                                              430

                                                                                                                              20frac14 15

                                                                                                                              40

                                                                                                                              20frac14 20 0224 (2) 65 72

                                                                                                                              Note sc frac14 mv0H frac14 0220 5 frac14 110 (mm) (0 frac14 )

                                                                                                                              78

                                                                                                                              Due to the thickness of the clay layer relative to the size of the foundation there will besignificant lateral strain in the clay and the SkemptonndashBjerrum method is appropriateThe clay is divided into six sublayers (Figure Q78) for the calculation of consolidationsettlement

                                                                                                                              (a) Immediate settlement

                                                                                                                              H

                                                                                                                              Bfrac14 30

                                                                                                                              35frac14 086

                                                                                                                              D

                                                                                                                              Bfrac14 2

                                                                                                                              35frac14 006

                                                                                                                              Figure Q78

                                                                                                                              Consolidation theory 57

                                                                                                                              From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                                              si frac14 130131qB

                                                                                                                              Eufrac14 10 032 105 35

                                                                                                                              40frac14 30mm

                                                                                                                              (b) Consolidation settlement

                                                                                                                              Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                                              1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                                              3150

                                                                                                                              Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                                              Now

                                                                                                                              H

                                                                                                                              Bfrac14 30

                                                                                                                              35frac14 086 and A frac14 065

                                                                                                                              from Figure 712 13 frac14 079

                                                                                                                              sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                                              Total settlement

                                                                                                                              s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                                              79

                                                                                                                              Without sand drains

                                                                                                                              Uv frac14 025

                                                                                                                              Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                                              t frac14 Tvd2

                                                                                                                              cvfrac14 0049 82

                                                                                                                              cvWith sand drains

                                                                                                                              R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                                              n frac14 Rrfrac14 169

                                                                                                                              015frac14 113

                                                                                                                              Tr frac14 cht

                                                                                                                              4R2frac14 ch

                                                                                                                              4 1692 0049 82

                                                                                                                              cvethand ch frac14 cvTHORN

                                                                                                                              frac14 0275

                                                                                                                              Ur frac14 073 (from Figure 730)

                                                                                                                              58 Consolidation theory

                                                                                                                              Using Equation 740

                                                                                                                              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                                              U frac14 080

                                                                                                                              710

                                                                                                                              Without sand drains

                                                                                                                              Uv frac14 090

                                                                                                                              Tv frac14 0848

                                                                                                                              t frac14 Tvd2

                                                                                                                              cvfrac14 0848 102

                                                                                                                              96frac14 88 years

                                                                                                                              With sand drains

                                                                                                                              R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                                              n frac14 Rrfrac14 226

                                                                                                                              015frac14 15

                                                                                                                              Tr

                                                                                                                              Tvfrac14 chcv

                                                                                                                              d2

                                                                                                                              4R2ethsame tTHORN

                                                                                                                              Tr

                                                                                                                              Tvfrac14 140

                                                                                                                              96 102

                                                                                                                              4 2262frac14 714 eth1THORN

                                                                                                                              Using Equation 740

                                                                                                                              eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                                              An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                                              Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                                              040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                                              Thus

                                                                                                                              Uv frac14 0295 and Ur frac14 086

                                                                                                                              t frac14 88 00683

                                                                                                                              0848frac14 07 years

                                                                                                                              Consolidation theory 59

                                                                                                                              Chapter 8

                                                                                                                              Bearing capacity

                                                                                                                              81

                                                                                                                              (a) The ultimate bearing capacity is given by Equation 83

                                                                                                                              qf frac14 cNc thorn DNq thorn 1

                                                                                                                              2BN

                                                                                                                              For u frac14 0

                                                                                                                              Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                              qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                                              The net ultimate bearing capacity is

                                                                                                                              qnf frac14 qf D frac14 540 kN=m2

                                                                                                                              The net foundation pressure is

                                                                                                                              qn frac14 q D frac14 425

                                                                                                                              2 eth21 1THORN frac14 192 kN=m2

                                                                                                                              The factor of safety (Equation 86) is

                                                                                                                              F frac14 qnfqnfrac14 540

                                                                                                                              192frac14 28

                                                                                                                              (b) For 0 frac14 28

                                                                                                                              Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                                              qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                                              2 112 2 13

                                                                                                                              frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                                              qnf frac14 574 112 frac14 563 kN=m2

                                                                                                                              F frac14 563

                                                                                                                              192frac14 29

                                                                                                                              (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                                              82

                                                                                                                              For 0 frac14 38

                                                                                                                              Nq frac14 49 N frac14 67

                                                                                                                              qnf frac14 DethNq 1THORN thorn 1

                                                                                                                              2BN ethfrom Equation 83THORN

                                                                                                                              frac14 eth18 075 48THORN thorn 1

                                                                                                                              2 18 15 67

                                                                                                                              frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                                              qn frac14 500

                                                                                                                              15 eth18 075THORN frac14 320 kN=m2

                                                                                                                              F frac14 qnfqnfrac14 1553

                                                                                                                              320frac14 48

                                                                                                                              0d frac14 tan1tan 38

                                                                                                                              125

                                                                                                                              frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                                              Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                                              2 18 15 25

                                                                                                                              frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                                              Design load (action) Vd frac14 500 kN=m

                                                                                                                              The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                                              83

                                                                                                                              D

                                                                                                                              Bfrac14 350

                                                                                                                              225frac14 155

                                                                                                                              From Figure 85 for a square foundation

                                                                                                                              Nc frac14 81

                                                                                                                              Bearing capacity 61

                                                                                                                              For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                                              Nc frac14 084thorn 016B

                                                                                                                              L

                                                                                                                              81 frac14 745

                                                                                                                              Using Equation 810

                                                                                                                              qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                                              For F frac14 3

                                                                                                                              qn frac14 1006

                                                                                                                              3frac14 335 kN=m2

                                                                                                                              q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                                              Design load frac14 405 450 225 frac14 4100 kN

                                                                                                                              Design undrained strength cud frac14 135

                                                                                                                              14frac14 96 kN=m2

                                                                                                                              Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                                              frac14 7241 kN

                                                                                                                              Design load Vd frac14 4100 kN

                                                                                                                              Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                              84

                                                                                                                              For 0 frac14 40

                                                                                                                              Nq frac14 64 N frac14 95

                                                                                                                              qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                                              (a) Water table 5m below ground level

                                                                                                                              qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                                              qn frac14 400 17 frac14 383 kN=m2

                                                                                                                              F frac14 2686

                                                                                                                              383frac14 70

                                                                                                                              (b) Water table 1m below ground level (ie at foundation level)

                                                                                                                              0 frac14 20 98 frac14 102 kN=m3

                                                                                                                              62 Bearing capacity

                                                                                                                              qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                                              F frac14 2040

                                                                                                                              383frac14 53

                                                                                                                              (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                                              eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                                              qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                                              F frac14 1296

                                                                                                                              392frac14 33

                                                                                                                              85

                                                                                                                              The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                                              Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                                              Design value of 0 frac14 tan1tan 39

                                                                                                                              125

                                                                                                                              frac14 33

                                                                                                                              For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                                              Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                                              Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                              86

                                                                                                                              (a) Undrained shear for u frac14 0

                                                                                                                              Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                              qnf frac14 12cuNc

                                                                                                                              frac14 12 100 514 frac14 617 kN=m2

                                                                                                                              qn frac14 qnfFfrac14 617

                                                                                                                              3frac14 206 kN=m2

                                                                                                                              q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                                              Bearing capacity 63

                                                                                                                              Drained shear for 0 frac14 32

                                                                                                                              Nq frac14 23 N frac14 25

                                                                                                                              0 frac14 21 98 frac14 112 kN=m3

                                                                                                                              qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                              frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                              frac14 694 kN=m2

                                                                                                                              q frac14 694

                                                                                                                              3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                              Design load frac14 42 227 frac14 3632 kN

                                                                                                                              (b) Design undrained strength cud frac14 100

                                                                                                                              14frac14 71 kNm2

                                                                                                                              Design bearing resistance Rd frac14 12cudNe area

                                                                                                                              frac14 12 71 514 42

                                                                                                                              frac14 7007 kN

                                                                                                                              For drained shear 0d frac14 tan1tan 32

                                                                                                                              125

                                                                                                                              frac14 26

                                                                                                                              Nq frac14 12 N frac14 10

                                                                                                                              Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                              (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                              Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                              1 2 100 0175 0700qn 0182qn

                                                                                                                              2 6 033 0044 0176qn 0046qn

                                                                                                                              3 10 020 0017 0068qn 0018qn

                                                                                                                              0246qn

                                                                                                                              Diameter of equivalent circle B frac14 45m

                                                                                                                              H

                                                                                                                              Bfrac14 12

                                                                                                                              45frac14 27 and A frac14 042

                                                                                                                              13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                              64 Bearing capacity

                                                                                                                              For sc frac14 30mm

                                                                                                                              qn frac14 30

                                                                                                                              0147frac14 204 kN=m2

                                                                                                                              q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                              Design load frac14 42 225 frac14 3600 kN

                                                                                                                              The design load is 3600 kN settlement being the limiting criterion

                                                                                                                              87

                                                                                                                              D

                                                                                                                              Bfrac14 8

                                                                                                                              4frac14 20

                                                                                                                              From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                              F frac14 cuNc

                                                                                                                              Dfrac14 40 71

                                                                                                                              20 8frac14 18

                                                                                                                              88

                                                                                                                              Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                              Design value of 0 frac14 tan1tan 38

                                                                                                                              125

                                                                                                                              frac14 32

                                                                                                                              Figure Q86

                                                                                                                              Bearing capacity 65

                                                                                                                              For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                              Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                              The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                              Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                              For B frac14 250m qn frac14 3750

                                                                                                                              2502 17 frac14 583 kN=m2

                                                                                                                              From Figure 510 m frac14 n frac14 126

                                                                                                                              6frac14 021

                                                                                                                              Ir frac14 0019

                                                                                                                              Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                              Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                              The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                              89

                                                                                                                              Depth (m) N 0v (kNm2) CN N1

                                                                                                                              070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                              Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                              (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                              Cw frac14 05thorn 0530

                                                                                                                              47

                                                                                                                              frac14 082

                                                                                                                              66 Bearing capacity

                                                                                                                              Thus

                                                                                                                              qa frac14 150 082 frac14 120 kN=m2

                                                                                                                              (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                              Thus

                                                                                                                              qa frac14 90 15 frac14 135 kN=m2

                                                                                                                              (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                              Ic frac14 171

                                                                                                                              1014frac14 0068

                                                                                                                              From Equation 819(a) with s frac14 25mm

                                                                                                                              q frac14 25

                                                                                                                              3507 0068frac14 150 kN=m2

                                                                                                                              810

                                                                                                                              Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                              Izp frac14 05thorn 01130

                                                                                                                              16 27

                                                                                                                              05

                                                                                                                              frac14 067

                                                                                                                              Refer to Figure Q810

                                                                                                                              E frac14 25qc

                                                                                                                              Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                              Ez (mm3MN)

                                                                                                                              1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                              0203

                                                                                                                              C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                              130frac14 093

                                                                                                                              C2 frac14 1 ethsayTHORN

                                                                                                                              s frac14 C1C2qnX Iz

                                                                                                                              Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                              Bearing capacity 67

                                                                                                                              811

                                                                                                                              At pile base level

                                                                                                                              cu frac14 220 kN=m2

                                                                                                                              qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                              Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                              00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                              qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                              Then

                                                                                                                              Qf frac14 Abqb thorn Asqs

                                                                                                                              frac14

                                                                                                                              4 32 1980

                                                                                                                              thorn eth 105 139 86THORN

                                                                                                                              frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                              0 01 02 03 04 05 06 07

                                                                                                                              0 2 4 6 8 10 12 14

                                                                                                                              1

                                                                                                                              2

                                                                                                                              3

                                                                                                                              4

                                                                                                                              5

                                                                                                                              6

                                                                                                                              7

                                                                                                                              8

                                                                                                                              (1)

                                                                                                                              (2)

                                                                                                                              (3)

                                                                                                                              (4)

                                                                                                                              (5)

                                                                                                                              qc

                                                                                                                              qc

                                                                                                                              Iz

                                                                                                                              Iz

                                                                                                                              (MNm2)

                                                                                                                              z (m)

                                                                                                                              Figure Q810

                                                                                                                              68 Bearing capacity

                                                                                                                              Allowable load

                                                                                                                              ethaTHORN Qf

                                                                                                                              2frac14 17 937

                                                                                                                              2frac14 8968 kN

                                                                                                                              ethbTHORN Abqb

                                                                                                                              3thorn Asqs frac14 13 996

                                                                                                                              3thorn 3941 frac14 8606 kN

                                                                                                                              ie allowable load frac14 8600 kN

                                                                                                                              Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                              According to the limit state method

                                                                                                                              Characteristic undrained strength at base level cuk frac14 220

                                                                                                                              150kN=m2

                                                                                                                              Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                              150frac14 1320 kN=m2

                                                                                                                              Characteristic shaft resistance qsk frac14 00150

                                                                                                                              frac14 86

                                                                                                                              150frac14 57 kN=m2

                                                                                                                              Characteristic base and shaft resistances

                                                                                                                              Rbk frac14

                                                                                                                              4 32 1320 frac14 9330 kN

                                                                                                                              Rsk frac14 105 139 86

                                                                                                                              150frac14 2629 kN

                                                                                                                              For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                              Design bearing resistance Rcd frac14 9330

                                                                                                                              160thorn 2629

                                                                                                                              130

                                                                                                                              frac14 5831thorn 2022

                                                                                                                              frac14 7850 kN

                                                                                                                              Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                              812

                                                                                                                              ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                              qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                              For a single pile

                                                                                                                              Qf frac14 Abqb thorn Asqs

                                                                                                                              frac14

                                                                                                                              4 062 1305

                                                                                                                              thorn eth 06 15 42THORN

                                                                                                                              frac14 369thorn 1187 frac14 1556 kN

                                                                                                                              Bearing capacity 69

                                                                                                                              Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                              eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                              (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                              (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                              qbkfrac14 9cuk frac14 9 220

                                                                                                                              150frac14 1320 kN=m2

                                                                                                                              qskfrac14cuk frac14 040 105

                                                                                                                              150frac14 28 kN=m2

                                                                                                                              Rbkfrac14

                                                                                                                              4 0602 1320 frac14 373 kN

                                                                                                                              Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                              Rcdfrac14 373

                                                                                                                              160thorn 791

                                                                                                                              130frac14 233thorn 608 frac14 841 kN

                                                                                                                              Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                              Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                              (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                              q frac14 21 000

                                                                                                                              1762frac14 68 kN=m2

                                                                                                                              Immediate settlement

                                                                                                                              H

                                                                                                                              Bfrac14 15

                                                                                                                              176frac14 085

                                                                                                                              D

                                                                                                                              Bfrac14 13

                                                                                                                              176frac14 074

                                                                                                                              L

                                                                                                                              Bfrac14 1

                                                                                                                              Hence from Figure 515

                                                                                                                              130 frac14 078 and 131 frac14 041

                                                                                                                              70 Bearing capacity

                                                                                                                              Thus using Equation 528

                                                                                                                              si frac14 078 041 68 176

                                                                                                                              65frac14 6mm

                                                                                                                              Consolidation settlement

                                                                                                                              Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                              1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                              434 (sod)

                                                                                                                              Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                              sc frac14 056 434 frac14 24mm

                                                                                                                              The total settlement is (6thorn 24) frac14 30mm

                                                                                                                              813

                                                                                                                              At base level N frac14 26 Then using Equation 830

                                                                                                                              qb frac14 40NDb

                                                                                                                              Bfrac14 40 26 2

                                                                                                                              025frac14 8320 kN=m2

                                                                                                                              ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                              Figure Q812

                                                                                                                              Bearing capacity 71

                                                                                                                              Over the length embedded in sand

                                                                                                                              N frac14 21 ie18thorn 24

                                                                                                                              2

                                                                                                                              Using Equation 831

                                                                                                                              qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                              For a single pile

                                                                                                                              Qf frac14 Abqb thorn Asqs

                                                                                                                              frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                              For the pile group assuming a group efficiency of 12

                                                                                                                              XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                              Then the load factor is

                                                                                                                              F frac14 6523

                                                                                                                              2000thorn 1000frac14 21

                                                                                                                              (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                              Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                              150frac14 5547 kNm2

                                                                                                                              Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                              150frac14 28 kNm2

                                                                                                                              Characteristic base and shaft resistances for a single pile

                                                                                                                              Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                              Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                              For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                              Design bearing resistance Rcd frac14 347

                                                                                                                              130thorn 56

                                                                                                                              130frac14 310 kN

                                                                                                                              For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                              Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                              (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                              From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                              72 Bearing capacity

                                                                                                                              N frac14 24thorn 26thorn 34

                                                                                                                              3frac14 28

                                                                                                                              Ic frac14 171

                                                                                                                              2814frac14 0016 ethEquation 818THORN

                                                                                                                              s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                              814

                                                                                                                              Using Equation 841

                                                                                                                              Tf frac14 DLcu thorn

                                                                                                                              4ethD2 d2THORNcuNc

                                                                                                                              frac14 eth 02 5 06 110THORN thorn

                                                                                                                              4eth022 012THORN110 9

                                                                                                                              frac14 207thorn 23 frac14 230 kN

                                                                                                                              Figure Q813

                                                                                                                              Bearing capacity 73

                                                                                                                              Chapter 9

                                                                                                                              Stability of slopes

                                                                                                                              91

                                                                                                                              Referring to Figure Q91

                                                                                                                              W frac14 417 19 frac14 792 kN=m

                                                                                                                              Q frac14 20 28 frac14 56 kN=m

                                                                                                                              Arc lengthAB frac14

                                                                                                                              180 73 90 frac14 115m

                                                                                                                              Arc length BC frac14

                                                                                                                              180 28 90 frac14 44m

                                                                                                                              The factor of safety is given by

                                                                                                                              F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                              90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                              Depth of tension crack z0 frac14 2cu

                                                                                                                              frac14 2 20

                                                                                                                              19frac14 21m

                                                                                                                              Arc length BD frac14

                                                                                                                              180 13

                                                                                                                              1

                                                                                                                              2 90 frac14 21m

                                                                                                                              F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                              The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                              Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                              14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                              Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                              The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                              92

                                                                                                                              u frac14 0

                                                                                                                              Depth factor D frac14 11

                                                                                                                              9frac14 122

                                                                                                                              Using Equation 92 with F frac14 10

                                                                                                                              Ns frac14 cu

                                                                                                                              FHfrac14 30

                                                                                                                              10 19 9frac14 0175

                                                                                                                              Hence from Figure 93

                                                                                                                              frac14 50

                                                                                                                              For F frac14 12

                                                                                                                              Ns frac14 30

                                                                                                                              12 19 9frac14 0146

                                                                                                                              frac14 27

                                                                                                                              93

                                                                                                                              Refer to Figure Q93

                                                                                                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                              1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                              74 m

                                                                                                                              214 1deg

                                                                                                                              213 1deg

                                                                                                                              39 m

                                                                                                                              WB

                                                                                                                              D

                                                                                                                              C

                                                                                                                              28 m

                                                                                                                              21 m

                                                                                                                              A

                                                                                                                              Q

                                                                                                                              Soil (1)Soil (2)

                                                                                                                              73deg

                                                                                                                              Figure Q91

                                                                                                                              Stability of slopes 75

                                                                                                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                              9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                              599 256 328 1372

                                                                                                                              Figure Q93

                                                                                                                              76 Stability of slopes

                                                                                                                              XW cos frac14 b

                                                                                                                              Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                              W sin frac14 bX

                                                                                                                              h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                              Arc length La frac14

                                                                                                                              180 57

                                                                                                                              1

                                                                                                                              2 326 frac14 327m

                                                                                                                              The factor of safety is given by

                                                                                                                              F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                              W sin

                                                                                                                              frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                              frac14 091

                                                                                                                              According to the limit state method

                                                                                                                              0d frac14 tan1tan 32

                                                                                                                              125

                                                                                                                              frac14 265

                                                                                                                              c0 frac14 8

                                                                                                                              160frac14 5 kN=m2

                                                                                                                              Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                              Design disturbing moment frac14 1075 kN=m

                                                                                                                              The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                              94

                                                                                                                              F frac14 1

                                                                                                                              W sin

                                                                                                                              Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                              1thorn ethtan tan0=FTHORN

                                                                                                                              c0 frac14 8 kN=m2

                                                                                                                              0 frac14 32

                                                                                                                              c0b frac14 8 2 frac14 16 kN=m

                                                                                                                              W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                              Try F frac14 100

                                                                                                                              tan0

                                                                                                                              Ffrac14 0625

                                                                                                                              Stability of slopes 77

                                                                                                                              Values of u are as obtained in Figure Q93

                                                                                                                              SliceNo

                                                                                                                              h(m)

                                                                                                                              W frac14 bh(kNm)

                                                                                                                              W sin(kNm)

                                                                                                                              ub(kNm)

                                                                                                                              c0bthorn (W ub) tan0(kNm)

                                                                                                                              sec

                                                                                                                              1thorn (tan tan0)FProduct(kNm)

                                                                                                                              1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                              23 33 30 1042 31

                                                                                                                              3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                              224 92 72 0931 67

                                                                                                                              6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                              12 58 112 90 0889 80

                                                                                                                              8 60 252 1812

                                                                                                                              80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                              10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                              2154 88 116 0853 99

                                                                                                                              14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                              1074 1091

                                                                                                                              F frac14 1091

                                                                                                                              1074frac14 102 (assumed value 100)

                                                                                                                              Thus

                                                                                                                              F frac14 101

                                                                                                                              95

                                                                                                                              F frac14 1

                                                                                                                              W sin

                                                                                                                              XfWeth1 ruTHORN tan0g sec

                                                                                                                              1thorn ethtan tan0THORN=F

                                                                                                                              0 frac14 33

                                                                                                                              ru frac14 020

                                                                                                                              W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                              eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                              Try F frac14 110

                                                                                                                              tan 0

                                                                                                                              Ffrac14 tan 33

                                                                                                                              110frac14 0590

                                                                                                                              78 Stability of slopes

                                                                                                                              Referring to Figure Q95

                                                                                                                              SliceNo

                                                                                                                              h(m)

                                                                                                                              W frac14 bh(kNm)

                                                                                                                              W sin(kNm)

                                                                                                                              W(1 ru) tan0(kNm)

                                                                                                                              sec

                                                                                                                              1thorn ( tan tan0)FProduct(kNm)

                                                                                                                              1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                              2120 234 0892 209

                                                                                                                              4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                              1185 1271

                                                                                                                              Figure Q95

                                                                                                                              Stability of slopes 79

                                                                                                                              F frac14 1271

                                                                                                                              1185frac14 107

                                                                                                                              The trial value was 110 therefore take F to be 108

                                                                                                                              96

                                                                                                                              (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                              F frac14 0

                                                                                                                              sat

                                                                                                                              tan0

                                                                                                                              tan

                                                                                                                              ptie 15 frac14 92

                                                                                                                              19

                                                                                                                              tan 36

                                                                                                                              tan

                                                                                                                              tan frac14 0234

                                                                                                                              frac14 13

                                                                                                                              Water table well below surface the factor of safety is given by Equation 911

                                                                                                                              F frac14 tan0

                                                                                                                              tan

                                                                                                                              frac14 tan 36

                                                                                                                              tan 13

                                                                                                                              frac14 31

                                                                                                                              (b) 0d frac14 tan1tan 36

                                                                                                                              125

                                                                                                                              frac14 30

                                                                                                                              Depth of potential failure surface frac14 z

                                                                                                                              Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                              frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                              frac14 504z kN

                                                                                                                              Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                              frac14 19 z sin 13 cos 13

                                                                                                                              frac14 416z kN

                                                                                                                              Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                              80 Stability of slopes

                                                                                                                              • Book Cover
                                                                                                                              • Title
                                                                                                                              • Contents
                                                                                                                              • Basic characteristics of soils
                                                                                                                              • Seepage
                                                                                                                              • Effective stress
                                                                                                                              • Shear strength
                                                                                                                              • Stresses and displacements
                                                                                                                              • Lateral earth pressure
                                                                                                                              • Consolidation theory
                                                                                                                              • Bearing capacity
                                                                                                                              • Stability of slopes

                                                                                                                                From Figure 515 (circle) 131 frac14 032 and 130 frac14 10

                                                                                                                                si frac14 130131qB

                                                                                                                                Eufrac14 10 032 105 35

                                                                                                                                40frac14 30mm

                                                                                                                                (b) Consolidation settlement

                                                                                                                                Layer z (m) Dz Ic (kNm2) syod (mm)

                                                                                                                                1 25 14 0997 107 7352 75 467 0930 98 6863 125 280 0804 84 5884 175 200 0647 68 4765 225 155 0505 53 3716 275 127 0396 42 294

                                                                                                                                3150

                                                                                                                                Notes From Figure 59y sod frac14 mv0H frac14 0140 5 frac14 0700 (0 frac14 )

                                                                                                                                Now

                                                                                                                                H

                                                                                                                                Bfrac14 30

                                                                                                                                35frac14 086 and A frac14 065

                                                                                                                                from Figure 712 13 frac14 079

                                                                                                                                sc frac14 13sod frac14 079 315 frac14 250mm

                                                                                                                                Total settlement

                                                                                                                                s frac14 si thorn scfrac14 30thorn 250 frac14 280mm

                                                                                                                                79

                                                                                                                                Without sand drains

                                                                                                                                Uv frac14 025

                                                                                                                                Tv frac14 0049 ethfrom Figure 718THORN

                                                                                                                                t frac14 Tvd2

                                                                                                                                cvfrac14 0049 82

                                                                                                                                cvWith sand drains

                                                                                                                                R frac14 0564S frac14 0564 3 frac14 169m

                                                                                                                                n frac14 Rrfrac14 169

                                                                                                                                015frac14 113

                                                                                                                                Tr frac14 cht

                                                                                                                                4R2frac14 ch

                                                                                                                                4 1692 0049 82

                                                                                                                                cvethand ch frac14 cvTHORN

                                                                                                                                frac14 0275

                                                                                                                                Ur frac14 073 (from Figure 730)

                                                                                                                                58 Consolidation theory

                                                                                                                                Using Equation 740

                                                                                                                                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                                                U frac14 080

                                                                                                                                710

                                                                                                                                Without sand drains

                                                                                                                                Uv frac14 090

                                                                                                                                Tv frac14 0848

                                                                                                                                t frac14 Tvd2

                                                                                                                                cvfrac14 0848 102

                                                                                                                                96frac14 88 years

                                                                                                                                With sand drains

                                                                                                                                R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                                                n frac14 Rrfrac14 226

                                                                                                                                015frac14 15

                                                                                                                                Tr

                                                                                                                                Tvfrac14 chcv

                                                                                                                                d2

                                                                                                                                4R2ethsame tTHORN

                                                                                                                                Tr

                                                                                                                                Tvfrac14 140

                                                                                                                                96 102

                                                                                                                                4 2262frac14 714 eth1THORN

                                                                                                                                Using Equation 740

                                                                                                                                eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                                                An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                                                Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                                                040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                                                Thus

                                                                                                                                Uv frac14 0295 and Ur frac14 086

                                                                                                                                t frac14 88 00683

                                                                                                                                0848frac14 07 years

                                                                                                                                Consolidation theory 59

                                                                                                                                Chapter 8

                                                                                                                                Bearing capacity

                                                                                                                                81

                                                                                                                                (a) The ultimate bearing capacity is given by Equation 83

                                                                                                                                qf frac14 cNc thorn DNq thorn 1

                                                                                                                                2BN

                                                                                                                                For u frac14 0

                                                                                                                                Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                                qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                                                The net ultimate bearing capacity is

                                                                                                                                qnf frac14 qf D frac14 540 kN=m2

                                                                                                                                The net foundation pressure is

                                                                                                                                qn frac14 q D frac14 425

                                                                                                                                2 eth21 1THORN frac14 192 kN=m2

                                                                                                                                The factor of safety (Equation 86) is

                                                                                                                                F frac14 qnfqnfrac14 540

                                                                                                                                192frac14 28

                                                                                                                                (b) For 0 frac14 28

                                                                                                                                Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                                                qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                                                2 112 2 13

                                                                                                                                frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                                                qnf frac14 574 112 frac14 563 kN=m2

                                                                                                                                F frac14 563

                                                                                                                                192frac14 29

                                                                                                                                (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                                                82

                                                                                                                                For 0 frac14 38

                                                                                                                                Nq frac14 49 N frac14 67

                                                                                                                                qnf frac14 DethNq 1THORN thorn 1

                                                                                                                                2BN ethfrom Equation 83THORN

                                                                                                                                frac14 eth18 075 48THORN thorn 1

                                                                                                                                2 18 15 67

                                                                                                                                frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                                                qn frac14 500

                                                                                                                                15 eth18 075THORN frac14 320 kN=m2

                                                                                                                                F frac14 qnfqnfrac14 1553

                                                                                                                                320frac14 48

                                                                                                                                0d frac14 tan1tan 38

                                                                                                                                125

                                                                                                                                frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                                                Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                                                2 18 15 25

                                                                                                                                frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                                                Design load (action) Vd frac14 500 kN=m

                                                                                                                                The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                                                83

                                                                                                                                D

                                                                                                                                Bfrac14 350

                                                                                                                                225frac14 155

                                                                                                                                From Figure 85 for a square foundation

                                                                                                                                Nc frac14 81

                                                                                                                                Bearing capacity 61

                                                                                                                                For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                                                Nc frac14 084thorn 016B

                                                                                                                                L

                                                                                                                                81 frac14 745

                                                                                                                                Using Equation 810

                                                                                                                                qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                                                For F frac14 3

                                                                                                                                qn frac14 1006

                                                                                                                                3frac14 335 kN=m2

                                                                                                                                q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                                                Design load frac14 405 450 225 frac14 4100 kN

                                                                                                                                Design undrained strength cud frac14 135

                                                                                                                                14frac14 96 kN=m2

                                                                                                                                Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                                                frac14 7241 kN

                                                                                                                                Design load Vd frac14 4100 kN

                                                                                                                                Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                                84

                                                                                                                                For 0 frac14 40

                                                                                                                                Nq frac14 64 N frac14 95

                                                                                                                                qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                                                (a) Water table 5m below ground level

                                                                                                                                qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                                                qn frac14 400 17 frac14 383 kN=m2

                                                                                                                                F frac14 2686

                                                                                                                                383frac14 70

                                                                                                                                (b) Water table 1m below ground level (ie at foundation level)

                                                                                                                                0 frac14 20 98 frac14 102 kN=m3

                                                                                                                                62 Bearing capacity

                                                                                                                                qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                                                F frac14 2040

                                                                                                                                383frac14 53

                                                                                                                                (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                                                eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                                                qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                                                F frac14 1296

                                                                                                                                392frac14 33

                                                                                                                                85

                                                                                                                                The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                                                Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                                                Design value of 0 frac14 tan1tan 39

                                                                                                                                125

                                                                                                                                frac14 33

                                                                                                                                For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                                                Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                                                Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                                86

                                                                                                                                (a) Undrained shear for u frac14 0

                                                                                                                                Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                                qnf frac14 12cuNc

                                                                                                                                frac14 12 100 514 frac14 617 kN=m2

                                                                                                                                qn frac14 qnfFfrac14 617

                                                                                                                                3frac14 206 kN=m2

                                                                                                                                q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                                                Bearing capacity 63

                                                                                                                                Drained shear for 0 frac14 32

                                                                                                                                Nq frac14 23 N frac14 25

                                                                                                                                0 frac14 21 98 frac14 112 kN=m3

                                                                                                                                qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                                frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                                frac14 694 kN=m2

                                                                                                                                q frac14 694

                                                                                                                                3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                                Design load frac14 42 227 frac14 3632 kN

                                                                                                                                (b) Design undrained strength cud frac14 100

                                                                                                                                14frac14 71 kNm2

                                                                                                                                Design bearing resistance Rd frac14 12cudNe area

                                                                                                                                frac14 12 71 514 42

                                                                                                                                frac14 7007 kN

                                                                                                                                For drained shear 0d frac14 tan1tan 32

                                                                                                                                125

                                                                                                                                frac14 26

                                                                                                                                Nq frac14 12 N frac14 10

                                                                                                                                Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                                (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                                Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                                1 2 100 0175 0700qn 0182qn

                                                                                                                                2 6 033 0044 0176qn 0046qn

                                                                                                                                3 10 020 0017 0068qn 0018qn

                                                                                                                                0246qn

                                                                                                                                Diameter of equivalent circle B frac14 45m

                                                                                                                                H

                                                                                                                                Bfrac14 12

                                                                                                                                45frac14 27 and A frac14 042

                                                                                                                                13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                                64 Bearing capacity

                                                                                                                                For sc frac14 30mm

                                                                                                                                qn frac14 30

                                                                                                                                0147frac14 204 kN=m2

                                                                                                                                q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                                Design load frac14 42 225 frac14 3600 kN

                                                                                                                                The design load is 3600 kN settlement being the limiting criterion

                                                                                                                                87

                                                                                                                                D

                                                                                                                                Bfrac14 8

                                                                                                                                4frac14 20

                                                                                                                                From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                                F frac14 cuNc

                                                                                                                                Dfrac14 40 71

                                                                                                                                20 8frac14 18

                                                                                                                                88

                                                                                                                                Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                                Design value of 0 frac14 tan1tan 38

                                                                                                                                125

                                                                                                                                frac14 32

                                                                                                                                Figure Q86

                                                                                                                                Bearing capacity 65

                                                                                                                                For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                                Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                                The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                                Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                                For B frac14 250m qn frac14 3750

                                                                                                                                2502 17 frac14 583 kN=m2

                                                                                                                                From Figure 510 m frac14 n frac14 126

                                                                                                                                6frac14 021

                                                                                                                                Ir frac14 0019

                                                                                                                                Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                                Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                                The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                89

                                                                                                                                Depth (m) N 0v (kNm2) CN N1

                                                                                                                                070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                                Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                                (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                                Cw frac14 05thorn 0530

                                                                                                                                47

                                                                                                                                frac14 082

                                                                                                                                66 Bearing capacity

                                                                                                                                Thus

                                                                                                                                qa frac14 150 082 frac14 120 kN=m2

                                                                                                                                (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                                Thus

                                                                                                                                qa frac14 90 15 frac14 135 kN=m2

                                                                                                                                (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                                Ic frac14 171

                                                                                                                                1014frac14 0068

                                                                                                                                From Equation 819(a) with s frac14 25mm

                                                                                                                                q frac14 25

                                                                                                                                3507 0068frac14 150 kN=m2

                                                                                                                                810

                                                                                                                                Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                                Izp frac14 05thorn 01130

                                                                                                                                16 27

                                                                                                                                05

                                                                                                                                frac14 067

                                                                                                                                Refer to Figure Q810

                                                                                                                                E frac14 25qc

                                                                                                                                Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                                Ez (mm3MN)

                                                                                                                                1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                                0203

                                                                                                                                C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                                130frac14 093

                                                                                                                                C2 frac14 1 ethsayTHORN

                                                                                                                                s frac14 C1C2qnX Iz

                                                                                                                                Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                                Bearing capacity 67

                                                                                                                                811

                                                                                                                                At pile base level

                                                                                                                                cu frac14 220 kN=m2

                                                                                                                                qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                                Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                                00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                                qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                                Then

                                                                                                                                Qf frac14 Abqb thorn Asqs

                                                                                                                                frac14

                                                                                                                                4 32 1980

                                                                                                                                thorn eth 105 139 86THORN

                                                                                                                                frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                                0 01 02 03 04 05 06 07

                                                                                                                                0 2 4 6 8 10 12 14

                                                                                                                                1

                                                                                                                                2

                                                                                                                                3

                                                                                                                                4

                                                                                                                                5

                                                                                                                                6

                                                                                                                                7

                                                                                                                                8

                                                                                                                                (1)

                                                                                                                                (2)

                                                                                                                                (3)

                                                                                                                                (4)

                                                                                                                                (5)

                                                                                                                                qc

                                                                                                                                qc

                                                                                                                                Iz

                                                                                                                                Iz

                                                                                                                                (MNm2)

                                                                                                                                z (m)

                                                                                                                                Figure Q810

                                                                                                                                68 Bearing capacity

                                                                                                                                Allowable load

                                                                                                                                ethaTHORN Qf

                                                                                                                                2frac14 17 937

                                                                                                                                2frac14 8968 kN

                                                                                                                                ethbTHORN Abqb

                                                                                                                                3thorn Asqs frac14 13 996

                                                                                                                                3thorn 3941 frac14 8606 kN

                                                                                                                                ie allowable load frac14 8600 kN

                                                                                                                                Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                                According to the limit state method

                                                                                                                                Characteristic undrained strength at base level cuk frac14 220

                                                                                                                                150kN=m2

                                                                                                                                Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                                150frac14 1320 kN=m2

                                                                                                                                Characteristic shaft resistance qsk frac14 00150

                                                                                                                                frac14 86

                                                                                                                                150frac14 57 kN=m2

                                                                                                                                Characteristic base and shaft resistances

                                                                                                                                Rbk frac14

                                                                                                                                4 32 1320 frac14 9330 kN

                                                                                                                                Rsk frac14 105 139 86

                                                                                                                                150frac14 2629 kN

                                                                                                                                For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                                Design bearing resistance Rcd frac14 9330

                                                                                                                                160thorn 2629

                                                                                                                                130

                                                                                                                                frac14 5831thorn 2022

                                                                                                                                frac14 7850 kN

                                                                                                                                Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                                812

                                                                                                                                ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                                qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                                For a single pile

                                                                                                                                Qf frac14 Abqb thorn Asqs

                                                                                                                                frac14

                                                                                                                                4 062 1305

                                                                                                                                thorn eth 06 15 42THORN

                                                                                                                                frac14 369thorn 1187 frac14 1556 kN

                                                                                                                                Bearing capacity 69

                                                                                                                                Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                                eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                                (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                                (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                                qbkfrac14 9cuk frac14 9 220

                                                                                                                                150frac14 1320 kN=m2

                                                                                                                                qskfrac14cuk frac14 040 105

                                                                                                                                150frac14 28 kN=m2

                                                                                                                                Rbkfrac14

                                                                                                                                4 0602 1320 frac14 373 kN

                                                                                                                                Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                                Rcdfrac14 373

                                                                                                                                160thorn 791

                                                                                                                                130frac14 233thorn 608 frac14 841 kN

                                                                                                                                Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                                Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                                (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                                q frac14 21 000

                                                                                                                                1762frac14 68 kN=m2

                                                                                                                                Immediate settlement

                                                                                                                                H

                                                                                                                                Bfrac14 15

                                                                                                                                176frac14 085

                                                                                                                                D

                                                                                                                                Bfrac14 13

                                                                                                                                176frac14 074

                                                                                                                                L

                                                                                                                                Bfrac14 1

                                                                                                                                Hence from Figure 515

                                                                                                                                130 frac14 078 and 131 frac14 041

                                                                                                                                70 Bearing capacity

                                                                                                                                Thus using Equation 528

                                                                                                                                si frac14 078 041 68 176

                                                                                                                                65frac14 6mm

                                                                                                                                Consolidation settlement

                                                                                                                                Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                                1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                                434 (sod)

                                                                                                                                Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                                sc frac14 056 434 frac14 24mm

                                                                                                                                The total settlement is (6thorn 24) frac14 30mm

                                                                                                                                813

                                                                                                                                At base level N frac14 26 Then using Equation 830

                                                                                                                                qb frac14 40NDb

                                                                                                                                Bfrac14 40 26 2

                                                                                                                                025frac14 8320 kN=m2

                                                                                                                                ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                                Figure Q812

                                                                                                                                Bearing capacity 71

                                                                                                                                Over the length embedded in sand

                                                                                                                                N frac14 21 ie18thorn 24

                                                                                                                                2

                                                                                                                                Using Equation 831

                                                                                                                                qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                For a single pile

                                                                                                                                Qf frac14 Abqb thorn Asqs

                                                                                                                                frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                For the pile group assuming a group efficiency of 12

                                                                                                                                XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                Then the load factor is

                                                                                                                                F frac14 6523

                                                                                                                                2000thorn 1000frac14 21

                                                                                                                                (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                150frac14 5547 kNm2

                                                                                                                                Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                150frac14 28 kNm2

                                                                                                                                Characteristic base and shaft resistances for a single pile

                                                                                                                                Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                Design bearing resistance Rcd frac14 347

                                                                                                                                130thorn 56

                                                                                                                                130frac14 310 kN

                                                                                                                                For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                72 Bearing capacity

                                                                                                                                N frac14 24thorn 26thorn 34

                                                                                                                                3frac14 28

                                                                                                                                Ic frac14 171

                                                                                                                                2814frac14 0016 ethEquation 818THORN

                                                                                                                                s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                814

                                                                                                                                Using Equation 841

                                                                                                                                Tf frac14 DLcu thorn

                                                                                                                                4ethD2 d2THORNcuNc

                                                                                                                                frac14 eth 02 5 06 110THORN thorn

                                                                                                                                4eth022 012THORN110 9

                                                                                                                                frac14 207thorn 23 frac14 230 kN

                                                                                                                                Figure Q813

                                                                                                                                Bearing capacity 73

                                                                                                                                Chapter 9

                                                                                                                                Stability of slopes

                                                                                                                                91

                                                                                                                                Referring to Figure Q91

                                                                                                                                W frac14 417 19 frac14 792 kN=m

                                                                                                                                Q frac14 20 28 frac14 56 kN=m

                                                                                                                                Arc lengthAB frac14

                                                                                                                                180 73 90 frac14 115m

                                                                                                                                Arc length BC frac14

                                                                                                                                180 28 90 frac14 44m

                                                                                                                                The factor of safety is given by

                                                                                                                                F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                Depth of tension crack z0 frac14 2cu

                                                                                                                                frac14 2 20

                                                                                                                                19frac14 21m

                                                                                                                                Arc length BD frac14

                                                                                                                                180 13

                                                                                                                                1

                                                                                                                                2 90 frac14 21m

                                                                                                                                F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                92

                                                                                                                                u frac14 0

                                                                                                                                Depth factor D frac14 11

                                                                                                                                9frac14 122

                                                                                                                                Using Equation 92 with F frac14 10

                                                                                                                                Ns frac14 cu

                                                                                                                                FHfrac14 30

                                                                                                                                10 19 9frac14 0175

                                                                                                                                Hence from Figure 93

                                                                                                                                frac14 50

                                                                                                                                For F frac14 12

                                                                                                                                Ns frac14 30

                                                                                                                                12 19 9frac14 0146

                                                                                                                                frac14 27

                                                                                                                                93

                                                                                                                                Refer to Figure Q93

                                                                                                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                74 m

                                                                                                                                214 1deg

                                                                                                                                213 1deg

                                                                                                                                39 m

                                                                                                                                WB

                                                                                                                                D

                                                                                                                                C

                                                                                                                                28 m

                                                                                                                                21 m

                                                                                                                                A

                                                                                                                                Q

                                                                                                                                Soil (1)Soil (2)

                                                                                                                                73deg

                                                                                                                                Figure Q91

                                                                                                                                Stability of slopes 75

                                                                                                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                599 256 328 1372

                                                                                                                                Figure Q93

                                                                                                                                76 Stability of slopes

                                                                                                                                XW cos frac14 b

                                                                                                                                Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                W sin frac14 bX

                                                                                                                                h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                Arc length La frac14

                                                                                                                                180 57

                                                                                                                                1

                                                                                                                                2 326 frac14 327m

                                                                                                                                The factor of safety is given by

                                                                                                                                F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                W sin

                                                                                                                                frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                frac14 091

                                                                                                                                According to the limit state method

                                                                                                                                0d frac14 tan1tan 32

                                                                                                                                125

                                                                                                                                frac14 265

                                                                                                                                c0 frac14 8

                                                                                                                                160frac14 5 kN=m2

                                                                                                                                Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                Design disturbing moment frac14 1075 kN=m

                                                                                                                                The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                94

                                                                                                                                F frac14 1

                                                                                                                                W sin

                                                                                                                                Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                1thorn ethtan tan0=FTHORN

                                                                                                                                c0 frac14 8 kN=m2

                                                                                                                                0 frac14 32

                                                                                                                                c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                Try F frac14 100

                                                                                                                                tan0

                                                                                                                                Ffrac14 0625

                                                                                                                                Stability of slopes 77

                                                                                                                                Values of u are as obtained in Figure Q93

                                                                                                                                SliceNo

                                                                                                                                h(m)

                                                                                                                                W frac14 bh(kNm)

                                                                                                                                W sin(kNm)

                                                                                                                                ub(kNm)

                                                                                                                                c0bthorn (W ub) tan0(kNm)

                                                                                                                                sec

                                                                                                                                1thorn (tan tan0)FProduct(kNm)

                                                                                                                                1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                23 33 30 1042 31

                                                                                                                                3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                224 92 72 0931 67

                                                                                                                                6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                12 58 112 90 0889 80

                                                                                                                                8 60 252 1812

                                                                                                                                80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                2154 88 116 0853 99

                                                                                                                                14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                1074 1091

                                                                                                                                F frac14 1091

                                                                                                                                1074frac14 102 (assumed value 100)

                                                                                                                                Thus

                                                                                                                                F frac14 101

                                                                                                                                95

                                                                                                                                F frac14 1

                                                                                                                                W sin

                                                                                                                                XfWeth1 ruTHORN tan0g sec

                                                                                                                                1thorn ethtan tan0THORN=F

                                                                                                                                0 frac14 33

                                                                                                                                ru frac14 020

                                                                                                                                W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                Try F frac14 110

                                                                                                                                tan 0

                                                                                                                                Ffrac14 tan 33

                                                                                                                                110frac14 0590

                                                                                                                                78 Stability of slopes

                                                                                                                                Referring to Figure Q95

                                                                                                                                SliceNo

                                                                                                                                h(m)

                                                                                                                                W frac14 bh(kNm)

                                                                                                                                W sin(kNm)

                                                                                                                                W(1 ru) tan0(kNm)

                                                                                                                                sec

                                                                                                                                1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                2120 234 0892 209

                                                                                                                                4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                1185 1271

                                                                                                                                Figure Q95

                                                                                                                                Stability of slopes 79

                                                                                                                                F frac14 1271

                                                                                                                                1185frac14 107

                                                                                                                                The trial value was 110 therefore take F to be 108

                                                                                                                                96

                                                                                                                                (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                F frac14 0

                                                                                                                                sat

                                                                                                                                tan0

                                                                                                                                tan

                                                                                                                                ptie 15 frac14 92

                                                                                                                                19

                                                                                                                                tan 36

                                                                                                                                tan

                                                                                                                                tan frac14 0234

                                                                                                                                frac14 13

                                                                                                                                Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                F frac14 tan0

                                                                                                                                tan

                                                                                                                                frac14 tan 36

                                                                                                                                tan 13

                                                                                                                                frac14 31

                                                                                                                                (b) 0d frac14 tan1tan 36

                                                                                                                                125

                                                                                                                                frac14 30

                                                                                                                                Depth of potential failure surface frac14 z

                                                                                                                                Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                frac14 504z kN

                                                                                                                                Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                frac14 19 z sin 13 cos 13

                                                                                                                                frac14 416z kN

                                                                                                                                Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                80 Stability of slopes

                                                                                                                                • Book Cover
                                                                                                                                • Title
                                                                                                                                • Contents
                                                                                                                                • Basic characteristics of soils
                                                                                                                                • Seepage
                                                                                                                                • Effective stress
                                                                                                                                • Shear strength
                                                                                                                                • Stresses and displacements
                                                                                                                                • Lateral earth pressure
                                                                                                                                • Consolidation theory
                                                                                                                                • Bearing capacity
                                                                                                                                • Stability of slopes

                                                                                                                                  Using Equation 740

                                                                                                                                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORNfrac14 eth1 025THORNeth1 073THORN frac14 020

                                                                                                                                  U frac14 080

                                                                                                                                  710

                                                                                                                                  Without sand drains

                                                                                                                                  Uv frac14 090

                                                                                                                                  Tv frac14 0848

                                                                                                                                  t frac14 Tvd2

                                                                                                                                  cvfrac14 0848 102

                                                                                                                                  96frac14 88 years

                                                                                                                                  With sand drains

                                                                                                                                  R frac14 0564S frac14 0564 4 frac14 226m

                                                                                                                                  n frac14 Rrfrac14 226

                                                                                                                                  015frac14 15

                                                                                                                                  Tr

                                                                                                                                  Tvfrac14 chcv

                                                                                                                                  d2

                                                                                                                                  4R2ethsame tTHORN

                                                                                                                                  Tr

                                                                                                                                  Tvfrac14 140

                                                                                                                                  96 102

                                                                                                                                  4 2262frac14 714 eth1THORN

                                                                                                                                  Using Equation 740

                                                                                                                                  eth1UTHORN frac14 eth1UvTHORNeth1UrTHORN eth1 090THORN frac14 eth1UvTHORNeth1UrTHORN eth1UvTHORNeth1UrTHORN frac14 010 eth2THORN

                                                                                                                                  An iterative solution is required using (1) and (2) an initial value ofUv being estimated

                                                                                                                                  Uv Tv Tr frac14 714Tv Ur (1 Uv)(1 Ur)

                                                                                                                                  040 01256 0897 097 060 003 frac14 0018030 00707 0505 087 070 013 frac14 0091029 00660 0471 085 071 015 frac14 01070295 00683 0488 086 0705 014 frac14 0099

                                                                                                                                  Thus

                                                                                                                                  Uv frac14 0295 and Ur frac14 086

                                                                                                                                  t frac14 88 00683

                                                                                                                                  0848frac14 07 years

                                                                                                                                  Consolidation theory 59

                                                                                                                                  Chapter 8

                                                                                                                                  Bearing capacity

                                                                                                                                  81

                                                                                                                                  (a) The ultimate bearing capacity is given by Equation 83

                                                                                                                                  qf frac14 cNc thorn DNq thorn 1

                                                                                                                                  2BN

                                                                                                                                  For u frac14 0

                                                                                                                                  Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                                  qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                                                  The net ultimate bearing capacity is

                                                                                                                                  qnf frac14 qf D frac14 540 kN=m2

                                                                                                                                  The net foundation pressure is

                                                                                                                                  qn frac14 q D frac14 425

                                                                                                                                  2 eth21 1THORN frac14 192 kN=m2

                                                                                                                                  The factor of safety (Equation 86) is

                                                                                                                                  F frac14 qnfqnfrac14 540

                                                                                                                                  192frac14 28

                                                                                                                                  (b) For 0 frac14 28

                                                                                                                                  Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                                                  qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                                                  2 112 2 13

                                                                                                                                  frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                                                  qnf frac14 574 112 frac14 563 kN=m2

                                                                                                                                  F frac14 563

                                                                                                                                  192frac14 29

                                                                                                                                  (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                                                  82

                                                                                                                                  For 0 frac14 38

                                                                                                                                  Nq frac14 49 N frac14 67

                                                                                                                                  qnf frac14 DethNq 1THORN thorn 1

                                                                                                                                  2BN ethfrom Equation 83THORN

                                                                                                                                  frac14 eth18 075 48THORN thorn 1

                                                                                                                                  2 18 15 67

                                                                                                                                  frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                                                  qn frac14 500

                                                                                                                                  15 eth18 075THORN frac14 320 kN=m2

                                                                                                                                  F frac14 qnfqnfrac14 1553

                                                                                                                                  320frac14 48

                                                                                                                                  0d frac14 tan1tan 38

                                                                                                                                  125

                                                                                                                                  frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                                                  Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                                                  2 18 15 25

                                                                                                                                  frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                                                  Design load (action) Vd frac14 500 kN=m

                                                                                                                                  The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                                                  83

                                                                                                                                  D

                                                                                                                                  Bfrac14 350

                                                                                                                                  225frac14 155

                                                                                                                                  From Figure 85 for a square foundation

                                                                                                                                  Nc frac14 81

                                                                                                                                  Bearing capacity 61

                                                                                                                                  For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                                                  Nc frac14 084thorn 016B

                                                                                                                                  L

                                                                                                                                  81 frac14 745

                                                                                                                                  Using Equation 810

                                                                                                                                  qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                                                  For F frac14 3

                                                                                                                                  qn frac14 1006

                                                                                                                                  3frac14 335 kN=m2

                                                                                                                                  q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                                                  Design load frac14 405 450 225 frac14 4100 kN

                                                                                                                                  Design undrained strength cud frac14 135

                                                                                                                                  14frac14 96 kN=m2

                                                                                                                                  Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                                                  frac14 7241 kN

                                                                                                                                  Design load Vd frac14 4100 kN

                                                                                                                                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                                  84

                                                                                                                                  For 0 frac14 40

                                                                                                                                  Nq frac14 64 N frac14 95

                                                                                                                                  qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                                                  (a) Water table 5m below ground level

                                                                                                                                  qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                                                  qn frac14 400 17 frac14 383 kN=m2

                                                                                                                                  F frac14 2686

                                                                                                                                  383frac14 70

                                                                                                                                  (b) Water table 1m below ground level (ie at foundation level)

                                                                                                                                  0 frac14 20 98 frac14 102 kN=m3

                                                                                                                                  62 Bearing capacity

                                                                                                                                  qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                                                  F frac14 2040

                                                                                                                                  383frac14 53

                                                                                                                                  (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                                                  eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                                                  qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                                                  F frac14 1296

                                                                                                                                  392frac14 33

                                                                                                                                  85

                                                                                                                                  The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                                                  Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                                                  Design value of 0 frac14 tan1tan 39

                                                                                                                                  125

                                                                                                                                  frac14 33

                                                                                                                                  For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                                                  Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                                                  Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                                  86

                                                                                                                                  (a) Undrained shear for u frac14 0

                                                                                                                                  Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                                  qnf frac14 12cuNc

                                                                                                                                  frac14 12 100 514 frac14 617 kN=m2

                                                                                                                                  qn frac14 qnfFfrac14 617

                                                                                                                                  3frac14 206 kN=m2

                                                                                                                                  q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                                                  Bearing capacity 63

                                                                                                                                  Drained shear for 0 frac14 32

                                                                                                                                  Nq frac14 23 N frac14 25

                                                                                                                                  0 frac14 21 98 frac14 112 kN=m3

                                                                                                                                  qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                                  frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                                  frac14 694 kN=m2

                                                                                                                                  q frac14 694

                                                                                                                                  3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                                  Design load frac14 42 227 frac14 3632 kN

                                                                                                                                  (b) Design undrained strength cud frac14 100

                                                                                                                                  14frac14 71 kNm2

                                                                                                                                  Design bearing resistance Rd frac14 12cudNe area

                                                                                                                                  frac14 12 71 514 42

                                                                                                                                  frac14 7007 kN

                                                                                                                                  For drained shear 0d frac14 tan1tan 32

                                                                                                                                  125

                                                                                                                                  frac14 26

                                                                                                                                  Nq frac14 12 N frac14 10

                                                                                                                                  Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                                  (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                                  Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                                  1 2 100 0175 0700qn 0182qn

                                                                                                                                  2 6 033 0044 0176qn 0046qn

                                                                                                                                  3 10 020 0017 0068qn 0018qn

                                                                                                                                  0246qn

                                                                                                                                  Diameter of equivalent circle B frac14 45m

                                                                                                                                  H

                                                                                                                                  Bfrac14 12

                                                                                                                                  45frac14 27 and A frac14 042

                                                                                                                                  13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                                  64 Bearing capacity

                                                                                                                                  For sc frac14 30mm

                                                                                                                                  qn frac14 30

                                                                                                                                  0147frac14 204 kN=m2

                                                                                                                                  q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                                  Design load frac14 42 225 frac14 3600 kN

                                                                                                                                  The design load is 3600 kN settlement being the limiting criterion

                                                                                                                                  87

                                                                                                                                  D

                                                                                                                                  Bfrac14 8

                                                                                                                                  4frac14 20

                                                                                                                                  From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                                  F frac14 cuNc

                                                                                                                                  Dfrac14 40 71

                                                                                                                                  20 8frac14 18

                                                                                                                                  88

                                                                                                                                  Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                                  Design value of 0 frac14 tan1tan 38

                                                                                                                                  125

                                                                                                                                  frac14 32

                                                                                                                                  Figure Q86

                                                                                                                                  Bearing capacity 65

                                                                                                                                  For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                                  Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                                  The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                                  Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                                  For B frac14 250m qn frac14 3750

                                                                                                                                  2502 17 frac14 583 kN=m2

                                                                                                                                  From Figure 510 m frac14 n frac14 126

                                                                                                                                  6frac14 021

                                                                                                                                  Ir frac14 0019

                                                                                                                                  Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                                  Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                                  The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                  89

                                                                                                                                  Depth (m) N 0v (kNm2) CN N1

                                                                                                                                  070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                                  Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                                  (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                                  Cw frac14 05thorn 0530

                                                                                                                                  47

                                                                                                                                  frac14 082

                                                                                                                                  66 Bearing capacity

                                                                                                                                  Thus

                                                                                                                                  qa frac14 150 082 frac14 120 kN=m2

                                                                                                                                  (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                                  Thus

                                                                                                                                  qa frac14 90 15 frac14 135 kN=m2

                                                                                                                                  (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                                  Ic frac14 171

                                                                                                                                  1014frac14 0068

                                                                                                                                  From Equation 819(a) with s frac14 25mm

                                                                                                                                  q frac14 25

                                                                                                                                  3507 0068frac14 150 kN=m2

                                                                                                                                  810

                                                                                                                                  Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                                  Izp frac14 05thorn 01130

                                                                                                                                  16 27

                                                                                                                                  05

                                                                                                                                  frac14 067

                                                                                                                                  Refer to Figure Q810

                                                                                                                                  E frac14 25qc

                                                                                                                                  Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                                  Ez (mm3MN)

                                                                                                                                  1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                                  0203

                                                                                                                                  C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                                  130frac14 093

                                                                                                                                  C2 frac14 1 ethsayTHORN

                                                                                                                                  s frac14 C1C2qnX Iz

                                                                                                                                  Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                                  Bearing capacity 67

                                                                                                                                  811

                                                                                                                                  At pile base level

                                                                                                                                  cu frac14 220 kN=m2

                                                                                                                                  qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                                  Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                                  00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                                  qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                                  Then

                                                                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                                                                  frac14

                                                                                                                                  4 32 1980

                                                                                                                                  thorn eth 105 139 86THORN

                                                                                                                                  frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                                  0 01 02 03 04 05 06 07

                                                                                                                                  0 2 4 6 8 10 12 14

                                                                                                                                  1

                                                                                                                                  2

                                                                                                                                  3

                                                                                                                                  4

                                                                                                                                  5

                                                                                                                                  6

                                                                                                                                  7

                                                                                                                                  8

                                                                                                                                  (1)

                                                                                                                                  (2)

                                                                                                                                  (3)

                                                                                                                                  (4)

                                                                                                                                  (5)

                                                                                                                                  qc

                                                                                                                                  qc

                                                                                                                                  Iz

                                                                                                                                  Iz

                                                                                                                                  (MNm2)

                                                                                                                                  z (m)

                                                                                                                                  Figure Q810

                                                                                                                                  68 Bearing capacity

                                                                                                                                  Allowable load

                                                                                                                                  ethaTHORN Qf

                                                                                                                                  2frac14 17 937

                                                                                                                                  2frac14 8968 kN

                                                                                                                                  ethbTHORN Abqb

                                                                                                                                  3thorn Asqs frac14 13 996

                                                                                                                                  3thorn 3941 frac14 8606 kN

                                                                                                                                  ie allowable load frac14 8600 kN

                                                                                                                                  Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                                  According to the limit state method

                                                                                                                                  Characteristic undrained strength at base level cuk frac14 220

                                                                                                                                  150kN=m2

                                                                                                                                  Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                                  150frac14 1320 kN=m2

                                                                                                                                  Characteristic shaft resistance qsk frac14 00150

                                                                                                                                  frac14 86

                                                                                                                                  150frac14 57 kN=m2

                                                                                                                                  Characteristic base and shaft resistances

                                                                                                                                  Rbk frac14

                                                                                                                                  4 32 1320 frac14 9330 kN

                                                                                                                                  Rsk frac14 105 139 86

                                                                                                                                  150frac14 2629 kN

                                                                                                                                  For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                                  Design bearing resistance Rcd frac14 9330

                                                                                                                                  160thorn 2629

                                                                                                                                  130

                                                                                                                                  frac14 5831thorn 2022

                                                                                                                                  frac14 7850 kN

                                                                                                                                  Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                                  812

                                                                                                                                  ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                                  qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                                  For a single pile

                                                                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                                                                  frac14

                                                                                                                                  4 062 1305

                                                                                                                                  thorn eth 06 15 42THORN

                                                                                                                                  frac14 369thorn 1187 frac14 1556 kN

                                                                                                                                  Bearing capacity 69

                                                                                                                                  Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                                  eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                                  (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                                  (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                                  qbkfrac14 9cuk frac14 9 220

                                                                                                                                  150frac14 1320 kN=m2

                                                                                                                                  qskfrac14cuk frac14 040 105

                                                                                                                                  150frac14 28 kN=m2

                                                                                                                                  Rbkfrac14

                                                                                                                                  4 0602 1320 frac14 373 kN

                                                                                                                                  Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                                  Rcdfrac14 373

                                                                                                                                  160thorn 791

                                                                                                                                  130frac14 233thorn 608 frac14 841 kN

                                                                                                                                  Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                                  Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                                  (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                                  q frac14 21 000

                                                                                                                                  1762frac14 68 kN=m2

                                                                                                                                  Immediate settlement

                                                                                                                                  H

                                                                                                                                  Bfrac14 15

                                                                                                                                  176frac14 085

                                                                                                                                  D

                                                                                                                                  Bfrac14 13

                                                                                                                                  176frac14 074

                                                                                                                                  L

                                                                                                                                  Bfrac14 1

                                                                                                                                  Hence from Figure 515

                                                                                                                                  130 frac14 078 and 131 frac14 041

                                                                                                                                  70 Bearing capacity

                                                                                                                                  Thus using Equation 528

                                                                                                                                  si frac14 078 041 68 176

                                                                                                                                  65frac14 6mm

                                                                                                                                  Consolidation settlement

                                                                                                                                  Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                                  1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                                  434 (sod)

                                                                                                                                  Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                                  sc frac14 056 434 frac14 24mm

                                                                                                                                  The total settlement is (6thorn 24) frac14 30mm

                                                                                                                                  813

                                                                                                                                  At base level N frac14 26 Then using Equation 830

                                                                                                                                  qb frac14 40NDb

                                                                                                                                  Bfrac14 40 26 2

                                                                                                                                  025frac14 8320 kN=m2

                                                                                                                                  ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                                  Figure Q812

                                                                                                                                  Bearing capacity 71

                                                                                                                                  Over the length embedded in sand

                                                                                                                                  N frac14 21 ie18thorn 24

                                                                                                                                  2

                                                                                                                                  Using Equation 831

                                                                                                                                  qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                  For a single pile

                                                                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                                                                  frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                  For the pile group assuming a group efficiency of 12

                                                                                                                                  XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                  Then the load factor is

                                                                                                                                  F frac14 6523

                                                                                                                                  2000thorn 1000frac14 21

                                                                                                                                  (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                  Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                  150frac14 5547 kNm2

                                                                                                                                  Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                  150frac14 28 kNm2

                                                                                                                                  Characteristic base and shaft resistances for a single pile

                                                                                                                                  Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                  Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                  For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                  Design bearing resistance Rcd frac14 347

                                                                                                                                  130thorn 56

                                                                                                                                  130frac14 310 kN

                                                                                                                                  For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                  Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                  (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                  From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                  72 Bearing capacity

                                                                                                                                  N frac14 24thorn 26thorn 34

                                                                                                                                  3frac14 28

                                                                                                                                  Ic frac14 171

                                                                                                                                  2814frac14 0016 ethEquation 818THORN

                                                                                                                                  s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                  814

                                                                                                                                  Using Equation 841

                                                                                                                                  Tf frac14 DLcu thorn

                                                                                                                                  4ethD2 d2THORNcuNc

                                                                                                                                  frac14 eth 02 5 06 110THORN thorn

                                                                                                                                  4eth022 012THORN110 9

                                                                                                                                  frac14 207thorn 23 frac14 230 kN

                                                                                                                                  Figure Q813

                                                                                                                                  Bearing capacity 73

                                                                                                                                  Chapter 9

                                                                                                                                  Stability of slopes

                                                                                                                                  91

                                                                                                                                  Referring to Figure Q91

                                                                                                                                  W frac14 417 19 frac14 792 kN=m

                                                                                                                                  Q frac14 20 28 frac14 56 kN=m

                                                                                                                                  Arc lengthAB frac14

                                                                                                                                  180 73 90 frac14 115m

                                                                                                                                  Arc length BC frac14

                                                                                                                                  180 28 90 frac14 44m

                                                                                                                                  The factor of safety is given by

                                                                                                                                  F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                  90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                  Depth of tension crack z0 frac14 2cu

                                                                                                                                  frac14 2 20

                                                                                                                                  19frac14 21m

                                                                                                                                  Arc length BD frac14

                                                                                                                                  180 13

                                                                                                                                  1

                                                                                                                                  2 90 frac14 21m

                                                                                                                                  F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                  The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                  Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                  14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                  Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                  The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                  92

                                                                                                                                  u frac14 0

                                                                                                                                  Depth factor D frac14 11

                                                                                                                                  9frac14 122

                                                                                                                                  Using Equation 92 with F frac14 10

                                                                                                                                  Ns frac14 cu

                                                                                                                                  FHfrac14 30

                                                                                                                                  10 19 9frac14 0175

                                                                                                                                  Hence from Figure 93

                                                                                                                                  frac14 50

                                                                                                                                  For F frac14 12

                                                                                                                                  Ns frac14 30

                                                                                                                                  12 19 9frac14 0146

                                                                                                                                  frac14 27

                                                                                                                                  93

                                                                                                                                  Refer to Figure Q93

                                                                                                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                  1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                  74 m

                                                                                                                                  214 1deg

                                                                                                                                  213 1deg

                                                                                                                                  39 m

                                                                                                                                  WB

                                                                                                                                  D

                                                                                                                                  C

                                                                                                                                  28 m

                                                                                                                                  21 m

                                                                                                                                  A

                                                                                                                                  Q

                                                                                                                                  Soil (1)Soil (2)

                                                                                                                                  73deg

                                                                                                                                  Figure Q91

                                                                                                                                  Stability of slopes 75

                                                                                                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                  9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                  599 256 328 1372

                                                                                                                                  Figure Q93

                                                                                                                                  76 Stability of slopes

                                                                                                                                  XW cos frac14 b

                                                                                                                                  Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                  W sin frac14 bX

                                                                                                                                  h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                  Arc length La frac14

                                                                                                                                  180 57

                                                                                                                                  1

                                                                                                                                  2 326 frac14 327m

                                                                                                                                  The factor of safety is given by

                                                                                                                                  F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                  W sin

                                                                                                                                  frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                  frac14 091

                                                                                                                                  According to the limit state method

                                                                                                                                  0d frac14 tan1tan 32

                                                                                                                                  125

                                                                                                                                  frac14 265

                                                                                                                                  c0 frac14 8

                                                                                                                                  160frac14 5 kN=m2

                                                                                                                                  Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                  Design disturbing moment frac14 1075 kN=m

                                                                                                                                  The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                  94

                                                                                                                                  F frac14 1

                                                                                                                                  W sin

                                                                                                                                  Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                  1thorn ethtan tan0=FTHORN

                                                                                                                                  c0 frac14 8 kN=m2

                                                                                                                                  0 frac14 32

                                                                                                                                  c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                  W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                  Try F frac14 100

                                                                                                                                  tan0

                                                                                                                                  Ffrac14 0625

                                                                                                                                  Stability of slopes 77

                                                                                                                                  Values of u are as obtained in Figure Q93

                                                                                                                                  SliceNo

                                                                                                                                  h(m)

                                                                                                                                  W frac14 bh(kNm)

                                                                                                                                  W sin(kNm)

                                                                                                                                  ub(kNm)

                                                                                                                                  c0bthorn (W ub) tan0(kNm)

                                                                                                                                  sec

                                                                                                                                  1thorn (tan tan0)FProduct(kNm)

                                                                                                                                  1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                  23 33 30 1042 31

                                                                                                                                  3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                  224 92 72 0931 67

                                                                                                                                  6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                  12 58 112 90 0889 80

                                                                                                                                  8 60 252 1812

                                                                                                                                  80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                  10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                  2154 88 116 0853 99

                                                                                                                                  14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                  1074 1091

                                                                                                                                  F frac14 1091

                                                                                                                                  1074frac14 102 (assumed value 100)

                                                                                                                                  Thus

                                                                                                                                  F frac14 101

                                                                                                                                  95

                                                                                                                                  F frac14 1

                                                                                                                                  W sin

                                                                                                                                  XfWeth1 ruTHORN tan0g sec

                                                                                                                                  1thorn ethtan tan0THORN=F

                                                                                                                                  0 frac14 33

                                                                                                                                  ru frac14 020

                                                                                                                                  W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                  eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                  Try F frac14 110

                                                                                                                                  tan 0

                                                                                                                                  Ffrac14 tan 33

                                                                                                                                  110frac14 0590

                                                                                                                                  78 Stability of slopes

                                                                                                                                  Referring to Figure Q95

                                                                                                                                  SliceNo

                                                                                                                                  h(m)

                                                                                                                                  W frac14 bh(kNm)

                                                                                                                                  W sin(kNm)

                                                                                                                                  W(1 ru) tan0(kNm)

                                                                                                                                  sec

                                                                                                                                  1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                  1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                  2120 234 0892 209

                                                                                                                                  4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                  1185 1271

                                                                                                                                  Figure Q95

                                                                                                                                  Stability of slopes 79

                                                                                                                                  F frac14 1271

                                                                                                                                  1185frac14 107

                                                                                                                                  The trial value was 110 therefore take F to be 108

                                                                                                                                  96

                                                                                                                                  (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                  F frac14 0

                                                                                                                                  sat

                                                                                                                                  tan0

                                                                                                                                  tan

                                                                                                                                  ptie 15 frac14 92

                                                                                                                                  19

                                                                                                                                  tan 36

                                                                                                                                  tan

                                                                                                                                  tan frac14 0234

                                                                                                                                  frac14 13

                                                                                                                                  Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                  F frac14 tan0

                                                                                                                                  tan

                                                                                                                                  frac14 tan 36

                                                                                                                                  tan 13

                                                                                                                                  frac14 31

                                                                                                                                  (b) 0d frac14 tan1tan 36

                                                                                                                                  125

                                                                                                                                  frac14 30

                                                                                                                                  Depth of potential failure surface frac14 z

                                                                                                                                  Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                  frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                  frac14 504z kN

                                                                                                                                  Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                  frac14 19 z sin 13 cos 13

                                                                                                                                  frac14 416z kN

                                                                                                                                  Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                  80 Stability of slopes

                                                                                                                                  • Book Cover
                                                                                                                                  • Title
                                                                                                                                  • Contents
                                                                                                                                  • Basic characteristics of soils
                                                                                                                                  • Seepage
                                                                                                                                  • Effective stress
                                                                                                                                  • Shear strength
                                                                                                                                  • Stresses and displacements
                                                                                                                                  • Lateral earth pressure
                                                                                                                                  • Consolidation theory
                                                                                                                                  • Bearing capacity
                                                                                                                                  • Stability of slopes

                                                                                                                                    Chapter 8

                                                                                                                                    Bearing capacity

                                                                                                                                    81

                                                                                                                                    (a) The ultimate bearing capacity is given by Equation 83

                                                                                                                                    qf frac14 cNc thorn DNq thorn 1

                                                                                                                                    2BN

                                                                                                                                    For u frac14 0

                                                                                                                                    Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                                    qf frac14 eth105 514THORN thorn eth21 1 1THORN frac14 540thorn 21 kN=m2

                                                                                                                                    The net ultimate bearing capacity is

                                                                                                                                    qnf frac14 qf D frac14 540 kN=m2

                                                                                                                                    The net foundation pressure is

                                                                                                                                    qn frac14 q D frac14 425

                                                                                                                                    2 eth21 1THORN frac14 192 kN=m2

                                                                                                                                    The factor of safety (Equation 86) is

                                                                                                                                    F frac14 qnfqnfrac14 540

                                                                                                                                    192frac14 28

                                                                                                                                    (b) For 0 frac14 28

                                                                                                                                    Nc frac14 26 Nq frac14 15 N frac14 13 ethfrom Figure 84THORN0 frac14 21 98 frac14 112 kN=m3

                                                                                                                                    qf frac14 eth10 26THORN thorn eth112 1 15THORN thorn 1

                                                                                                                                    2 112 2 13

                                                                                                                                    frac14 260thorn 168thorn 146 frac14 574 kN=m2

                                                                                                                                    qnf frac14 574 112 frac14 563 kN=m2

                                                                                                                                    F frac14 563

                                                                                                                                    192frac14 29

                                                                                                                                    (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                                                    82

                                                                                                                                    For 0 frac14 38

                                                                                                                                    Nq frac14 49 N frac14 67

                                                                                                                                    qnf frac14 DethNq 1THORN thorn 1

                                                                                                                                    2BN ethfrom Equation 83THORN

                                                                                                                                    frac14 eth18 075 48THORN thorn 1

                                                                                                                                    2 18 15 67

                                                                                                                                    frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                                                    qn frac14 500

                                                                                                                                    15 eth18 075THORN frac14 320 kN=m2

                                                                                                                                    F frac14 qnfqnfrac14 1553

                                                                                                                                    320frac14 48

                                                                                                                                    0d frac14 tan1tan 38

                                                                                                                                    125

                                                                                                                                    frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                                                    Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                                                    2 18 15 25

                                                                                                                                    frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                                                    Design load (action) Vd frac14 500 kN=m

                                                                                                                                    The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                                                    83

                                                                                                                                    D

                                                                                                                                    Bfrac14 350

                                                                                                                                    225frac14 155

                                                                                                                                    From Figure 85 for a square foundation

                                                                                                                                    Nc frac14 81

                                                                                                                                    Bearing capacity 61

                                                                                                                                    For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                                                    Nc frac14 084thorn 016B

                                                                                                                                    L

                                                                                                                                    81 frac14 745

                                                                                                                                    Using Equation 810

                                                                                                                                    qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                                                    For F frac14 3

                                                                                                                                    qn frac14 1006

                                                                                                                                    3frac14 335 kN=m2

                                                                                                                                    q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                                                    Design load frac14 405 450 225 frac14 4100 kN

                                                                                                                                    Design undrained strength cud frac14 135

                                                                                                                                    14frac14 96 kN=m2

                                                                                                                                    Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                                                    frac14 7241 kN

                                                                                                                                    Design load Vd frac14 4100 kN

                                                                                                                                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                                    84

                                                                                                                                    For 0 frac14 40

                                                                                                                                    Nq frac14 64 N frac14 95

                                                                                                                                    qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                                                    (a) Water table 5m below ground level

                                                                                                                                    qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                                                    qn frac14 400 17 frac14 383 kN=m2

                                                                                                                                    F frac14 2686

                                                                                                                                    383frac14 70

                                                                                                                                    (b) Water table 1m below ground level (ie at foundation level)

                                                                                                                                    0 frac14 20 98 frac14 102 kN=m3

                                                                                                                                    62 Bearing capacity

                                                                                                                                    qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                                                    F frac14 2040

                                                                                                                                    383frac14 53

                                                                                                                                    (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                                                    eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                                                    qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                                                    F frac14 1296

                                                                                                                                    392frac14 33

                                                                                                                                    85

                                                                                                                                    The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                                                    Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                                                    Design value of 0 frac14 tan1tan 39

                                                                                                                                    125

                                                                                                                                    frac14 33

                                                                                                                                    For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                                                    Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                                                    Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                                    86

                                                                                                                                    (a) Undrained shear for u frac14 0

                                                                                                                                    Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                                    qnf frac14 12cuNc

                                                                                                                                    frac14 12 100 514 frac14 617 kN=m2

                                                                                                                                    qn frac14 qnfFfrac14 617

                                                                                                                                    3frac14 206 kN=m2

                                                                                                                                    q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                                                    Bearing capacity 63

                                                                                                                                    Drained shear for 0 frac14 32

                                                                                                                                    Nq frac14 23 N frac14 25

                                                                                                                                    0 frac14 21 98 frac14 112 kN=m3

                                                                                                                                    qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                                    frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                                    frac14 694 kN=m2

                                                                                                                                    q frac14 694

                                                                                                                                    3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                                    Design load frac14 42 227 frac14 3632 kN

                                                                                                                                    (b) Design undrained strength cud frac14 100

                                                                                                                                    14frac14 71 kNm2

                                                                                                                                    Design bearing resistance Rd frac14 12cudNe area

                                                                                                                                    frac14 12 71 514 42

                                                                                                                                    frac14 7007 kN

                                                                                                                                    For drained shear 0d frac14 tan1tan 32

                                                                                                                                    125

                                                                                                                                    frac14 26

                                                                                                                                    Nq frac14 12 N frac14 10

                                                                                                                                    Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                                    (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                                    Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                                    1 2 100 0175 0700qn 0182qn

                                                                                                                                    2 6 033 0044 0176qn 0046qn

                                                                                                                                    3 10 020 0017 0068qn 0018qn

                                                                                                                                    0246qn

                                                                                                                                    Diameter of equivalent circle B frac14 45m

                                                                                                                                    H

                                                                                                                                    Bfrac14 12

                                                                                                                                    45frac14 27 and A frac14 042

                                                                                                                                    13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                                    64 Bearing capacity

                                                                                                                                    For sc frac14 30mm

                                                                                                                                    qn frac14 30

                                                                                                                                    0147frac14 204 kN=m2

                                                                                                                                    q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                                    Design load frac14 42 225 frac14 3600 kN

                                                                                                                                    The design load is 3600 kN settlement being the limiting criterion

                                                                                                                                    87

                                                                                                                                    D

                                                                                                                                    Bfrac14 8

                                                                                                                                    4frac14 20

                                                                                                                                    From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                                    F frac14 cuNc

                                                                                                                                    Dfrac14 40 71

                                                                                                                                    20 8frac14 18

                                                                                                                                    88

                                                                                                                                    Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                                    Design value of 0 frac14 tan1tan 38

                                                                                                                                    125

                                                                                                                                    frac14 32

                                                                                                                                    Figure Q86

                                                                                                                                    Bearing capacity 65

                                                                                                                                    For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                                    Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                                    The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                                    Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                                    For B frac14 250m qn frac14 3750

                                                                                                                                    2502 17 frac14 583 kN=m2

                                                                                                                                    From Figure 510 m frac14 n frac14 126

                                                                                                                                    6frac14 021

                                                                                                                                    Ir frac14 0019

                                                                                                                                    Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                                    Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                                    The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                    89

                                                                                                                                    Depth (m) N 0v (kNm2) CN N1

                                                                                                                                    070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                                    Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                                    (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                                    Cw frac14 05thorn 0530

                                                                                                                                    47

                                                                                                                                    frac14 082

                                                                                                                                    66 Bearing capacity

                                                                                                                                    Thus

                                                                                                                                    qa frac14 150 082 frac14 120 kN=m2

                                                                                                                                    (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                                    Thus

                                                                                                                                    qa frac14 90 15 frac14 135 kN=m2

                                                                                                                                    (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                                    Ic frac14 171

                                                                                                                                    1014frac14 0068

                                                                                                                                    From Equation 819(a) with s frac14 25mm

                                                                                                                                    q frac14 25

                                                                                                                                    3507 0068frac14 150 kN=m2

                                                                                                                                    810

                                                                                                                                    Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                                    Izp frac14 05thorn 01130

                                                                                                                                    16 27

                                                                                                                                    05

                                                                                                                                    frac14 067

                                                                                                                                    Refer to Figure Q810

                                                                                                                                    E frac14 25qc

                                                                                                                                    Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                                    Ez (mm3MN)

                                                                                                                                    1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                                    0203

                                                                                                                                    C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                                    130frac14 093

                                                                                                                                    C2 frac14 1 ethsayTHORN

                                                                                                                                    s frac14 C1C2qnX Iz

                                                                                                                                    Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                                    Bearing capacity 67

                                                                                                                                    811

                                                                                                                                    At pile base level

                                                                                                                                    cu frac14 220 kN=m2

                                                                                                                                    qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                                    Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                                    00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                                    qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                                    Then

                                                                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                                                                    frac14

                                                                                                                                    4 32 1980

                                                                                                                                    thorn eth 105 139 86THORN

                                                                                                                                    frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                                    0 01 02 03 04 05 06 07

                                                                                                                                    0 2 4 6 8 10 12 14

                                                                                                                                    1

                                                                                                                                    2

                                                                                                                                    3

                                                                                                                                    4

                                                                                                                                    5

                                                                                                                                    6

                                                                                                                                    7

                                                                                                                                    8

                                                                                                                                    (1)

                                                                                                                                    (2)

                                                                                                                                    (3)

                                                                                                                                    (4)

                                                                                                                                    (5)

                                                                                                                                    qc

                                                                                                                                    qc

                                                                                                                                    Iz

                                                                                                                                    Iz

                                                                                                                                    (MNm2)

                                                                                                                                    z (m)

                                                                                                                                    Figure Q810

                                                                                                                                    68 Bearing capacity

                                                                                                                                    Allowable load

                                                                                                                                    ethaTHORN Qf

                                                                                                                                    2frac14 17 937

                                                                                                                                    2frac14 8968 kN

                                                                                                                                    ethbTHORN Abqb

                                                                                                                                    3thorn Asqs frac14 13 996

                                                                                                                                    3thorn 3941 frac14 8606 kN

                                                                                                                                    ie allowable load frac14 8600 kN

                                                                                                                                    Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                                    According to the limit state method

                                                                                                                                    Characteristic undrained strength at base level cuk frac14 220

                                                                                                                                    150kN=m2

                                                                                                                                    Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                                    150frac14 1320 kN=m2

                                                                                                                                    Characteristic shaft resistance qsk frac14 00150

                                                                                                                                    frac14 86

                                                                                                                                    150frac14 57 kN=m2

                                                                                                                                    Characteristic base and shaft resistances

                                                                                                                                    Rbk frac14

                                                                                                                                    4 32 1320 frac14 9330 kN

                                                                                                                                    Rsk frac14 105 139 86

                                                                                                                                    150frac14 2629 kN

                                                                                                                                    For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                                    Design bearing resistance Rcd frac14 9330

                                                                                                                                    160thorn 2629

                                                                                                                                    130

                                                                                                                                    frac14 5831thorn 2022

                                                                                                                                    frac14 7850 kN

                                                                                                                                    Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                                    812

                                                                                                                                    ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                                    qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                                    For a single pile

                                                                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                                                                    frac14

                                                                                                                                    4 062 1305

                                                                                                                                    thorn eth 06 15 42THORN

                                                                                                                                    frac14 369thorn 1187 frac14 1556 kN

                                                                                                                                    Bearing capacity 69

                                                                                                                                    Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                                    eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                                    (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                                    (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                                    qbkfrac14 9cuk frac14 9 220

                                                                                                                                    150frac14 1320 kN=m2

                                                                                                                                    qskfrac14cuk frac14 040 105

                                                                                                                                    150frac14 28 kN=m2

                                                                                                                                    Rbkfrac14

                                                                                                                                    4 0602 1320 frac14 373 kN

                                                                                                                                    Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                                    Rcdfrac14 373

                                                                                                                                    160thorn 791

                                                                                                                                    130frac14 233thorn 608 frac14 841 kN

                                                                                                                                    Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                                    Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                                    (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                                    q frac14 21 000

                                                                                                                                    1762frac14 68 kN=m2

                                                                                                                                    Immediate settlement

                                                                                                                                    H

                                                                                                                                    Bfrac14 15

                                                                                                                                    176frac14 085

                                                                                                                                    D

                                                                                                                                    Bfrac14 13

                                                                                                                                    176frac14 074

                                                                                                                                    L

                                                                                                                                    Bfrac14 1

                                                                                                                                    Hence from Figure 515

                                                                                                                                    130 frac14 078 and 131 frac14 041

                                                                                                                                    70 Bearing capacity

                                                                                                                                    Thus using Equation 528

                                                                                                                                    si frac14 078 041 68 176

                                                                                                                                    65frac14 6mm

                                                                                                                                    Consolidation settlement

                                                                                                                                    Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                                    1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                                    434 (sod)

                                                                                                                                    Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                                    sc frac14 056 434 frac14 24mm

                                                                                                                                    The total settlement is (6thorn 24) frac14 30mm

                                                                                                                                    813

                                                                                                                                    At base level N frac14 26 Then using Equation 830

                                                                                                                                    qb frac14 40NDb

                                                                                                                                    Bfrac14 40 26 2

                                                                                                                                    025frac14 8320 kN=m2

                                                                                                                                    ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                                    Figure Q812

                                                                                                                                    Bearing capacity 71

                                                                                                                                    Over the length embedded in sand

                                                                                                                                    N frac14 21 ie18thorn 24

                                                                                                                                    2

                                                                                                                                    Using Equation 831

                                                                                                                                    qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                    For a single pile

                                                                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                                                                    frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                    For the pile group assuming a group efficiency of 12

                                                                                                                                    XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                    Then the load factor is

                                                                                                                                    F frac14 6523

                                                                                                                                    2000thorn 1000frac14 21

                                                                                                                                    (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                    Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                    150frac14 5547 kNm2

                                                                                                                                    Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                    150frac14 28 kNm2

                                                                                                                                    Characteristic base and shaft resistances for a single pile

                                                                                                                                    Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                    Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                    For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                    Design bearing resistance Rcd frac14 347

                                                                                                                                    130thorn 56

                                                                                                                                    130frac14 310 kN

                                                                                                                                    For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                    Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                    (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                    From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                    72 Bearing capacity

                                                                                                                                    N frac14 24thorn 26thorn 34

                                                                                                                                    3frac14 28

                                                                                                                                    Ic frac14 171

                                                                                                                                    2814frac14 0016 ethEquation 818THORN

                                                                                                                                    s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                    814

                                                                                                                                    Using Equation 841

                                                                                                                                    Tf frac14 DLcu thorn

                                                                                                                                    4ethD2 d2THORNcuNc

                                                                                                                                    frac14 eth 02 5 06 110THORN thorn

                                                                                                                                    4eth022 012THORN110 9

                                                                                                                                    frac14 207thorn 23 frac14 230 kN

                                                                                                                                    Figure Q813

                                                                                                                                    Bearing capacity 73

                                                                                                                                    Chapter 9

                                                                                                                                    Stability of slopes

                                                                                                                                    91

                                                                                                                                    Referring to Figure Q91

                                                                                                                                    W frac14 417 19 frac14 792 kN=m

                                                                                                                                    Q frac14 20 28 frac14 56 kN=m

                                                                                                                                    Arc lengthAB frac14

                                                                                                                                    180 73 90 frac14 115m

                                                                                                                                    Arc length BC frac14

                                                                                                                                    180 28 90 frac14 44m

                                                                                                                                    The factor of safety is given by

                                                                                                                                    F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                    90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                    Depth of tension crack z0 frac14 2cu

                                                                                                                                    frac14 2 20

                                                                                                                                    19frac14 21m

                                                                                                                                    Arc length BD frac14

                                                                                                                                    180 13

                                                                                                                                    1

                                                                                                                                    2 90 frac14 21m

                                                                                                                                    F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                    The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                    Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                    14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                    Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                    The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                    92

                                                                                                                                    u frac14 0

                                                                                                                                    Depth factor D frac14 11

                                                                                                                                    9frac14 122

                                                                                                                                    Using Equation 92 with F frac14 10

                                                                                                                                    Ns frac14 cu

                                                                                                                                    FHfrac14 30

                                                                                                                                    10 19 9frac14 0175

                                                                                                                                    Hence from Figure 93

                                                                                                                                    frac14 50

                                                                                                                                    For F frac14 12

                                                                                                                                    Ns frac14 30

                                                                                                                                    12 19 9frac14 0146

                                                                                                                                    frac14 27

                                                                                                                                    93

                                                                                                                                    Refer to Figure Q93

                                                                                                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                    1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                    74 m

                                                                                                                                    214 1deg

                                                                                                                                    213 1deg

                                                                                                                                    39 m

                                                                                                                                    WB

                                                                                                                                    D

                                                                                                                                    C

                                                                                                                                    28 m

                                                                                                                                    21 m

                                                                                                                                    A

                                                                                                                                    Q

                                                                                                                                    Soil (1)Soil (2)

                                                                                                                                    73deg

                                                                                                                                    Figure Q91

                                                                                                                                    Stability of slopes 75

                                                                                                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                    9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                    599 256 328 1372

                                                                                                                                    Figure Q93

                                                                                                                                    76 Stability of slopes

                                                                                                                                    XW cos frac14 b

                                                                                                                                    Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                    W sin frac14 bX

                                                                                                                                    h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                    Arc length La frac14

                                                                                                                                    180 57

                                                                                                                                    1

                                                                                                                                    2 326 frac14 327m

                                                                                                                                    The factor of safety is given by

                                                                                                                                    F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                    W sin

                                                                                                                                    frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                    frac14 091

                                                                                                                                    According to the limit state method

                                                                                                                                    0d frac14 tan1tan 32

                                                                                                                                    125

                                                                                                                                    frac14 265

                                                                                                                                    c0 frac14 8

                                                                                                                                    160frac14 5 kN=m2

                                                                                                                                    Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                    Design disturbing moment frac14 1075 kN=m

                                                                                                                                    The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                    94

                                                                                                                                    F frac14 1

                                                                                                                                    W sin

                                                                                                                                    Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                    1thorn ethtan tan0=FTHORN

                                                                                                                                    c0 frac14 8 kN=m2

                                                                                                                                    0 frac14 32

                                                                                                                                    c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                    W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                    Try F frac14 100

                                                                                                                                    tan0

                                                                                                                                    Ffrac14 0625

                                                                                                                                    Stability of slopes 77

                                                                                                                                    Values of u are as obtained in Figure Q93

                                                                                                                                    SliceNo

                                                                                                                                    h(m)

                                                                                                                                    W frac14 bh(kNm)

                                                                                                                                    W sin(kNm)

                                                                                                                                    ub(kNm)

                                                                                                                                    c0bthorn (W ub) tan0(kNm)

                                                                                                                                    sec

                                                                                                                                    1thorn (tan tan0)FProduct(kNm)

                                                                                                                                    1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                    23 33 30 1042 31

                                                                                                                                    3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                    224 92 72 0931 67

                                                                                                                                    6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                    12 58 112 90 0889 80

                                                                                                                                    8 60 252 1812

                                                                                                                                    80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                    10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                    2154 88 116 0853 99

                                                                                                                                    14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                    1074 1091

                                                                                                                                    F frac14 1091

                                                                                                                                    1074frac14 102 (assumed value 100)

                                                                                                                                    Thus

                                                                                                                                    F frac14 101

                                                                                                                                    95

                                                                                                                                    F frac14 1

                                                                                                                                    W sin

                                                                                                                                    XfWeth1 ruTHORN tan0g sec

                                                                                                                                    1thorn ethtan tan0THORN=F

                                                                                                                                    0 frac14 33

                                                                                                                                    ru frac14 020

                                                                                                                                    W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                    eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                    Try F frac14 110

                                                                                                                                    tan 0

                                                                                                                                    Ffrac14 tan 33

                                                                                                                                    110frac14 0590

                                                                                                                                    78 Stability of slopes

                                                                                                                                    Referring to Figure Q95

                                                                                                                                    SliceNo

                                                                                                                                    h(m)

                                                                                                                                    W frac14 bh(kNm)

                                                                                                                                    W sin(kNm)

                                                                                                                                    W(1 ru) tan0(kNm)

                                                                                                                                    sec

                                                                                                                                    1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                    1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                    2120 234 0892 209

                                                                                                                                    4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                    1185 1271

                                                                                                                                    Figure Q95

                                                                                                                                    Stability of slopes 79

                                                                                                                                    F frac14 1271

                                                                                                                                    1185frac14 107

                                                                                                                                    The trial value was 110 therefore take F to be 108

                                                                                                                                    96

                                                                                                                                    (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                    F frac14 0

                                                                                                                                    sat

                                                                                                                                    tan0

                                                                                                                                    tan

                                                                                                                                    ptie 15 frac14 92

                                                                                                                                    19

                                                                                                                                    tan 36

                                                                                                                                    tan

                                                                                                                                    tan frac14 0234

                                                                                                                                    frac14 13

                                                                                                                                    Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                    F frac14 tan0

                                                                                                                                    tan

                                                                                                                                    frac14 tan 36

                                                                                                                                    tan 13

                                                                                                                                    frac14 31

                                                                                                                                    (b) 0d frac14 tan1tan 36

                                                                                                                                    125

                                                                                                                                    frac14 30

                                                                                                                                    Depth of potential failure surface frac14 z

                                                                                                                                    Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                    frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                    frac14 504z kN

                                                                                                                                    Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                    frac14 19 z sin 13 cos 13

                                                                                                                                    frac14 416z kN

                                                                                                                                    Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                    80 Stability of slopes

                                                                                                                                    • Book Cover
                                                                                                                                    • Title
                                                                                                                                    • Contents
                                                                                                                                    • Basic characteristics of soils
                                                                                                                                    • Seepage
                                                                                                                                    • Effective stress
                                                                                                                                    • Shear strength
                                                                                                                                    • Stresses and displacements
                                                                                                                                    • Lateral earth pressure
                                                                                                                                    • Consolidation theory
                                                                                                                                    • Bearing capacity
                                                                                                                                    • Stability of slopes

                                                                                                                                      qnf frac14 574 112 frac14 563 kN=m2

                                                                                                                                      F frac14 563

                                                                                                                                      192frac14 29

                                                                                                                                      (qn frac14 192 kNm2 assumes that backfilled soil on the footing slab is included in the loadof 425 kNm)

                                                                                                                                      82

                                                                                                                                      For 0 frac14 38

                                                                                                                                      Nq frac14 49 N frac14 67

                                                                                                                                      qnf frac14 DethNq 1THORN thorn 1

                                                                                                                                      2BN ethfrom Equation 83THORN

                                                                                                                                      frac14 eth18 075 48THORN thorn 1

                                                                                                                                      2 18 15 67

                                                                                                                                      frac14 648thorn 905 frac14 1553 kN=m2

                                                                                                                                      qn frac14 500

                                                                                                                                      15 eth18 075THORN frac14 320 kN=m2

                                                                                                                                      F frac14 qnfqnfrac14 1553

                                                                                                                                      320frac14 48

                                                                                                                                      0d frac14 tan1tan 38

                                                                                                                                      125

                                                                                                                                      frac14 32 therefore Nq frac14 23 and N frac14 25

                                                                                                                                      Design bearing resistance Rd frac14 15 eth18 075 23THORN thorn 1

                                                                                                                                      2 18 15 25

                                                                                                                                      frac14 15eth310thorn 337THORNfrac14 970 kN=m

                                                                                                                                      Design load (action) Vd frac14 500 kN=m

                                                                                                                                      The design bearing resistance is greater than the design load therefore the bearingresistance limit state is satisfied

                                                                                                                                      83

                                                                                                                                      D

                                                                                                                                      Bfrac14 350

                                                                                                                                      225frac14 155

                                                                                                                                      From Figure 85 for a square foundation

                                                                                                                                      Nc frac14 81

                                                                                                                                      Bearing capacity 61

                                                                                                                                      For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                                                      Nc frac14 084thorn 016B

                                                                                                                                      L

                                                                                                                                      81 frac14 745

                                                                                                                                      Using Equation 810

                                                                                                                                      qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                                                      For F frac14 3

                                                                                                                                      qn frac14 1006

                                                                                                                                      3frac14 335 kN=m2

                                                                                                                                      q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                                                      Design load frac14 405 450 225 frac14 4100 kN

                                                                                                                                      Design undrained strength cud frac14 135

                                                                                                                                      14frac14 96 kN=m2

                                                                                                                                      Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                                                      frac14 7241 kN

                                                                                                                                      Design load Vd frac14 4100 kN

                                                                                                                                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                                      84

                                                                                                                                      For 0 frac14 40

                                                                                                                                      Nq frac14 64 N frac14 95

                                                                                                                                      qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                                                      (a) Water table 5m below ground level

                                                                                                                                      qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                                                      qn frac14 400 17 frac14 383 kN=m2

                                                                                                                                      F frac14 2686

                                                                                                                                      383frac14 70

                                                                                                                                      (b) Water table 1m below ground level (ie at foundation level)

                                                                                                                                      0 frac14 20 98 frac14 102 kN=m3

                                                                                                                                      62 Bearing capacity

                                                                                                                                      qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                                                      F frac14 2040

                                                                                                                                      383frac14 53

                                                                                                                                      (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                                                      eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                                                      qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                                                      F frac14 1296

                                                                                                                                      392frac14 33

                                                                                                                                      85

                                                                                                                                      The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                                                      Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                                                      Design value of 0 frac14 tan1tan 39

                                                                                                                                      125

                                                                                                                                      frac14 33

                                                                                                                                      For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                                                      Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                                                      Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                                      86

                                                                                                                                      (a) Undrained shear for u frac14 0

                                                                                                                                      Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                                      qnf frac14 12cuNc

                                                                                                                                      frac14 12 100 514 frac14 617 kN=m2

                                                                                                                                      qn frac14 qnfFfrac14 617

                                                                                                                                      3frac14 206 kN=m2

                                                                                                                                      q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                                                      Bearing capacity 63

                                                                                                                                      Drained shear for 0 frac14 32

                                                                                                                                      Nq frac14 23 N frac14 25

                                                                                                                                      0 frac14 21 98 frac14 112 kN=m3

                                                                                                                                      qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                                      frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                                      frac14 694 kN=m2

                                                                                                                                      q frac14 694

                                                                                                                                      3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                                      Design load frac14 42 227 frac14 3632 kN

                                                                                                                                      (b) Design undrained strength cud frac14 100

                                                                                                                                      14frac14 71 kNm2

                                                                                                                                      Design bearing resistance Rd frac14 12cudNe area

                                                                                                                                      frac14 12 71 514 42

                                                                                                                                      frac14 7007 kN

                                                                                                                                      For drained shear 0d frac14 tan1tan 32

                                                                                                                                      125

                                                                                                                                      frac14 26

                                                                                                                                      Nq frac14 12 N frac14 10

                                                                                                                                      Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                                      (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                                      Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                                      1 2 100 0175 0700qn 0182qn

                                                                                                                                      2 6 033 0044 0176qn 0046qn

                                                                                                                                      3 10 020 0017 0068qn 0018qn

                                                                                                                                      0246qn

                                                                                                                                      Diameter of equivalent circle B frac14 45m

                                                                                                                                      H

                                                                                                                                      Bfrac14 12

                                                                                                                                      45frac14 27 and A frac14 042

                                                                                                                                      13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                                      64 Bearing capacity

                                                                                                                                      For sc frac14 30mm

                                                                                                                                      qn frac14 30

                                                                                                                                      0147frac14 204 kN=m2

                                                                                                                                      q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                                      Design load frac14 42 225 frac14 3600 kN

                                                                                                                                      The design load is 3600 kN settlement being the limiting criterion

                                                                                                                                      87

                                                                                                                                      D

                                                                                                                                      Bfrac14 8

                                                                                                                                      4frac14 20

                                                                                                                                      From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                                      F frac14 cuNc

                                                                                                                                      Dfrac14 40 71

                                                                                                                                      20 8frac14 18

                                                                                                                                      88

                                                                                                                                      Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                                      Design value of 0 frac14 tan1tan 38

                                                                                                                                      125

                                                                                                                                      frac14 32

                                                                                                                                      Figure Q86

                                                                                                                                      Bearing capacity 65

                                                                                                                                      For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                                      Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                                      The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                                      Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                                      For B frac14 250m qn frac14 3750

                                                                                                                                      2502 17 frac14 583 kN=m2

                                                                                                                                      From Figure 510 m frac14 n frac14 126

                                                                                                                                      6frac14 021

                                                                                                                                      Ir frac14 0019

                                                                                                                                      Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                                      Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                                      The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                      89

                                                                                                                                      Depth (m) N 0v (kNm2) CN N1

                                                                                                                                      070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                                      Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                                      (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                                      Cw frac14 05thorn 0530

                                                                                                                                      47

                                                                                                                                      frac14 082

                                                                                                                                      66 Bearing capacity

                                                                                                                                      Thus

                                                                                                                                      qa frac14 150 082 frac14 120 kN=m2

                                                                                                                                      (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                                      Thus

                                                                                                                                      qa frac14 90 15 frac14 135 kN=m2

                                                                                                                                      (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                                      Ic frac14 171

                                                                                                                                      1014frac14 0068

                                                                                                                                      From Equation 819(a) with s frac14 25mm

                                                                                                                                      q frac14 25

                                                                                                                                      3507 0068frac14 150 kN=m2

                                                                                                                                      810

                                                                                                                                      Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                                      Izp frac14 05thorn 01130

                                                                                                                                      16 27

                                                                                                                                      05

                                                                                                                                      frac14 067

                                                                                                                                      Refer to Figure Q810

                                                                                                                                      E frac14 25qc

                                                                                                                                      Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                                      Ez (mm3MN)

                                                                                                                                      1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                                      0203

                                                                                                                                      C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                                      130frac14 093

                                                                                                                                      C2 frac14 1 ethsayTHORN

                                                                                                                                      s frac14 C1C2qnX Iz

                                                                                                                                      Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                                      Bearing capacity 67

                                                                                                                                      811

                                                                                                                                      At pile base level

                                                                                                                                      cu frac14 220 kN=m2

                                                                                                                                      qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                                      Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                                      00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                                      qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                                      Then

                                                                                                                                      Qf frac14 Abqb thorn Asqs

                                                                                                                                      frac14

                                                                                                                                      4 32 1980

                                                                                                                                      thorn eth 105 139 86THORN

                                                                                                                                      frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                                      0 01 02 03 04 05 06 07

                                                                                                                                      0 2 4 6 8 10 12 14

                                                                                                                                      1

                                                                                                                                      2

                                                                                                                                      3

                                                                                                                                      4

                                                                                                                                      5

                                                                                                                                      6

                                                                                                                                      7

                                                                                                                                      8

                                                                                                                                      (1)

                                                                                                                                      (2)

                                                                                                                                      (3)

                                                                                                                                      (4)

                                                                                                                                      (5)

                                                                                                                                      qc

                                                                                                                                      qc

                                                                                                                                      Iz

                                                                                                                                      Iz

                                                                                                                                      (MNm2)

                                                                                                                                      z (m)

                                                                                                                                      Figure Q810

                                                                                                                                      68 Bearing capacity

                                                                                                                                      Allowable load

                                                                                                                                      ethaTHORN Qf

                                                                                                                                      2frac14 17 937

                                                                                                                                      2frac14 8968 kN

                                                                                                                                      ethbTHORN Abqb

                                                                                                                                      3thorn Asqs frac14 13 996

                                                                                                                                      3thorn 3941 frac14 8606 kN

                                                                                                                                      ie allowable load frac14 8600 kN

                                                                                                                                      Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                                      According to the limit state method

                                                                                                                                      Characteristic undrained strength at base level cuk frac14 220

                                                                                                                                      150kN=m2

                                                                                                                                      Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                                      150frac14 1320 kN=m2

                                                                                                                                      Characteristic shaft resistance qsk frac14 00150

                                                                                                                                      frac14 86

                                                                                                                                      150frac14 57 kN=m2

                                                                                                                                      Characteristic base and shaft resistances

                                                                                                                                      Rbk frac14

                                                                                                                                      4 32 1320 frac14 9330 kN

                                                                                                                                      Rsk frac14 105 139 86

                                                                                                                                      150frac14 2629 kN

                                                                                                                                      For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                                      Design bearing resistance Rcd frac14 9330

                                                                                                                                      160thorn 2629

                                                                                                                                      130

                                                                                                                                      frac14 5831thorn 2022

                                                                                                                                      frac14 7850 kN

                                                                                                                                      Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                                      812

                                                                                                                                      ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                                      qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                                      For a single pile

                                                                                                                                      Qf frac14 Abqb thorn Asqs

                                                                                                                                      frac14

                                                                                                                                      4 062 1305

                                                                                                                                      thorn eth 06 15 42THORN

                                                                                                                                      frac14 369thorn 1187 frac14 1556 kN

                                                                                                                                      Bearing capacity 69

                                                                                                                                      Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                                      eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                                      (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                                      (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                                      qbkfrac14 9cuk frac14 9 220

                                                                                                                                      150frac14 1320 kN=m2

                                                                                                                                      qskfrac14cuk frac14 040 105

                                                                                                                                      150frac14 28 kN=m2

                                                                                                                                      Rbkfrac14

                                                                                                                                      4 0602 1320 frac14 373 kN

                                                                                                                                      Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                                      Rcdfrac14 373

                                                                                                                                      160thorn 791

                                                                                                                                      130frac14 233thorn 608 frac14 841 kN

                                                                                                                                      Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                                      Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                                      (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                                      q frac14 21 000

                                                                                                                                      1762frac14 68 kN=m2

                                                                                                                                      Immediate settlement

                                                                                                                                      H

                                                                                                                                      Bfrac14 15

                                                                                                                                      176frac14 085

                                                                                                                                      D

                                                                                                                                      Bfrac14 13

                                                                                                                                      176frac14 074

                                                                                                                                      L

                                                                                                                                      Bfrac14 1

                                                                                                                                      Hence from Figure 515

                                                                                                                                      130 frac14 078 and 131 frac14 041

                                                                                                                                      70 Bearing capacity

                                                                                                                                      Thus using Equation 528

                                                                                                                                      si frac14 078 041 68 176

                                                                                                                                      65frac14 6mm

                                                                                                                                      Consolidation settlement

                                                                                                                                      Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                                      1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                                      434 (sod)

                                                                                                                                      Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                                      sc frac14 056 434 frac14 24mm

                                                                                                                                      The total settlement is (6thorn 24) frac14 30mm

                                                                                                                                      813

                                                                                                                                      At base level N frac14 26 Then using Equation 830

                                                                                                                                      qb frac14 40NDb

                                                                                                                                      Bfrac14 40 26 2

                                                                                                                                      025frac14 8320 kN=m2

                                                                                                                                      ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                                      Figure Q812

                                                                                                                                      Bearing capacity 71

                                                                                                                                      Over the length embedded in sand

                                                                                                                                      N frac14 21 ie18thorn 24

                                                                                                                                      2

                                                                                                                                      Using Equation 831

                                                                                                                                      qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                      For a single pile

                                                                                                                                      Qf frac14 Abqb thorn Asqs

                                                                                                                                      frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                      For the pile group assuming a group efficiency of 12

                                                                                                                                      XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                      Then the load factor is

                                                                                                                                      F frac14 6523

                                                                                                                                      2000thorn 1000frac14 21

                                                                                                                                      (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                      Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                      150frac14 5547 kNm2

                                                                                                                                      Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                      150frac14 28 kNm2

                                                                                                                                      Characteristic base and shaft resistances for a single pile

                                                                                                                                      Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                      Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                      For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                      Design bearing resistance Rcd frac14 347

                                                                                                                                      130thorn 56

                                                                                                                                      130frac14 310 kN

                                                                                                                                      For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                      Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                      (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                      From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                      72 Bearing capacity

                                                                                                                                      N frac14 24thorn 26thorn 34

                                                                                                                                      3frac14 28

                                                                                                                                      Ic frac14 171

                                                                                                                                      2814frac14 0016 ethEquation 818THORN

                                                                                                                                      s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                      814

                                                                                                                                      Using Equation 841

                                                                                                                                      Tf frac14 DLcu thorn

                                                                                                                                      4ethD2 d2THORNcuNc

                                                                                                                                      frac14 eth 02 5 06 110THORN thorn

                                                                                                                                      4eth022 012THORN110 9

                                                                                                                                      frac14 207thorn 23 frac14 230 kN

                                                                                                                                      Figure Q813

                                                                                                                                      Bearing capacity 73

                                                                                                                                      Chapter 9

                                                                                                                                      Stability of slopes

                                                                                                                                      91

                                                                                                                                      Referring to Figure Q91

                                                                                                                                      W frac14 417 19 frac14 792 kN=m

                                                                                                                                      Q frac14 20 28 frac14 56 kN=m

                                                                                                                                      Arc lengthAB frac14

                                                                                                                                      180 73 90 frac14 115m

                                                                                                                                      Arc length BC frac14

                                                                                                                                      180 28 90 frac14 44m

                                                                                                                                      The factor of safety is given by

                                                                                                                                      F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                      90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                      Depth of tension crack z0 frac14 2cu

                                                                                                                                      frac14 2 20

                                                                                                                                      19frac14 21m

                                                                                                                                      Arc length BD frac14

                                                                                                                                      180 13

                                                                                                                                      1

                                                                                                                                      2 90 frac14 21m

                                                                                                                                      F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                      The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                      Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                      14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                      Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                      The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                      92

                                                                                                                                      u frac14 0

                                                                                                                                      Depth factor D frac14 11

                                                                                                                                      9frac14 122

                                                                                                                                      Using Equation 92 with F frac14 10

                                                                                                                                      Ns frac14 cu

                                                                                                                                      FHfrac14 30

                                                                                                                                      10 19 9frac14 0175

                                                                                                                                      Hence from Figure 93

                                                                                                                                      frac14 50

                                                                                                                                      For F frac14 12

                                                                                                                                      Ns frac14 30

                                                                                                                                      12 19 9frac14 0146

                                                                                                                                      frac14 27

                                                                                                                                      93

                                                                                                                                      Refer to Figure Q93

                                                                                                                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                      1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                      74 m

                                                                                                                                      214 1deg

                                                                                                                                      213 1deg

                                                                                                                                      39 m

                                                                                                                                      WB

                                                                                                                                      D

                                                                                                                                      C

                                                                                                                                      28 m

                                                                                                                                      21 m

                                                                                                                                      A

                                                                                                                                      Q

                                                                                                                                      Soil (1)Soil (2)

                                                                                                                                      73deg

                                                                                                                                      Figure Q91

                                                                                                                                      Stability of slopes 75

                                                                                                                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                      9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                      599 256 328 1372

                                                                                                                                      Figure Q93

                                                                                                                                      76 Stability of slopes

                                                                                                                                      XW cos frac14 b

                                                                                                                                      Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                      W sin frac14 bX

                                                                                                                                      h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                      Arc length La frac14

                                                                                                                                      180 57

                                                                                                                                      1

                                                                                                                                      2 326 frac14 327m

                                                                                                                                      The factor of safety is given by

                                                                                                                                      F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                      W sin

                                                                                                                                      frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                      frac14 091

                                                                                                                                      According to the limit state method

                                                                                                                                      0d frac14 tan1tan 32

                                                                                                                                      125

                                                                                                                                      frac14 265

                                                                                                                                      c0 frac14 8

                                                                                                                                      160frac14 5 kN=m2

                                                                                                                                      Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                      Design disturbing moment frac14 1075 kN=m

                                                                                                                                      The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                      94

                                                                                                                                      F frac14 1

                                                                                                                                      W sin

                                                                                                                                      Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                      1thorn ethtan tan0=FTHORN

                                                                                                                                      c0 frac14 8 kN=m2

                                                                                                                                      0 frac14 32

                                                                                                                                      c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                      W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                      Try F frac14 100

                                                                                                                                      tan0

                                                                                                                                      Ffrac14 0625

                                                                                                                                      Stability of slopes 77

                                                                                                                                      Values of u are as obtained in Figure Q93

                                                                                                                                      SliceNo

                                                                                                                                      h(m)

                                                                                                                                      W frac14 bh(kNm)

                                                                                                                                      W sin(kNm)

                                                                                                                                      ub(kNm)

                                                                                                                                      c0bthorn (W ub) tan0(kNm)

                                                                                                                                      sec

                                                                                                                                      1thorn (tan tan0)FProduct(kNm)

                                                                                                                                      1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                      23 33 30 1042 31

                                                                                                                                      3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                      224 92 72 0931 67

                                                                                                                                      6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                      12 58 112 90 0889 80

                                                                                                                                      8 60 252 1812

                                                                                                                                      80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                      10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                      2154 88 116 0853 99

                                                                                                                                      14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                      1074 1091

                                                                                                                                      F frac14 1091

                                                                                                                                      1074frac14 102 (assumed value 100)

                                                                                                                                      Thus

                                                                                                                                      F frac14 101

                                                                                                                                      95

                                                                                                                                      F frac14 1

                                                                                                                                      W sin

                                                                                                                                      XfWeth1 ruTHORN tan0g sec

                                                                                                                                      1thorn ethtan tan0THORN=F

                                                                                                                                      0 frac14 33

                                                                                                                                      ru frac14 020

                                                                                                                                      W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                      eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                      Try F frac14 110

                                                                                                                                      tan 0

                                                                                                                                      Ffrac14 tan 33

                                                                                                                                      110frac14 0590

                                                                                                                                      78 Stability of slopes

                                                                                                                                      Referring to Figure Q95

                                                                                                                                      SliceNo

                                                                                                                                      h(m)

                                                                                                                                      W frac14 bh(kNm)

                                                                                                                                      W sin(kNm)

                                                                                                                                      W(1 ru) tan0(kNm)

                                                                                                                                      sec

                                                                                                                                      1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                      1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                      2120 234 0892 209

                                                                                                                                      4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                      1185 1271

                                                                                                                                      Figure Q95

                                                                                                                                      Stability of slopes 79

                                                                                                                                      F frac14 1271

                                                                                                                                      1185frac14 107

                                                                                                                                      The trial value was 110 therefore take F to be 108

                                                                                                                                      96

                                                                                                                                      (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                      F frac14 0

                                                                                                                                      sat

                                                                                                                                      tan0

                                                                                                                                      tan

                                                                                                                                      ptie 15 frac14 92

                                                                                                                                      19

                                                                                                                                      tan 36

                                                                                                                                      tan

                                                                                                                                      tan frac14 0234

                                                                                                                                      frac14 13

                                                                                                                                      Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                      F frac14 tan0

                                                                                                                                      tan

                                                                                                                                      frac14 tan 36

                                                                                                                                      tan 13

                                                                                                                                      frac14 31

                                                                                                                                      (b) 0d frac14 tan1tan 36

                                                                                                                                      125

                                                                                                                                      frac14 30

                                                                                                                                      Depth of potential failure surface frac14 z

                                                                                                                                      Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                      frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                      frac14 504z kN

                                                                                                                                      Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                      frac14 19 z sin 13 cos 13

                                                                                                                                      frac14 416z kN

                                                                                                                                      Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                      80 Stability of slopes

                                                                                                                                      • Book Cover
                                                                                                                                      • Title
                                                                                                                                      • Contents
                                                                                                                                      • Basic characteristics of soils
                                                                                                                                      • Seepage
                                                                                                                                      • Effective stress
                                                                                                                                      • Shear strength
                                                                                                                                      • Stresses and displacements
                                                                                                                                      • Lateral earth pressure
                                                                                                                                      • Consolidation theory
                                                                                                                                      • Bearing capacity
                                                                                                                                      • Stability of slopes

                                                                                                                                        For a rectangular foundation (L frac14 450m B frac14 225m)

                                                                                                                                        Nc frac14 084thorn 016B

                                                                                                                                        L

                                                                                                                                        81 frac14 745

                                                                                                                                        Using Equation 810

                                                                                                                                        qnf frac14 qf D frac14 cuNc frac14 135 745 frac14 1006 kN=m2

                                                                                                                                        For F frac14 3

                                                                                                                                        qn frac14 1006

                                                                                                                                        3frac14 335 kN=m2

                                                                                                                                        q frac14 qn thorn D frac14 335thorn eth20 350THORN frac14 405 kN=m2

                                                                                                                                        Design load frac14 405 450 225 frac14 4100 kN

                                                                                                                                        Design undrained strength cud frac14 135

                                                                                                                                        14frac14 96 kN=m2

                                                                                                                                        Design bearing resistance Rd frac14 cudNc area frac14 96 745 450 225

                                                                                                                                        frac14 7241 kN

                                                                                                                                        Design load Vd frac14 4100 kN

                                                                                                                                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                                        84

                                                                                                                                        For 0 frac14 40

                                                                                                                                        Nq frac14 64 N frac14 95

                                                                                                                                        qnf frac14 DethNq 1THORN thorn 04BN

                                                                                                                                        (a) Water table 5m below ground level

                                                                                                                                        qnf frac14 eth17 1 63THORN thorn eth04 17 25 95THORNfrac14 1071thorn 1615 frac14 2686 kN=m2

                                                                                                                                        qn frac14 400 17 frac14 383 kN=m2

                                                                                                                                        F frac14 2686

                                                                                                                                        383frac14 70

                                                                                                                                        (b) Water table 1m below ground level (ie at foundation level)

                                                                                                                                        0 frac14 20 98 frac14 102 kN=m3

                                                                                                                                        62 Bearing capacity

                                                                                                                                        qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                                                        F frac14 2040

                                                                                                                                        383frac14 53

                                                                                                                                        (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                                                        eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                                                        qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                                                        F frac14 1296

                                                                                                                                        392frac14 33

                                                                                                                                        85

                                                                                                                                        The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                                                        Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                                                        Design value of 0 frac14 tan1tan 39

                                                                                                                                        125

                                                                                                                                        frac14 33

                                                                                                                                        For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                                                        Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                                                        Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                                        86

                                                                                                                                        (a) Undrained shear for u frac14 0

                                                                                                                                        Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                                        qnf frac14 12cuNc

                                                                                                                                        frac14 12 100 514 frac14 617 kN=m2

                                                                                                                                        qn frac14 qnfFfrac14 617

                                                                                                                                        3frac14 206 kN=m2

                                                                                                                                        q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                                                        Bearing capacity 63

                                                                                                                                        Drained shear for 0 frac14 32

                                                                                                                                        Nq frac14 23 N frac14 25

                                                                                                                                        0 frac14 21 98 frac14 112 kN=m3

                                                                                                                                        qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                                        frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                                        frac14 694 kN=m2

                                                                                                                                        q frac14 694

                                                                                                                                        3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                                        Design load frac14 42 227 frac14 3632 kN

                                                                                                                                        (b) Design undrained strength cud frac14 100

                                                                                                                                        14frac14 71 kNm2

                                                                                                                                        Design bearing resistance Rd frac14 12cudNe area

                                                                                                                                        frac14 12 71 514 42

                                                                                                                                        frac14 7007 kN

                                                                                                                                        For drained shear 0d frac14 tan1tan 32

                                                                                                                                        125

                                                                                                                                        frac14 26

                                                                                                                                        Nq frac14 12 N frac14 10

                                                                                                                                        Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                                        (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                                        Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                                        1 2 100 0175 0700qn 0182qn

                                                                                                                                        2 6 033 0044 0176qn 0046qn

                                                                                                                                        3 10 020 0017 0068qn 0018qn

                                                                                                                                        0246qn

                                                                                                                                        Diameter of equivalent circle B frac14 45m

                                                                                                                                        H

                                                                                                                                        Bfrac14 12

                                                                                                                                        45frac14 27 and A frac14 042

                                                                                                                                        13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                                        64 Bearing capacity

                                                                                                                                        For sc frac14 30mm

                                                                                                                                        qn frac14 30

                                                                                                                                        0147frac14 204 kN=m2

                                                                                                                                        q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                                        Design load frac14 42 225 frac14 3600 kN

                                                                                                                                        The design load is 3600 kN settlement being the limiting criterion

                                                                                                                                        87

                                                                                                                                        D

                                                                                                                                        Bfrac14 8

                                                                                                                                        4frac14 20

                                                                                                                                        From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                                        F frac14 cuNc

                                                                                                                                        Dfrac14 40 71

                                                                                                                                        20 8frac14 18

                                                                                                                                        88

                                                                                                                                        Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                                        Design value of 0 frac14 tan1tan 38

                                                                                                                                        125

                                                                                                                                        frac14 32

                                                                                                                                        Figure Q86

                                                                                                                                        Bearing capacity 65

                                                                                                                                        For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                                        Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                                        The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                                        Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                                        For B frac14 250m qn frac14 3750

                                                                                                                                        2502 17 frac14 583 kN=m2

                                                                                                                                        From Figure 510 m frac14 n frac14 126

                                                                                                                                        6frac14 021

                                                                                                                                        Ir frac14 0019

                                                                                                                                        Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                                        Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                                        The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                        89

                                                                                                                                        Depth (m) N 0v (kNm2) CN N1

                                                                                                                                        070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                                        Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                                        (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                                        Cw frac14 05thorn 0530

                                                                                                                                        47

                                                                                                                                        frac14 082

                                                                                                                                        66 Bearing capacity

                                                                                                                                        Thus

                                                                                                                                        qa frac14 150 082 frac14 120 kN=m2

                                                                                                                                        (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                                        Thus

                                                                                                                                        qa frac14 90 15 frac14 135 kN=m2

                                                                                                                                        (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                                        Ic frac14 171

                                                                                                                                        1014frac14 0068

                                                                                                                                        From Equation 819(a) with s frac14 25mm

                                                                                                                                        q frac14 25

                                                                                                                                        3507 0068frac14 150 kN=m2

                                                                                                                                        810

                                                                                                                                        Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                                        Izp frac14 05thorn 01130

                                                                                                                                        16 27

                                                                                                                                        05

                                                                                                                                        frac14 067

                                                                                                                                        Refer to Figure Q810

                                                                                                                                        E frac14 25qc

                                                                                                                                        Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                                        Ez (mm3MN)

                                                                                                                                        1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                                        0203

                                                                                                                                        C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                                        130frac14 093

                                                                                                                                        C2 frac14 1 ethsayTHORN

                                                                                                                                        s frac14 C1C2qnX Iz

                                                                                                                                        Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                                        Bearing capacity 67

                                                                                                                                        811

                                                                                                                                        At pile base level

                                                                                                                                        cu frac14 220 kN=m2

                                                                                                                                        qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                                        Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                                        00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                                        qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                                        Then

                                                                                                                                        Qf frac14 Abqb thorn Asqs

                                                                                                                                        frac14

                                                                                                                                        4 32 1980

                                                                                                                                        thorn eth 105 139 86THORN

                                                                                                                                        frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                                        0 01 02 03 04 05 06 07

                                                                                                                                        0 2 4 6 8 10 12 14

                                                                                                                                        1

                                                                                                                                        2

                                                                                                                                        3

                                                                                                                                        4

                                                                                                                                        5

                                                                                                                                        6

                                                                                                                                        7

                                                                                                                                        8

                                                                                                                                        (1)

                                                                                                                                        (2)

                                                                                                                                        (3)

                                                                                                                                        (4)

                                                                                                                                        (5)

                                                                                                                                        qc

                                                                                                                                        qc

                                                                                                                                        Iz

                                                                                                                                        Iz

                                                                                                                                        (MNm2)

                                                                                                                                        z (m)

                                                                                                                                        Figure Q810

                                                                                                                                        68 Bearing capacity

                                                                                                                                        Allowable load

                                                                                                                                        ethaTHORN Qf

                                                                                                                                        2frac14 17 937

                                                                                                                                        2frac14 8968 kN

                                                                                                                                        ethbTHORN Abqb

                                                                                                                                        3thorn Asqs frac14 13 996

                                                                                                                                        3thorn 3941 frac14 8606 kN

                                                                                                                                        ie allowable load frac14 8600 kN

                                                                                                                                        Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                                        According to the limit state method

                                                                                                                                        Characteristic undrained strength at base level cuk frac14 220

                                                                                                                                        150kN=m2

                                                                                                                                        Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                                        150frac14 1320 kN=m2

                                                                                                                                        Characteristic shaft resistance qsk frac14 00150

                                                                                                                                        frac14 86

                                                                                                                                        150frac14 57 kN=m2

                                                                                                                                        Characteristic base and shaft resistances

                                                                                                                                        Rbk frac14

                                                                                                                                        4 32 1320 frac14 9330 kN

                                                                                                                                        Rsk frac14 105 139 86

                                                                                                                                        150frac14 2629 kN

                                                                                                                                        For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                                        Design bearing resistance Rcd frac14 9330

                                                                                                                                        160thorn 2629

                                                                                                                                        130

                                                                                                                                        frac14 5831thorn 2022

                                                                                                                                        frac14 7850 kN

                                                                                                                                        Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                                        812

                                                                                                                                        ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                                        qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                                        For a single pile

                                                                                                                                        Qf frac14 Abqb thorn Asqs

                                                                                                                                        frac14

                                                                                                                                        4 062 1305

                                                                                                                                        thorn eth 06 15 42THORN

                                                                                                                                        frac14 369thorn 1187 frac14 1556 kN

                                                                                                                                        Bearing capacity 69

                                                                                                                                        Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                                        eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                                        (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                                        (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                                        qbkfrac14 9cuk frac14 9 220

                                                                                                                                        150frac14 1320 kN=m2

                                                                                                                                        qskfrac14cuk frac14 040 105

                                                                                                                                        150frac14 28 kN=m2

                                                                                                                                        Rbkfrac14

                                                                                                                                        4 0602 1320 frac14 373 kN

                                                                                                                                        Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                                        Rcdfrac14 373

                                                                                                                                        160thorn 791

                                                                                                                                        130frac14 233thorn 608 frac14 841 kN

                                                                                                                                        Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                                        Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                                        (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                                        q frac14 21 000

                                                                                                                                        1762frac14 68 kN=m2

                                                                                                                                        Immediate settlement

                                                                                                                                        H

                                                                                                                                        Bfrac14 15

                                                                                                                                        176frac14 085

                                                                                                                                        D

                                                                                                                                        Bfrac14 13

                                                                                                                                        176frac14 074

                                                                                                                                        L

                                                                                                                                        Bfrac14 1

                                                                                                                                        Hence from Figure 515

                                                                                                                                        130 frac14 078 and 131 frac14 041

                                                                                                                                        70 Bearing capacity

                                                                                                                                        Thus using Equation 528

                                                                                                                                        si frac14 078 041 68 176

                                                                                                                                        65frac14 6mm

                                                                                                                                        Consolidation settlement

                                                                                                                                        Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                                        1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                                        434 (sod)

                                                                                                                                        Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                                        sc frac14 056 434 frac14 24mm

                                                                                                                                        The total settlement is (6thorn 24) frac14 30mm

                                                                                                                                        813

                                                                                                                                        At base level N frac14 26 Then using Equation 830

                                                                                                                                        qb frac14 40NDb

                                                                                                                                        Bfrac14 40 26 2

                                                                                                                                        025frac14 8320 kN=m2

                                                                                                                                        ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                                        Figure Q812

                                                                                                                                        Bearing capacity 71

                                                                                                                                        Over the length embedded in sand

                                                                                                                                        N frac14 21 ie18thorn 24

                                                                                                                                        2

                                                                                                                                        Using Equation 831

                                                                                                                                        qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                        For a single pile

                                                                                                                                        Qf frac14 Abqb thorn Asqs

                                                                                                                                        frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                        For the pile group assuming a group efficiency of 12

                                                                                                                                        XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                        Then the load factor is

                                                                                                                                        F frac14 6523

                                                                                                                                        2000thorn 1000frac14 21

                                                                                                                                        (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                        Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                        150frac14 5547 kNm2

                                                                                                                                        Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                        150frac14 28 kNm2

                                                                                                                                        Characteristic base and shaft resistances for a single pile

                                                                                                                                        Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                        Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                        For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                        Design bearing resistance Rcd frac14 347

                                                                                                                                        130thorn 56

                                                                                                                                        130frac14 310 kN

                                                                                                                                        For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                        Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                        (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                        From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                        72 Bearing capacity

                                                                                                                                        N frac14 24thorn 26thorn 34

                                                                                                                                        3frac14 28

                                                                                                                                        Ic frac14 171

                                                                                                                                        2814frac14 0016 ethEquation 818THORN

                                                                                                                                        s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                        814

                                                                                                                                        Using Equation 841

                                                                                                                                        Tf frac14 DLcu thorn

                                                                                                                                        4ethD2 d2THORNcuNc

                                                                                                                                        frac14 eth 02 5 06 110THORN thorn

                                                                                                                                        4eth022 012THORN110 9

                                                                                                                                        frac14 207thorn 23 frac14 230 kN

                                                                                                                                        Figure Q813

                                                                                                                                        Bearing capacity 73

                                                                                                                                        Chapter 9

                                                                                                                                        Stability of slopes

                                                                                                                                        91

                                                                                                                                        Referring to Figure Q91

                                                                                                                                        W frac14 417 19 frac14 792 kN=m

                                                                                                                                        Q frac14 20 28 frac14 56 kN=m

                                                                                                                                        Arc lengthAB frac14

                                                                                                                                        180 73 90 frac14 115m

                                                                                                                                        Arc length BC frac14

                                                                                                                                        180 28 90 frac14 44m

                                                                                                                                        The factor of safety is given by

                                                                                                                                        F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                        90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                        Depth of tension crack z0 frac14 2cu

                                                                                                                                        frac14 2 20

                                                                                                                                        19frac14 21m

                                                                                                                                        Arc length BD frac14

                                                                                                                                        180 13

                                                                                                                                        1

                                                                                                                                        2 90 frac14 21m

                                                                                                                                        F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                        The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                        Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                        14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                        Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                        The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                        92

                                                                                                                                        u frac14 0

                                                                                                                                        Depth factor D frac14 11

                                                                                                                                        9frac14 122

                                                                                                                                        Using Equation 92 with F frac14 10

                                                                                                                                        Ns frac14 cu

                                                                                                                                        FHfrac14 30

                                                                                                                                        10 19 9frac14 0175

                                                                                                                                        Hence from Figure 93

                                                                                                                                        frac14 50

                                                                                                                                        For F frac14 12

                                                                                                                                        Ns frac14 30

                                                                                                                                        12 19 9frac14 0146

                                                                                                                                        frac14 27

                                                                                                                                        93

                                                                                                                                        Refer to Figure Q93

                                                                                                                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                        1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                        74 m

                                                                                                                                        214 1deg

                                                                                                                                        213 1deg

                                                                                                                                        39 m

                                                                                                                                        WB

                                                                                                                                        D

                                                                                                                                        C

                                                                                                                                        28 m

                                                                                                                                        21 m

                                                                                                                                        A

                                                                                                                                        Q

                                                                                                                                        Soil (1)Soil (2)

                                                                                                                                        73deg

                                                                                                                                        Figure Q91

                                                                                                                                        Stability of slopes 75

                                                                                                                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                        9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                        599 256 328 1372

                                                                                                                                        Figure Q93

                                                                                                                                        76 Stability of slopes

                                                                                                                                        XW cos frac14 b

                                                                                                                                        Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                        W sin frac14 bX

                                                                                                                                        h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                        Arc length La frac14

                                                                                                                                        180 57

                                                                                                                                        1

                                                                                                                                        2 326 frac14 327m

                                                                                                                                        The factor of safety is given by

                                                                                                                                        F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                        W sin

                                                                                                                                        frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                        frac14 091

                                                                                                                                        According to the limit state method

                                                                                                                                        0d frac14 tan1tan 32

                                                                                                                                        125

                                                                                                                                        frac14 265

                                                                                                                                        c0 frac14 8

                                                                                                                                        160frac14 5 kN=m2

                                                                                                                                        Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                        Design disturbing moment frac14 1075 kN=m

                                                                                                                                        The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                        94

                                                                                                                                        F frac14 1

                                                                                                                                        W sin

                                                                                                                                        Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                        1thorn ethtan tan0=FTHORN

                                                                                                                                        c0 frac14 8 kN=m2

                                                                                                                                        0 frac14 32

                                                                                                                                        c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                        W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                        Try F frac14 100

                                                                                                                                        tan0

                                                                                                                                        Ffrac14 0625

                                                                                                                                        Stability of slopes 77

                                                                                                                                        Values of u are as obtained in Figure Q93

                                                                                                                                        SliceNo

                                                                                                                                        h(m)

                                                                                                                                        W frac14 bh(kNm)

                                                                                                                                        W sin(kNm)

                                                                                                                                        ub(kNm)

                                                                                                                                        c0bthorn (W ub) tan0(kNm)

                                                                                                                                        sec

                                                                                                                                        1thorn (tan tan0)FProduct(kNm)

                                                                                                                                        1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                        23 33 30 1042 31

                                                                                                                                        3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                        224 92 72 0931 67

                                                                                                                                        6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                        12 58 112 90 0889 80

                                                                                                                                        8 60 252 1812

                                                                                                                                        80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                        10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                        2154 88 116 0853 99

                                                                                                                                        14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                        1074 1091

                                                                                                                                        F frac14 1091

                                                                                                                                        1074frac14 102 (assumed value 100)

                                                                                                                                        Thus

                                                                                                                                        F frac14 101

                                                                                                                                        95

                                                                                                                                        F frac14 1

                                                                                                                                        W sin

                                                                                                                                        XfWeth1 ruTHORN tan0g sec

                                                                                                                                        1thorn ethtan tan0THORN=F

                                                                                                                                        0 frac14 33

                                                                                                                                        ru frac14 020

                                                                                                                                        W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                        eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                        Try F frac14 110

                                                                                                                                        tan 0

                                                                                                                                        Ffrac14 tan 33

                                                                                                                                        110frac14 0590

                                                                                                                                        78 Stability of slopes

                                                                                                                                        Referring to Figure Q95

                                                                                                                                        SliceNo

                                                                                                                                        h(m)

                                                                                                                                        W frac14 bh(kNm)

                                                                                                                                        W sin(kNm)

                                                                                                                                        W(1 ru) tan0(kNm)

                                                                                                                                        sec

                                                                                                                                        1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                        1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                        2120 234 0892 209

                                                                                                                                        4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                        1185 1271

                                                                                                                                        Figure Q95

                                                                                                                                        Stability of slopes 79

                                                                                                                                        F frac14 1271

                                                                                                                                        1185frac14 107

                                                                                                                                        The trial value was 110 therefore take F to be 108

                                                                                                                                        96

                                                                                                                                        (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                        F frac14 0

                                                                                                                                        sat

                                                                                                                                        tan0

                                                                                                                                        tan

                                                                                                                                        ptie 15 frac14 92

                                                                                                                                        19

                                                                                                                                        tan 36

                                                                                                                                        tan

                                                                                                                                        tan frac14 0234

                                                                                                                                        frac14 13

                                                                                                                                        Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                        F frac14 tan0

                                                                                                                                        tan

                                                                                                                                        frac14 tan 36

                                                                                                                                        tan 13

                                                                                                                                        frac14 31

                                                                                                                                        (b) 0d frac14 tan1tan 36

                                                                                                                                        125

                                                                                                                                        frac14 30

                                                                                                                                        Depth of potential failure surface frac14 z

                                                                                                                                        Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                        frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                        frac14 504z kN

                                                                                                                                        Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                        frac14 19 z sin 13 cos 13

                                                                                                                                        frac14 416z kN

                                                                                                                                        Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                        80 Stability of slopes

                                                                                                                                        • Book Cover
                                                                                                                                        • Title
                                                                                                                                        • Contents
                                                                                                                                        • Basic characteristics of soils
                                                                                                                                        • Seepage
                                                                                                                                        • Effective stress
                                                                                                                                        • Shear strength
                                                                                                                                        • Stresses and displacements
                                                                                                                                        • Lateral earth pressure
                                                                                                                                        • Consolidation theory
                                                                                                                                        • Bearing capacity
                                                                                                                                        • Stability of slopes

                                                                                                                                          qnf frac14 eth17 1 63THORN thorn eth04 102 25 95THORNfrac14 1071thorn 969 frac14 2040 kN=m2

                                                                                                                                          F frac14 2040

                                                                                                                                          383frac14 53

                                                                                                                                          (c) Water table at ground level with upward hydraulic gradient 02

                                                                                                                                          eth0 jTHORN frac14 102 eth02 98THORN frac14 82 kN=m3

                                                                                                                                          qnf frac14 eth82 1 63THORN thorn eth04 82 25 95THORNfrac14 517thorn 779 frac14 1296 kN=m2

                                                                                                                                          F frac14 1296

                                                                                                                                          392frac14 33

                                                                                                                                          85

                                                                                                                                          The following partial factors are used dead load 10 imposed load 13 shear strength( tan0) 125

                                                                                                                                          Design load Vd frac14 4000thorn eth13 1500THORN frac14 5950 kN

                                                                                                                                          Design value of 0 frac14 tan1tan 39

                                                                                                                                          125

                                                                                                                                          frac14 33

                                                                                                                                          For 0 frac14 33 Nq frac14 26 and N frac14 29

                                                                                                                                          Design bearing resistance Rd frac14 32frac12eth102 15 26THORN thorn eth04 102 30 29THORNfrac14 32eth398thorn 355THORNfrac14 6777 kN

                                                                                                                                          Rd gt Vd therefore the bearing resistance limit state is satisfied

                                                                                                                                          86

                                                                                                                                          (a) Undrained shear for u frac14 0

                                                                                                                                          Nc frac14 514 Nq frac14 1 N frac14 0

                                                                                                                                          qnf frac14 12cuNc

                                                                                                                                          frac14 12 100 514 frac14 617 kN=m2

                                                                                                                                          qn frac14 qnfFfrac14 617

                                                                                                                                          3frac14 206 kN=m2

                                                                                                                                          q frac14 qn thorn D frac14 206thorn 21 frac14 227 kN=m2

                                                                                                                                          Bearing capacity 63

                                                                                                                                          Drained shear for 0 frac14 32

                                                                                                                                          Nq frac14 23 N frac14 25

                                                                                                                                          0 frac14 21 98 frac14 112 kN=m3

                                                                                                                                          qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                                          frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                                          frac14 694 kN=m2

                                                                                                                                          q frac14 694

                                                                                                                                          3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                                          Design load frac14 42 227 frac14 3632 kN

                                                                                                                                          (b) Design undrained strength cud frac14 100

                                                                                                                                          14frac14 71 kNm2

                                                                                                                                          Design bearing resistance Rd frac14 12cudNe area

                                                                                                                                          frac14 12 71 514 42

                                                                                                                                          frac14 7007 kN

                                                                                                                                          For drained shear 0d frac14 tan1tan 32

                                                                                                                                          125

                                                                                                                                          frac14 26

                                                                                                                                          Nq frac14 12 N frac14 10

                                                                                                                                          Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                                          (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                                          Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                                          1 2 100 0175 0700qn 0182qn

                                                                                                                                          2 6 033 0044 0176qn 0046qn

                                                                                                                                          3 10 020 0017 0068qn 0018qn

                                                                                                                                          0246qn

                                                                                                                                          Diameter of equivalent circle B frac14 45m

                                                                                                                                          H

                                                                                                                                          Bfrac14 12

                                                                                                                                          45frac14 27 and A frac14 042

                                                                                                                                          13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                                          64 Bearing capacity

                                                                                                                                          For sc frac14 30mm

                                                                                                                                          qn frac14 30

                                                                                                                                          0147frac14 204 kN=m2

                                                                                                                                          q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                                          Design load frac14 42 225 frac14 3600 kN

                                                                                                                                          The design load is 3600 kN settlement being the limiting criterion

                                                                                                                                          87

                                                                                                                                          D

                                                                                                                                          Bfrac14 8

                                                                                                                                          4frac14 20

                                                                                                                                          From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                                          F frac14 cuNc

                                                                                                                                          Dfrac14 40 71

                                                                                                                                          20 8frac14 18

                                                                                                                                          88

                                                                                                                                          Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                                          Design value of 0 frac14 tan1tan 38

                                                                                                                                          125

                                                                                                                                          frac14 32

                                                                                                                                          Figure Q86

                                                                                                                                          Bearing capacity 65

                                                                                                                                          For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                                          Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                                          The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                                          Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                                          For B frac14 250m qn frac14 3750

                                                                                                                                          2502 17 frac14 583 kN=m2

                                                                                                                                          From Figure 510 m frac14 n frac14 126

                                                                                                                                          6frac14 021

                                                                                                                                          Ir frac14 0019

                                                                                                                                          Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                                          Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                                          The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                          89

                                                                                                                                          Depth (m) N 0v (kNm2) CN N1

                                                                                                                                          070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                                          Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                                          (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                                          Cw frac14 05thorn 0530

                                                                                                                                          47

                                                                                                                                          frac14 082

                                                                                                                                          66 Bearing capacity

                                                                                                                                          Thus

                                                                                                                                          qa frac14 150 082 frac14 120 kN=m2

                                                                                                                                          (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                                          Thus

                                                                                                                                          qa frac14 90 15 frac14 135 kN=m2

                                                                                                                                          (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                                          Ic frac14 171

                                                                                                                                          1014frac14 0068

                                                                                                                                          From Equation 819(a) with s frac14 25mm

                                                                                                                                          q frac14 25

                                                                                                                                          3507 0068frac14 150 kN=m2

                                                                                                                                          810

                                                                                                                                          Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                                          Izp frac14 05thorn 01130

                                                                                                                                          16 27

                                                                                                                                          05

                                                                                                                                          frac14 067

                                                                                                                                          Refer to Figure Q810

                                                                                                                                          E frac14 25qc

                                                                                                                                          Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                                          Ez (mm3MN)

                                                                                                                                          1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                                          0203

                                                                                                                                          C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                                          130frac14 093

                                                                                                                                          C2 frac14 1 ethsayTHORN

                                                                                                                                          s frac14 C1C2qnX Iz

                                                                                                                                          Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                                          Bearing capacity 67

                                                                                                                                          811

                                                                                                                                          At pile base level

                                                                                                                                          cu frac14 220 kN=m2

                                                                                                                                          qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                                          Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                                          00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                                          qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                                          Then

                                                                                                                                          Qf frac14 Abqb thorn Asqs

                                                                                                                                          frac14

                                                                                                                                          4 32 1980

                                                                                                                                          thorn eth 105 139 86THORN

                                                                                                                                          frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                                          0 01 02 03 04 05 06 07

                                                                                                                                          0 2 4 6 8 10 12 14

                                                                                                                                          1

                                                                                                                                          2

                                                                                                                                          3

                                                                                                                                          4

                                                                                                                                          5

                                                                                                                                          6

                                                                                                                                          7

                                                                                                                                          8

                                                                                                                                          (1)

                                                                                                                                          (2)

                                                                                                                                          (3)

                                                                                                                                          (4)

                                                                                                                                          (5)

                                                                                                                                          qc

                                                                                                                                          qc

                                                                                                                                          Iz

                                                                                                                                          Iz

                                                                                                                                          (MNm2)

                                                                                                                                          z (m)

                                                                                                                                          Figure Q810

                                                                                                                                          68 Bearing capacity

                                                                                                                                          Allowable load

                                                                                                                                          ethaTHORN Qf

                                                                                                                                          2frac14 17 937

                                                                                                                                          2frac14 8968 kN

                                                                                                                                          ethbTHORN Abqb

                                                                                                                                          3thorn Asqs frac14 13 996

                                                                                                                                          3thorn 3941 frac14 8606 kN

                                                                                                                                          ie allowable load frac14 8600 kN

                                                                                                                                          Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                                          According to the limit state method

                                                                                                                                          Characteristic undrained strength at base level cuk frac14 220

                                                                                                                                          150kN=m2

                                                                                                                                          Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                                          150frac14 1320 kN=m2

                                                                                                                                          Characteristic shaft resistance qsk frac14 00150

                                                                                                                                          frac14 86

                                                                                                                                          150frac14 57 kN=m2

                                                                                                                                          Characteristic base and shaft resistances

                                                                                                                                          Rbk frac14

                                                                                                                                          4 32 1320 frac14 9330 kN

                                                                                                                                          Rsk frac14 105 139 86

                                                                                                                                          150frac14 2629 kN

                                                                                                                                          For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                                          Design bearing resistance Rcd frac14 9330

                                                                                                                                          160thorn 2629

                                                                                                                                          130

                                                                                                                                          frac14 5831thorn 2022

                                                                                                                                          frac14 7850 kN

                                                                                                                                          Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                                          812

                                                                                                                                          ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                                          qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                                          For a single pile

                                                                                                                                          Qf frac14 Abqb thorn Asqs

                                                                                                                                          frac14

                                                                                                                                          4 062 1305

                                                                                                                                          thorn eth 06 15 42THORN

                                                                                                                                          frac14 369thorn 1187 frac14 1556 kN

                                                                                                                                          Bearing capacity 69

                                                                                                                                          Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                                          eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                                          (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                                          (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                                          qbkfrac14 9cuk frac14 9 220

                                                                                                                                          150frac14 1320 kN=m2

                                                                                                                                          qskfrac14cuk frac14 040 105

                                                                                                                                          150frac14 28 kN=m2

                                                                                                                                          Rbkfrac14

                                                                                                                                          4 0602 1320 frac14 373 kN

                                                                                                                                          Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                                          Rcdfrac14 373

                                                                                                                                          160thorn 791

                                                                                                                                          130frac14 233thorn 608 frac14 841 kN

                                                                                                                                          Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                                          Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                                          (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                                          q frac14 21 000

                                                                                                                                          1762frac14 68 kN=m2

                                                                                                                                          Immediate settlement

                                                                                                                                          H

                                                                                                                                          Bfrac14 15

                                                                                                                                          176frac14 085

                                                                                                                                          D

                                                                                                                                          Bfrac14 13

                                                                                                                                          176frac14 074

                                                                                                                                          L

                                                                                                                                          Bfrac14 1

                                                                                                                                          Hence from Figure 515

                                                                                                                                          130 frac14 078 and 131 frac14 041

                                                                                                                                          70 Bearing capacity

                                                                                                                                          Thus using Equation 528

                                                                                                                                          si frac14 078 041 68 176

                                                                                                                                          65frac14 6mm

                                                                                                                                          Consolidation settlement

                                                                                                                                          Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                                          1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                                          434 (sod)

                                                                                                                                          Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                                          sc frac14 056 434 frac14 24mm

                                                                                                                                          The total settlement is (6thorn 24) frac14 30mm

                                                                                                                                          813

                                                                                                                                          At base level N frac14 26 Then using Equation 830

                                                                                                                                          qb frac14 40NDb

                                                                                                                                          Bfrac14 40 26 2

                                                                                                                                          025frac14 8320 kN=m2

                                                                                                                                          ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                                          Figure Q812

                                                                                                                                          Bearing capacity 71

                                                                                                                                          Over the length embedded in sand

                                                                                                                                          N frac14 21 ie18thorn 24

                                                                                                                                          2

                                                                                                                                          Using Equation 831

                                                                                                                                          qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                          For a single pile

                                                                                                                                          Qf frac14 Abqb thorn Asqs

                                                                                                                                          frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                          For the pile group assuming a group efficiency of 12

                                                                                                                                          XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                          Then the load factor is

                                                                                                                                          F frac14 6523

                                                                                                                                          2000thorn 1000frac14 21

                                                                                                                                          (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                          Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                          150frac14 5547 kNm2

                                                                                                                                          Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                          150frac14 28 kNm2

                                                                                                                                          Characteristic base and shaft resistances for a single pile

                                                                                                                                          Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                          Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                          For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                          Design bearing resistance Rcd frac14 347

                                                                                                                                          130thorn 56

                                                                                                                                          130frac14 310 kN

                                                                                                                                          For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                          Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                          (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                          From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                          72 Bearing capacity

                                                                                                                                          N frac14 24thorn 26thorn 34

                                                                                                                                          3frac14 28

                                                                                                                                          Ic frac14 171

                                                                                                                                          2814frac14 0016 ethEquation 818THORN

                                                                                                                                          s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                          814

                                                                                                                                          Using Equation 841

                                                                                                                                          Tf frac14 DLcu thorn

                                                                                                                                          4ethD2 d2THORNcuNc

                                                                                                                                          frac14 eth 02 5 06 110THORN thorn

                                                                                                                                          4eth022 012THORN110 9

                                                                                                                                          frac14 207thorn 23 frac14 230 kN

                                                                                                                                          Figure Q813

                                                                                                                                          Bearing capacity 73

                                                                                                                                          Chapter 9

                                                                                                                                          Stability of slopes

                                                                                                                                          91

                                                                                                                                          Referring to Figure Q91

                                                                                                                                          W frac14 417 19 frac14 792 kN=m

                                                                                                                                          Q frac14 20 28 frac14 56 kN=m

                                                                                                                                          Arc lengthAB frac14

                                                                                                                                          180 73 90 frac14 115m

                                                                                                                                          Arc length BC frac14

                                                                                                                                          180 28 90 frac14 44m

                                                                                                                                          The factor of safety is given by

                                                                                                                                          F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                          90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                          Depth of tension crack z0 frac14 2cu

                                                                                                                                          frac14 2 20

                                                                                                                                          19frac14 21m

                                                                                                                                          Arc length BD frac14

                                                                                                                                          180 13

                                                                                                                                          1

                                                                                                                                          2 90 frac14 21m

                                                                                                                                          F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                          The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                          Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                          14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                          Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                          The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                          92

                                                                                                                                          u frac14 0

                                                                                                                                          Depth factor D frac14 11

                                                                                                                                          9frac14 122

                                                                                                                                          Using Equation 92 with F frac14 10

                                                                                                                                          Ns frac14 cu

                                                                                                                                          FHfrac14 30

                                                                                                                                          10 19 9frac14 0175

                                                                                                                                          Hence from Figure 93

                                                                                                                                          frac14 50

                                                                                                                                          For F frac14 12

                                                                                                                                          Ns frac14 30

                                                                                                                                          12 19 9frac14 0146

                                                                                                                                          frac14 27

                                                                                                                                          93

                                                                                                                                          Refer to Figure Q93

                                                                                                                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                          1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                          74 m

                                                                                                                                          214 1deg

                                                                                                                                          213 1deg

                                                                                                                                          39 m

                                                                                                                                          WB

                                                                                                                                          D

                                                                                                                                          C

                                                                                                                                          28 m

                                                                                                                                          21 m

                                                                                                                                          A

                                                                                                                                          Q

                                                                                                                                          Soil (1)Soil (2)

                                                                                                                                          73deg

                                                                                                                                          Figure Q91

                                                                                                                                          Stability of slopes 75

                                                                                                                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                          9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                          599 256 328 1372

                                                                                                                                          Figure Q93

                                                                                                                                          76 Stability of slopes

                                                                                                                                          XW cos frac14 b

                                                                                                                                          Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                          W sin frac14 bX

                                                                                                                                          h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                          Arc length La frac14

                                                                                                                                          180 57

                                                                                                                                          1

                                                                                                                                          2 326 frac14 327m

                                                                                                                                          The factor of safety is given by

                                                                                                                                          F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                          W sin

                                                                                                                                          frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                          frac14 091

                                                                                                                                          According to the limit state method

                                                                                                                                          0d frac14 tan1tan 32

                                                                                                                                          125

                                                                                                                                          frac14 265

                                                                                                                                          c0 frac14 8

                                                                                                                                          160frac14 5 kN=m2

                                                                                                                                          Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                          Design disturbing moment frac14 1075 kN=m

                                                                                                                                          The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                          94

                                                                                                                                          F frac14 1

                                                                                                                                          W sin

                                                                                                                                          Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                          1thorn ethtan tan0=FTHORN

                                                                                                                                          c0 frac14 8 kN=m2

                                                                                                                                          0 frac14 32

                                                                                                                                          c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                          W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                          Try F frac14 100

                                                                                                                                          tan0

                                                                                                                                          Ffrac14 0625

                                                                                                                                          Stability of slopes 77

                                                                                                                                          Values of u are as obtained in Figure Q93

                                                                                                                                          SliceNo

                                                                                                                                          h(m)

                                                                                                                                          W frac14 bh(kNm)

                                                                                                                                          W sin(kNm)

                                                                                                                                          ub(kNm)

                                                                                                                                          c0bthorn (W ub) tan0(kNm)

                                                                                                                                          sec

                                                                                                                                          1thorn (tan tan0)FProduct(kNm)

                                                                                                                                          1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                          23 33 30 1042 31

                                                                                                                                          3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                          224 92 72 0931 67

                                                                                                                                          6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                          12 58 112 90 0889 80

                                                                                                                                          8 60 252 1812

                                                                                                                                          80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                          10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                          2154 88 116 0853 99

                                                                                                                                          14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                          1074 1091

                                                                                                                                          F frac14 1091

                                                                                                                                          1074frac14 102 (assumed value 100)

                                                                                                                                          Thus

                                                                                                                                          F frac14 101

                                                                                                                                          95

                                                                                                                                          F frac14 1

                                                                                                                                          W sin

                                                                                                                                          XfWeth1 ruTHORN tan0g sec

                                                                                                                                          1thorn ethtan tan0THORN=F

                                                                                                                                          0 frac14 33

                                                                                                                                          ru frac14 020

                                                                                                                                          W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                          eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                          Try F frac14 110

                                                                                                                                          tan 0

                                                                                                                                          Ffrac14 tan 33

                                                                                                                                          110frac14 0590

                                                                                                                                          78 Stability of slopes

                                                                                                                                          Referring to Figure Q95

                                                                                                                                          SliceNo

                                                                                                                                          h(m)

                                                                                                                                          W frac14 bh(kNm)

                                                                                                                                          W sin(kNm)

                                                                                                                                          W(1 ru) tan0(kNm)

                                                                                                                                          sec

                                                                                                                                          1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                          1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                          2120 234 0892 209

                                                                                                                                          4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                          1185 1271

                                                                                                                                          Figure Q95

                                                                                                                                          Stability of slopes 79

                                                                                                                                          F frac14 1271

                                                                                                                                          1185frac14 107

                                                                                                                                          The trial value was 110 therefore take F to be 108

                                                                                                                                          96

                                                                                                                                          (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                          F frac14 0

                                                                                                                                          sat

                                                                                                                                          tan0

                                                                                                                                          tan

                                                                                                                                          ptie 15 frac14 92

                                                                                                                                          19

                                                                                                                                          tan 36

                                                                                                                                          tan

                                                                                                                                          tan frac14 0234

                                                                                                                                          frac14 13

                                                                                                                                          Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                          F frac14 tan0

                                                                                                                                          tan

                                                                                                                                          frac14 tan 36

                                                                                                                                          tan 13

                                                                                                                                          frac14 31

                                                                                                                                          (b) 0d frac14 tan1tan 36

                                                                                                                                          125

                                                                                                                                          frac14 30

                                                                                                                                          Depth of potential failure surface frac14 z

                                                                                                                                          Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                          frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                          frac14 504z kN

                                                                                                                                          Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                          frac14 19 z sin 13 cos 13

                                                                                                                                          frac14 416z kN

                                                                                                                                          Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                          80 Stability of slopes

                                                                                                                                          • Book Cover
                                                                                                                                          • Title
                                                                                                                                          • Contents
                                                                                                                                          • Basic characteristics of soils
                                                                                                                                          • Seepage
                                                                                                                                          • Effective stress
                                                                                                                                          • Shear strength
                                                                                                                                          • Stresses and displacements
                                                                                                                                          • Lateral earth pressure
                                                                                                                                          • Consolidation theory
                                                                                                                                          • Bearing capacity
                                                                                                                                          • Stability of slopes

                                                                                                                                            Drained shear for 0 frac14 32

                                                                                                                                            Nq frac14 23 N frac14 25

                                                                                                                                            0 frac14 21 98 frac14 112 kN=m3

                                                                                                                                            qnf frac14 0DethNq 1THORN thorn 040BN

                                                                                                                                            frac14 eth112 1 22THORN thorn eth04 112 4 25THORNfrac14 246thorn 448

                                                                                                                                            frac14 694 kN=m2

                                                                                                                                            q frac14 694

                                                                                                                                            3thorn 21 frac14 231thorn 21 frac14 252 kN=m2

                                                                                                                                            Design load frac14 42 227 frac14 3632 kN

                                                                                                                                            (b) Design undrained strength cud frac14 100

                                                                                                                                            14frac14 71 kNm2

                                                                                                                                            Design bearing resistance Rd frac14 12cudNe area

                                                                                                                                            frac14 12 71 514 42

                                                                                                                                            frac14 7007 kN

                                                                                                                                            For drained shear 0d frac14 tan1tan 32

                                                                                                                                            125

                                                                                                                                            frac14 26

                                                                                                                                            Nq frac14 12 N frac14 10

                                                                                                                                            Design bearing resistance Rd frac14 42frac12eth112 1 12THORN thorn eth04 112 4 10THORNfrac14 42eth134thorn 179THORNfrac14 5008 kN

                                                                                                                                            (c) Consolidation settlement the clay will be divided into three sublayers (Figure Q86)

                                                                                                                                            Layer z (m) m n Ir 0 (kNm2) sod (mm)

                                                                                                                                            1 2 100 0175 0700qn 0182qn

                                                                                                                                            2 6 033 0044 0176qn 0046qn

                                                                                                                                            3 10 020 0017 0068qn 0018qn

                                                                                                                                            0246qn

                                                                                                                                            Diameter of equivalent circle B frac14 45m

                                                                                                                                            H

                                                                                                                                            Bfrac14 12

                                                                                                                                            45frac14 27 and A frac14 042

                                                                                                                                            13 frac14 060 ethfrom Figure 712THORNsc frac14 060 0246qn frac14 0147qn ethmmTHORN

                                                                                                                                            64 Bearing capacity

                                                                                                                                            For sc frac14 30mm

                                                                                                                                            qn frac14 30

                                                                                                                                            0147frac14 204 kN=m2

                                                                                                                                            q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                                            Design load frac14 42 225 frac14 3600 kN

                                                                                                                                            The design load is 3600 kN settlement being the limiting criterion

                                                                                                                                            87

                                                                                                                                            D

                                                                                                                                            Bfrac14 8

                                                                                                                                            4frac14 20

                                                                                                                                            From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                                            F frac14 cuNc

                                                                                                                                            Dfrac14 40 71

                                                                                                                                            20 8frac14 18

                                                                                                                                            88

                                                                                                                                            Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                                            Design value of 0 frac14 tan1tan 38

                                                                                                                                            125

                                                                                                                                            frac14 32

                                                                                                                                            Figure Q86

                                                                                                                                            Bearing capacity 65

                                                                                                                                            For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                                            Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                                            The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                                            Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                                            For B frac14 250m qn frac14 3750

                                                                                                                                            2502 17 frac14 583 kN=m2

                                                                                                                                            From Figure 510 m frac14 n frac14 126

                                                                                                                                            6frac14 021

                                                                                                                                            Ir frac14 0019

                                                                                                                                            Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                                            Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                                            The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                            89

                                                                                                                                            Depth (m) N 0v (kNm2) CN N1

                                                                                                                                            070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                                            Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                                            (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                                            Cw frac14 05thorn 0530

                                                                                                                                            47

                                                                                                                                            frac14 082

                                                                                                                                            66 Bearing capacity

                                                                                                                                            Thus

                                                                                                                                            qa frac14 150 082 frac14 120 kN=m2

                                                                                                                                            (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                                            Thus

                                                                                                                                            qa frac14 90 15 frac14 135 kN=m2

                                                                                                                                            (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                                            Ic frac14 171

                                                                                                                                            1014frac14 0068

                                                                                                                                            From Equation 819(a) with s frac14 25mm

                                                                                                                                            q frac14 25

                                                                                                                                            3507 0068frac14 150 kN=m2

                                                                                                                                            810

                                                                                                                                            Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                                            Izp frac14 05thorn 01130

                                                                                                                                            16 27

                                                                                                                                            05

                                                                                                                                            frac14 067

                                                                                                                                            Refer to Figure Q810

                                                                                                                                            E frac14 25qc

                                                                                                                                            Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                                            Ez (mm3MN)

                                                                                                                                            1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                                            0203

                                                                                                                                            C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                                            130frac14 093

                                                                                                                                            C2 frac14 1 ethsayTHORN

                                                                                                                                            s frac14 C1C2qnX Iz

                                                                                                                                            Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                                            Bearing capacity 67

                                                                                                                                            811

                                                                                                                                            At pile base level

                                                                                                                                            cu frac14 220 kN=m2

                                                                                                                                            qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                                            Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                                            00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                                            qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                                            Then

                                                                                                                                            Qf frac14 Abqb thorn Asqs

                                                                                                                                            frac14

                                                                                                                                            4 32 1980

                                                                                                                                            thorn eth 105 139 86THORN

                                                                                                                                            frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                                            0 01 02 03 04 05 06 07

                                                                                                                                            0 2 4 6 8 10 12 14

                                                                                                                                            1

                                                                                                                                            2

                                                                                                                                            3

                                                                                                                                            4

                                                                                                                                            5

                                                                                                                                            6

                                                                                                                                            7

                                                                                                                                            8

                                                                                                                                            (1)

                                                                                                                                            (2)

                                                                                                                                            (3)

                                                                                                                                            (4)

                                                                                                                                            (5)

                                                                                                                                            qc

                                                                                                                                            qc

                                                                                                                                            Iz

                                                                                                                                            Iz

                                                                                                                                            (MNm2)

                                                                                                                                            z (m)

                                                                                                                                            Figure Q810

                                                                                                                                            68 Bearing capacity

                                                                                                                                            Allowable load

                                                                                                                                            ethaTHORN Qf

                                                                                                                                            2frac14 17 937

                                                                                                                                            2frac14 8968 kN

                                                                                                                                            ethbTHORN Abqb

                                                                                                                                            3thorn Asqs frac14 13 996

                                                                                                                                            3thorn 3941 frac14 8606 kN

                                                                                                                                            ie allowable load frac14 8600 kN

                                                                                                                                            Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                                            According to the limit state method

                                                                                                                                            Characteristic undrained strength at base level cuk frac14 220

                                                                                                                                            150kN=m2

                                                                                                                                            Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                                            150frac14 1320 kN=m2

                                                                                                                                            Characteristic shaft resistance qsk frac14 00150

                                                                                                                                            frac14 86

                                                                                                                                            150frac14 57 kN=m2

                                                                                                                                            Characteristic base and shaft resistances

                                                                                                                                            Rbk frac14

                                                                                                                                            4 32 1320 frac14 9330 kN

                                                                                                                                            Rsk frac14 105 139 86

                                                                                                                                            150frac14 2629 kN

                                                                                                                                            For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                                            Design bearing resistance Rcd frac14 9330

                                                                                                                                            160thorn 2629

                                                                                                                                            130

                                                                                                                                            frac14 5831thorn 2022

                                                                                                                                            frac14 7850 kN

                                                                                                                                            Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                                            812

                                                                                                                                            ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                                            qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                                            For a single pile

                                                                                                                                            Qf frac14 Abqb thorn Asqs

                                                                                                                                            frac14

                                                                                                                                            4 062 1305

                                                                                                                                            thorn eth 06 15 42THORN

                                                                                                                                            frac14 369thorn 1187 frac14 1556 kN

                                                                                                                                            Bearing capacity 69

                                                                                                                                            Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                                            eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                                            (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                                            (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                                            qbkfrac14 9cuk frac14 9 220

                                                                                                                                            150frac14 1320 kN=m2

                                                                                                                                            qskfrac14cuk frac14 040 105

                                                                                                                                            150frac14 28 kN=m2

                                                                                                                                            Rbkfrac14

                                                                                                                                            4 0602 1320 frac14 373 kN

                                                                                                                                            Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                                            Rcdfrac14 373

                                                                                                                                            160thorn 791

                                                                                                                                            130frac14 233thorn 608 frac14 841 kN

                                                                                                                                            Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                                            Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                                            (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                                            q frac14 21 000

                                                                                                                                            1762frac14 68 kN=m2

                                                                                                                                            Immediate settlement

                                                                                                                                            H

                                                                                                                                            Bfrac14 15

                                                                                                                                            176frac14 085

                                                                                                                                            D

                                                                                                                                            Bfrac14 13

                                                                                                                                            176frac14 074

                                                                                                                                            L

                                                                                                                                            Bfrac14 1

                                                                                                                                            Hence from Figure 515

                                                                                                                                            130 frac14 078 and 131 frac14 041

                                                                                                                                            70 Bearing capacity

                                                                                                                                            Thus using Equation 528

                                                                                                                                            si frac14 078 041 68 176

                                                                                                                                            65frac14 6mm

                                                                                                                                            Consolidation settlement

                                                                                                                                            Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                                            1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                                            434 (sod)

                                                                                                                                            Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                                            sc frac14 056 434 frac14 24mm

                                                                                                                                            The total settlement is (6thorn 24) frac14 30mm

                                                                                                                                            813

                                                                                                                                            At base level N frac14 26 Then using Equation 830

                                                                                                                                            qb frac14 40NDb

                                                                                                                                            Bfrac14 40 26 2

                                                                                                                                            025frac14 8320 kN=m2

                                                                                                                                            ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                                            Figure Q812

                                                                                                                                            Bearing capacity 71

                                                                                                                                            Over the length embedded in sand

                                                                                                                                            N frac14 21 ie18thorn 24

                                                                                                                                            2

                                                                                                                                            Using Equation 831

                                                                                                                                            qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                            For a single pile

                                                                                                                                            Qf frac14 Abqb thorn Asqs

                                                                                                                                            frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                            For the pile group assuming a group efficiency of 12

                                                                                                                                            XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                            Then the load factor is

                                                                                                                                            F frac14 6523

                                                                                                                                            2000thorn 1000frac14 21

                                                                                                                                            (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                            Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                            150frac14 5547 kNm2

                                                                                                                                            Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                            150frac14 28 kNm2

                                                                                                                                            Characteristic base and shaft resistances for a single pile

                                                                                                                                            Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                            Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                            For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                            Design bearing resistance Rcd frac14 347

                                                                                                                                            130thorn 56

                                                                                                                                            130frac14 310 kN

                                                                                                                                            For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                            Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                            (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                            From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                            72 Bearing capacity

                                                                                                                                            N frac14 24thorn 26thorn 34

                                                                                                                                            3frac14 28

                                                                                                                                            Ic frac14 171

                                                                                                                                            2814frac14 0016 ethEquation 818THORN

                                                                                                                                            s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                            814

                                                                                                                                            Using Equation 841

                                                                                                                                            Tf frac14 DLcu thorn

                                                                                                                                            4ethD2 d2THORNcuNc

                                                                                                                                            frac14 eth 02 5 06 110THORN thorn

                                                                                                                                            4eth022 012THORN110 9

                                                                                                                                            frac14 207thorn 23 frac14 230 kN

                                                                                                                                            Figure Q813

                                                                                                                                            Bearing capacity 73

                                                                                                                                            Chapter 9

                                                                                                                                            Stability of slopes

                                                                                                                                            91

                                                                                                                                            Referring to Figure Q91

                                                                                                                                            W frac14 417 19 frac14 792 kN=m

                                                                                                                                            Q frac14 20 28 frac14 56 kN=m

                                                                                                                                            Arc lengthAB frac14

                                                                                                                                            180 73 90 frac14 115m

                                                                                                                                            Arc length BC frac14

                                                                                                                                            180 28 90 frac14 44m

                                                                                                                                            The factor of safety is given by

                                                                                                                                            F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                            90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                            Depth of tension crack z0 frac14 2cu

                                                                                                                                            frac14 2 20

                                                                                                                                            19frac14 21m

                                                                                                                                            Arc length BD frac14

                                                                                                                                            180 13

                                                                                                                                            1

                                                                                                                                            2 90 frac14 21m

                                                                                                                                            F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                            The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                            Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                            14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                            Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                            The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                            92

                                                                                                                                            u frac14 0

                                                                                                                                            Depth factor D frac14 11

                                                                                                                                            9frac14 122

                                                                                                                                            Using Equation 92 with F frac14 10

                                                                                                                                            Ns frac14 cu

                                                                                                                                            FHfrac14 30

                                                                                                                                            10 19 9frac14 0175

                                                                                                                                            Hence from Figure 93

                                                                                                                                            frac14 50

                                                                                                                                            For F frac14 12

                                                                                                                                            Ns frac14 30

                                                                                                                                            12 19 9frac14 0146

                                                                                                                                            frac14 27

                                                                                                                                            93

                                                                                                                                            Refer to Figure Q93

                                                                                                                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                            1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                            74 m

                                                                                                                                            214 1deg

                                                                                                                                            213 1deg

                                                                                                                                            39 m

                                                                                                                                            WB

                                                                                                                                            D

                                                                                                                                            C

                                                                                                                                            28 m

                                                                                                                                            21 m

                                                                                                                                            A

                                                                                                                                            Q

                                                                                                                                            Soil (1)Soil (2)

                                                                                                                                            73deg

                                                                                                                                            Figure Q91

                                                                                                                                            Stability of slopes 75

                                                                                                                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                            9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                            599 256 328 1372

                                                                                                                                            Figure Q93

                                                                                                                                            76 Stability of slopes

                                                                                                                                            XW cos frac14 b

                                                                                                                                            Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                            W sin frac14 bX

                                                                                                                                            h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                            Arc length La frac14

                                                                                                                                            180 57

                                                                                                                                            1

                                                                                                                                            2 326 frac14 327m

                                                                                                                                            The factor of safety is given by

                                                                                                                                            F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                            W sin

                                                                                                                                            frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                            frac14 091

                                                                                                                                            According to the limit state method

                                                                                                                                            0d frac14 tan1tan 32

                                                                                                                                            125

                                                                                                                                            frac14 265

                                                                                                                                            c0 frac14 8

                                                                                                                                            160frac14 5 kN=m2

                                                                                                                                            Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                            Design disturbing moment frac14 1075 kN=m

                                                                                                                                            The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                            94

                                                                                                                                            F frac14 1

                                                                                                                                            W sin

                                                                                                                                            Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                            1thorn ethtan tan0=FTHORN

                                                                                                                                            c0 frac14 8 kN=m2

                                                                                                                                            0 frac14 32

                                                                                                                                            c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                            W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                            Try F frac14 100

                                                                                                                                            tan0

                                                                                                                                            Ffrac14 0625

                                                                                                                                            Stability of slopes 77

                                                                                                                                            Values of u are as obtained in Figure Q93

                                                                                                                                            SliceNo

                                                                                                                                            h(m)

                                                                                                                                            W frac14 bh(kNm)

                                                                                                                                            W sin(kNm)

                                                                                                                                            ub(kNm)

                                                                                                                                            c0bthorn (W ub) tan0(kNm)

                                                                                                                                            sec

                                                                                                                                            1thorn (tan tan0)FProduct(kNm)

                                                                                                                                            1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                            23 33 30 1042 31

                                                                                                                                            3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                            224 92 72 0931 67

                                                                                                                                            6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                            12 58 112 90 0889 80

                                                                                                                                            8 60 252 1812

                                                                                                                                            80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                            10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                            2154 88 116 0853 99

                                                                                                                                            14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                            1074 1091

                                                                                                                                            F frac14 1091

                                                                                                                                            1074frac14 102 (assumed value 100)

                                                                                                                                            Thus

                                                                                                                                            F frac14 101

                                                                                                                                            95

                                                                                                                                            F frac14 1

                                                                                                                                            W sin

                                                                                                                                            XfWeth1 ruTHORN tan0g sec

                                                                                                                                            1thorn ethtan tan0THORN=F

                                                                                                                                            0 frac14 33

                                                                                                                                            ru frac14 020

                                                                                                                                            W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                            eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                            Try F frac14 110

                                                                                                                                            tan 0

                                                                                                                                            Ffrac14 tan 33

                                                                                                                                            110frac14 0590

                                                                                                                                            78 Stability of slopes

                                                                                                                                            Referring to Figure Q95

                                                                                                                                            SliceNo

                                                                                                                                            h(m)

                                                                                                                                            W frac14 bh(kNm)

                                                                                                                                            W sin(kNm)

                                                                                                                                            W(1 ru) tan0(kNm)

                                                                                                                                            sec

                                                                                                                                            1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                            1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                            2120 234 0892 209

                                                                                                                                            4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                            1185 1271

                                                                                                                                            Figure Q95

                                                                                                                                            Stability of slopes 79

                                                                                                                                            F frac14 1271

                                                                                                                                            1185frac14 107

                                                                                                                                            The trial value was 110 therefore take F to be 108

                                                                                                                                            96

                                                                                                                                            (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                            F frac14 0

                                                                                                                                            sat

                                                                                                                                            tan0

                                                                                                                                            tan

                                                                                                                                            ptie 15 frac14 92

                                                                                                                                            19

                                                                                                                                            tan 36

                                                                                                                                            tan

                                                                                                                                            tan frac14 0234

                                                                                                                                            frac14 13

                                                                                                                                            Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                            F frac14 tan0

                                                                                                                                            tan

                                                                                                                                            frac14 tan 36

                                                                                                                                            tan 13

                                                                                                                                            frac14 31

                                                                                                                                            (b) 0d frac14 tan1tan 36

                                                                                                                                            125

                                                                                                                                            frac14 30

                                                                                                                                            Depth of potential failure surface frac14 z

                                                                                                                                            Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                            frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                            frac14 504z kN

                                                                                                                                            Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                            frac14 19 z sin 13 cos 13

                                                                                                                                            frac14 416z kN

                                                                                                                                            Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                            80 Stability of slopes

                                                                                                                                            • Book Cover
                                                                                                                                            • Title
                                                                                                                                            • Contents
                                                                                                                                            • Basic characteristics of soils
                                                                                                                                            • Seepage
                                                                                                                                            • Effective stress
                                                                                                                                            • Shear strength
                                                                                                                                            • Stresses and displacements
                                                                                                                                            • Lateral earth pressure
                                                                                                                                            • Consolidation theory
                                                                                                                                            • Bearing capacity
                                                                                                                                            • Stability of slopes

                                                                                                                                              For sc frac14 30mm

                                                                                                                                              qn frac14 30

                                                                                                                                              0147frac14 204 kN=m2

                                                                                                                                              q frac14 204thorn 21 frac14 225 kN=m2

                                                                                                                                              Design load frac14 42 225 frac14 3600 kN

                                                                                                                                              The design load is 3600 kN settlement being the limiting criterion

                                                                                                                                              87

                                                                                                                                              D

                                                                                                                                              Bfrac14 8

                                                                                                                                              4frac14 20

                                                                                                                                              From Figure 85 for a strip Nc frac14 71 For a depthbreadth ratio of 2 Equation 812should be used

                                                                                                                                              F frac14 cuNc

                                                                                                                                              Dfrac14 40 71

                                                                                                                                              20 8frac14 18

                                                                                                                                              88

                                                                                                                                              Design load for ultimate limit state Vd frac14 2500thorn (1250 130) frac14 4125 kN

                                                                                                                                              Design value of 0 frac14 tan1tan 38

                                                                                                                                              125

                                                                                                                                              frac14 32

                                                                                                                                              Figure Q86

                                                                                                                                              Bearing capacity 65

                                                                                                                                              For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                                              Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                                              The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                                              Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                                              For B frac14 250m qn frac14 3750

                                                                                                                                              2502 17 frac14 583 kN=m2

                                                                                                                                              From Figure 510 m frac14 n frac14 126

                                                                                                                                              6frac14 021

                                                                                                                                              Ir frac14 0019

                                                                                                                                              Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                                              Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                                              The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                              89

                                                                                                                                              Depth (m) N 0v (kNm2) CN N1

                                                                                                                                              070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                                              Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                                              (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                                              Cw frac14 05thorn 0530

                                                                                                                                              47

                                                                                                                                              frac14 082

                                                                                                                                              66 Bearing capacity

                                                                                                                                              Thus

                                                                                                                                              qa frac14 150 082 frac14 120 kN=m2

                                                                                                                                              (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                                              Thus

                                                                                                                                              qa frac14 90 15 frac14 135 kN=m2

                                                                                                                                              (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                                              Ic frac14 171

                                                                                                                                              1014frac14 0068

                                                                                                                                              From Equation 819(a) with s frac14 25mm

                                                                                                                                              q frac14 25

                                                                                                                                              3507 0068frac14 150 kN=m2

                                                                                                                                              810

                                                                                                                                              Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                                              Izp frac14 05thorn 01130

                                                                                                                                              16 27

                                                                                                                                              05

                                                                                                                                              frac14 067

                                                                                                                                              Refer to Figure Q810

                                                                                                                                              E frac14 25qc

                                                                                                                                              Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                                              Ez (mm3MN)

                                                                                                                                              1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                                              0203

                                                                                                                                              C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                                              130frac14 093

                                                                                                                                              C2 frac14 1 ethsayTHORN

                                                                                                                                              s frac14 C1C2qnX Iz

                                                                                                                                              Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                                              Bearing capacity 67

                                                                                                                                              811

                                                                                                                                              At pile base level

                                                                                                                                              cu frac14 220 kN=m2

                                                                                                                                              qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                                              Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                                              00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                                              qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                                              Then

                                                                                                                                              Qf frac14 Abqb thorn Asqs

                                                                                                                                              frac14

                                                                                                                                              4 32 1980

                                                                                                                                              thorn eth 105 139 86THORN

                                                                                                                                              frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                                              0 01 02 03 04 05 06 07

                                                                                                                                              0 2 4 6 8 10 12 14

                                                                                                                                              1

                                                                                                                                              2

                                                                                                                                              3

                                                                                                                                              4

                                                                                                                                              5

                                                                                                                                              6

                                                                                                                                              7

                                                                                                                                              8

                                                                                                                                              (1)

                                                                                                                                              (2)

                                                                                                                                              (3)

                                                                                                                                              (4)

                                                                                                                                              (5)

                                                                                                                                              qc

                                                                                                                                              qc

                                                                                                                                              Iz

                                                                                                                                              Iz

                                                                                                                                              (MNm2)

                                                                                                                                              z (m)

                                                                                                                                              Figure Q810

                                                                                                                                              68 Bearing capacity

                                                                                                                                              Allowable load

                                                                                                                                              ethaTHORN Qf

                                                                                                                                              2frac14 17 937

                                                                                                                                              2frac14 8968 kN

                                                                                                                                              ethbTHORN Abqb

                                                                                                                                              3thorn Asqs frac14 13 996

                                                                                                                                              3thorn 3941 frac14 8606 kN

                                                                                                                                              ie allowable load frac14 8600 kN

                                                                                                                                              Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                                              According to the limit state method

                                                                                                                                              Characteristic undrained strength at base level cuk frac14 220

                                                                                                                                              150kN=m2

                                                                                                                                              Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                                              150frac14 1320 kN=m2

                                                                                                                                              Characteristic shaft resistance qsk frac14 00150

                                                                                                                                              frac14 86

                                                                                                                                              150frac14 57 kN=m2

                                                                                                                                              Characteristic base and shaft resistances

                                                                                                                                              Rbk frac14

                                                                                                                                              4 32 1320 frac14 9330 kN

                                                                                                                                              Rsk frac14 105 139 86

                                                                                                                                              150frac14 2629 kN

                                                                                                                                              For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                                              Design bearing resistance Rcd frac14 9330

                                                                                                                                              160thorn 2629

                                                                                                                                              130

                                                                                                                                              frac14 5831thorn 2022

                                                                                                                                              frac14 7850 kN

                                                                                                                                              Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                                              812

                                                                                                                                              ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                                              qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                                              For a single pile

                                                                                                                                              Qf frac14 Abqb thorn Asqs

                                                                                                                                              frac14

                                                                                                                                              4 062 1305

                                                                                                                                              thorn eth 06 15 42THORN

                                                                                                                                              frac14 369thorn 1187 frac14 1556 kN

                                                                                                                                              Bearing capacity 69

                                                                                                                                              Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                                              eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                                              (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                                              (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                                              qbkfrac14 9cuk frac14 9 220

                                                                                                                                              150frac14 1320 kN=m2

                                                                                                                                              qskfrac14cuk frac14 040 105

                                                                                                                                              150frac14 28 kN=m2

                                                                                                                                              Rbkfrac14

                                                                                                                                              4 0602 1320 frac14 373 kN

                                                                                                                                              Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                                              Rcdfrac14 373

                                                                                                                                              160thorn 791

                                                                                                                                              130frac14 233thorn 608 frac14 841 kN

                                                                                                                                              Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                                              Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                                              (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                                              q frac14 21 000

                                                                                                                                              1762frac14 68 kN=m2

                                                                                                                                              Immediate settlement

                                                                                                                                              H

                                                                                                                                              Bfrac14 15

                                                                                                                                              176frac14 085

                                                                                                                                              D

                                                                                                                                              Bfrac14 13

                                                                                                                                              176frac14 074

                                                                                                                                              L

                                                                                                                                              Bfrac14 1

                                                                                                                                              Hence from Figure 515

                                                                                                                                              130 frac14 078 and 131 frac14 041

                                                                                                                                              70 Bearing capacity

                                                                                                                                              Thus using Equation 528

                                                                                                                                              si frac14 078 041 68 176

                                                                                                                                              65frac14 6mm

                                                                                                                                              Consolidation settlement

                                                                                                                                              Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                                              1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                                              434 (sod)

                                                                                                                                              Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                                              sc frac14 056 434 frac14 24mm

                                                                                                                                              The total settlement is (6thorn 24) frac14 30mm

                                                                                                                                              813

                                                                                                                                              At base level N frac14 26 Then using Equation 830

                                                                                                                                              qb frac14 40NDb

                                                                                                                                              Bfrac14 40 26 2

                                                                                                                                              025frac14 8320 kN=m2

                                                                                                                                              ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                                              Figure Q812

                                                                                                                                              Bearing capacity 71

                                                                                                                                              Over the length embedded in sand

                                                                                                                                              N frac14 21 ie18thorn 24

                                                                                                                                              2

                                                                                                                                              Using Equation 831

                                                                                                                                              qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                              For a single pile

                                                                                                                                              Qf frac14 Abqb thorn Asqs

                                                                                                                                              frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                              For the pile group assuming a group efficiency of 12

                                                                                                                                              XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                              Then the load factor is

                                                                                                                                              F frac14 6523

                                                                                                                                              2000thorn 1000frac14 21

                                                                                                                                              (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                              Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                              150frac14 5547 kNm2

                                                                                                                                              Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                              150frac14 28 kNm2

                                                                                                                                              Characteristic base and shaft resistances for a single pile

                                                                                                                                              Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                              Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                              For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                              Design bearing resistance Rcd frac14 347

                                                                                                                                              130thorn 56

                                                                                                                                              130frac14 310 kN

                                                                                                                                              For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                              Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                              (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                              From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                              72 Bearing capacity

                                                                                                                                              N frac14 24thorn 26thorn 34

                                                                                                                                              3frac14 28

                                                                                                                                              Ic frac14 171

                                                                                                                                              2814frac14 0016 ethEquation 818THORN

                                                                                                                                              s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                              814

                                                                                                                                              Using Equation 841

                                                                                                                                              Tf frac14 DLcu thorn

                                                                                                                                              4ethD2 d2THORNcuNc

                                                                                                                                              frac14 eth 02 5 06 110THORN thorn

                                                                                                                                              4eth022 012THORN110 9

                                                                                                                                              frac14 207thorn 23 frac14 230 kN

                                                                                                                                              Figure Q813

                                                                                                                                              Bearing capacity 73

                                                                                                                                              Chapter 9

                                                                                                                                              Stability of slopes

                                                                                                                                              91

                                                                                                                                              Referring to Figure Q91

                                                                                                                                              W frac14 417 19 frac14 792 kN=m

                                                                                                                                              Q frac14 20 28 frac14 56 kN=m

                                                                                                                                              Arc lengthAB frac14

                                                                                                                                              180 73 90 frac14 115m

                                                                                                                                              Arc length BC frac14

                                                                                                                                              180 28 90 frac14 44m

                                                                                                                                              The factor of safety is given by

                                                                                                                                              F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                              90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                              Depth of tension crack z0 frac14 2cu

                                                                                                                                              frac14 2 20

                                                                                                                                              19frac14 21m

                                                                                                                                              Arc length BD frac14

                                                                                                                                              180 13

                                                                                                                                              1

                                                                                                                                              2 90 frac14 21m

                                                                                                                                              F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                              The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                              Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                              14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                              Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                              The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                              92

                                                                                                                                              u frac14 0

                                                                                                                                              Depth factor D frac14 11

                                                                                                                                              9frac14 122

                                                                                                                                              Using Equation 92 with F frac14 10

                                                                                                                                              Ns frac14 cu

                                                                                                                                              FHfrac14 30

                                                                                                                                              10 19 9frac14 0175

                                                                                                                                              Hence from Figure 93

                                                                                                                                              frac14 50

                                                                                                                                              For F frac14 12

                                                                                                                                              Ns frac14 30

                                                                                                                                              12 19 9frac14 0146

                                                                                                                                              frac14 27

                                                                                                                                              93

                                                                                                                                              Refer to Figure Q93

                                                                                                                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                              1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                              74 m

                                                                                                                                              214 1deg

                                                                                                                                              213 1deg

                                                                                                                                              39 m

                                                                                                                                              WB

                                                                                                                                              D

                                                                                                                                              C

                                                                                                                                              28 m

                                                                                                                                              21 m

                                                                                                                                              A

                                                                                                                                              Q

                                                                                                                                              Soil (1)Soil (2)

                                                                                                                                              73deg

                                                                                                                                              Figure Q91

                                                                                                                                              Stability of slopes 75

                                                                                                                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                              9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                              599 256 328 1372

                                                                                                                                              Figure Q93

                                                                                                                                              76 Stability of slopes

                                                                                                                                              XW cos frac14 b

                                                                                                                                              Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                              W sin frac14 bX

                                                                                                                                              h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                              Arc length La frac14

                                                                                                                                              180 57

                                                                                                                                              1

                                                                                                                                              2 326 frac14 327m

                                                                                                                                              The factor of safety is given by

                                                                                                                                              F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                              W sin

                                                                                                                                              frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                              frac14 091

                                                                                                                                              According to the limit state method

                                                                                                                                              0d frac14 tan1tan 32

                                                                                                                                              125

                                                                                                                                              frac14 265

                                                                                                                                              c0 frac14 8

                                                                                                                                              160frac14 5 kN=m2

                                                                                                                                              Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                              Design disturbing moment frac14 1075 kN=m

                                                                                                                                              The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                              94

                                                                                                                                              F frac14 1

                                                                                                                                              W sin

                                                                                                                                              Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                              1thorn ethtan tan0=FTHORN

                                                                                                                                              c0 frac14 8 kN=m2

                                                                                                                                              0 frac14 32

                                                                                                                                              c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                              W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                              Try F frac14 100

                                                                                                                                              tan0

                                                                                                                                              Ffrac14 0625

                                                                                                                                              Stability of slopes 77

                                                                                                                                              Values of u are as obtained in Figure Q93

                                                                                                                                              SliceNo

                                                                                                                                              h(m)

                                                                                                                                              W frac14 bh(kNm)

                                                                                                                                              W sin(kNm)

                                                                                                                                              ub(kNm)

                                                                                                                                              c0bthorn (W ub) tan0(kNm)

                                                                                                                                              sec

                                                                                                                                              1thorn (tan tan0)FProduct(kNm)

                                                                                                                                              1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                              23 33 30 1042 31

                                                                                                                                              3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                              224 92 72 0931 67

                                                                                                                                              6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                              12 58 112 90 0889 80

                                                                                                                                              8 60 252 1812

                                                                                                                                              80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                              10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                              2154 88 116 0853 99

                                                                                                                                              14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                              1074 1091

                                                                                                                                              F frac14 1091

                                                                                                                                              1074frac14 102 (assumed value 100)

                                                                                                                                              Thus

                                                                                                                                              F frac14 101

                                                                                                                                              95

                                                                                                                                              F frac14 1

                                                                                                                                              W sin

                                                                                                                                              XfWeth1 ruTHORN tan0g sec

                                                                                                                                              1thorn ethtan tan0THORN=F

                                                                                                                                              0 frac14 33

                                                                                                                                              ru frac14 020

                                                                                                                                              W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                              eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                              Try F frac14 110

                                                                                                                                              tan 0

                                                                                                                                              Ffrac14 tan 33

                                                                                                                                              110frac14 0590

                                                                                                                                              78 Stability of slopes

                                                                                                                                              Referring to Figure Q95

                                                                                                                                              SliceNo

                                                                                                                                              h(m)

                                                                                                                                              W frac14 bh(kNm)

                                                                                                                                              W sin(kNm)

                                                                                                                                              W(1 ru) tan0(kNm)

                                                                                                                                              sec

                                                                                                                                              1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                              1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                              2120 234 0892 209

                                                                                                                                              4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                              1185 1271

                                                                                                                                              Figure Q95

                                                                                                                                              Stability of slopes 79

                                                                                                                                              F frac14 1271

                                                                                                                                              1185frac14 107

                                                                                                                                              The trial value was 110 therefore take F to be 108

                                                                                                                                              96

                                                                                                                                              (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                              F frac14 0

                                                                                                                                              sat

                                                                                                                                              tan0

                                                                                                                                              tan

                                                                                                                                              ptie 15 frac14 92

                                                                                                                                              19

                                                                                                                                              tan 36

                                                                                                                                              tan

                                                                                                                                              tan frac14 0234

                                                                                                                                              frac14 13

                                                                                                                                              Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                              F frac14 tan0

                                                                                                                                              tan

                                                                                                                                              frac14 tan 36

                                                                                                                                              tan 13

                                                                                                                                              frac14 31

                                                                                                                                              (b) 0d frac14 tan1tan 36

                                                                                                                                              125

                                                                                                                                              frac14 30

                                                                                                                                              Depth of potential failure surface frac14 z

                                                                                                                                              Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                              frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                              frac14 504z kN

                                                                                                                                              Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                              frac14 19 z sin 13 cos 13

                                                                                                                                              frac14 416z kN

                                                                                                                                              Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                              80 Stability of slopes

                                                                                                                                              • Book Cover
                                                                                                                                              • Title
                                                                                                                                              • Contents
                                                                                                                                              • Basic characteristics of soils
                                                                                                                                              • Seepage
                                                                                                                                              • Effective stress
                                                                                                                                              • Shear strength
                                                                                                                                              • Stresses and displacements
                                                                                                                                              • Lateral earth pressure
                                                                                                                                              • Consolidation theory
                                                                                                                                              • Bearing capacity
                                                                                                                                              • Stability of slopes

                                                                                                                                                For 0 frac14 32 Nq frac14 23 and N frac14 25

                                                                                                                                                Design bearing resistance Rd frac14 2502frac12eth17 10 23THORN thorn eth04 102 250 25THORNfrac14 2502eth391thorn 255THORNfrac14 4037 kN

                                                                                                                                                The design bearing resistance is (slightly) less than the design load therefore thebearing resistance limit state is not satisfied To satisfy the limit state the dimensionof the foundation should be increased to 253m

                                                                                                                                                Design load for serviceability limit state frac14 2500thorn 1250 frac14 3750 kN

                                                                                                                                                For B frac14 250m qn frac14 3750

                                                                                                                                                2502 17 frac14 583 kN=m2

                                                                                                                                                From Figure 510 m frac14 n frac14 126

                                                                                                                                                6frac14 021

                                                                                                                                                Ir frac14 0019

                                                                                                                                                Stress increment frac14 4 0019 583 frac14 44 kN=m2

                                                                                                                                                Consolidation settlement sc frac14 mvH frac14 015 44 2 frac14 13mm

                                                                                                                                                The settlement would be less than 13mm if an appropriate value of settlementcoefficient (Figure 712) was applied

                                                                                                                                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                                89

                                                                                                                                                Depth (m) N 0v (kNm2) CN N1

                                                                                                                                                070 6 ndash ndash ndash135 9 23 190 17220 10 37 155 15295 8 50 137 11365 12 58 128 15440 13 65 123 16515 17 ndash ndash ndash600 23 ndash ndash ndash

                                                                                                                                                Note Using frac14 17 kNm3 and 0 frac14 10 kNm3

                                                                                                                                                (a) Terzaghi and Peck Use N1 values between depths of 12 and 47m the averagevalue being 15 For B frac14 35m and N frac14 15 the provisional value of bearing capacityusing Figure 810 is 150 kNm2 The water table correction factor (Equation 816) is

                                                                                                                                                Cw frac14 05thorn 0530

                                                                                                                                                47

                                                                                                                                                frac14 082

                                                                                                                                                66 Bearing capacity

                                                                                                                                                Thus

                                                                                                                                                qa frac14 150 082 frac14 120 kN=m2

                                                                                                                                                (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                                                Thus

                                                                                                                                                qa frac14 90 15 frac14 135 kN=m2

                                                                                                                                                (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                                                Ic frac14 171

                                                                                                                                                1014frac14 0068

                                                                                                                                                From Equation 819(a) with s frac14 25mm

                                                                                                                                                q frac14 25

                                                                                                                                                3507 0068frac14 150 kN=m2

                                                                                                                                                810

                                                                                                                                                Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                                                Izp frac14 05thorn 01130

                                                                                                                                                16 27

                                                                                                                                                05

                                                                                                                                                frac14 067

                                                                                                                                                Refer to Figure Q810

                                                                                                                                                E frac14 25qc

                                                                                                                                                Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                                                Ez (mm3MN)

                                                                                                                                                1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                                                0203

                                                                                                                                                C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                                                130frac14 093

                                                                                                                                                C2 frac14 1 ethsayTHORN

                                                                                                                                                s frac14 C1C2qnX Iz

                                                                                                                                                Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                                                Bearing capacity 67

                                                                                                                                                811

                                                                                                                                                At pile base level

                                                                                                                                                cu frac14 220 kN=m2

                                                                                                                                                qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                                                Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                                                00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                                                qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                                                Then

                                                                                                                                                Qf frac14 Abqb thorn Asqs

                                                                                                                                                frac14

                                                                                                                                                4 32 1980

                                                                                                                                                thorn eth 105 139 86THORN

                                                                                                                                                frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                                                0 01 02 03 04 05 06 07

                                                                                                                                                0 2 4 6 8 10 12 14

                                                                                                                                                1

                                                                                                                                                2

                                                                                                                                                3

                                                                                                                                                4

                                                                                                                                                5

                                                                                                                                                6

                                                                                                                                                7

                                                                                                                                                8

                                                                                                                                                (1)

                                                                                                                                                (2)

                                                                                                                                                (3)

                                                                                                                                                (4)

                                                                                                                                                (5)

                                                                                                                                                qc

                                                                                                                                                qc

                                                                                                                                                Iz

                                                                                                                                                Iz

                                                                                                                                                (MNm2)

                                                                                                                                                z (m)

                                                                                                                                                Figure Q810

                                                                                                                                                68 Bearing capacity

                                                                                                                                                Allowable load

                                                                                                                                                ethaTHORN Qf

                                                                                                                                                2frac14 17 937

                                                                                                                                                2frac14 8968 kN

                                                                                                                                                ethbTHORN Abqb

                                                                                                                                                3thorn Asqs frac14 13 996

                                                                                                                                                3thorn 3941 frac14 8606 kN

                                                                                                                                                ie allowable load frac14 8600 kN

                                                                                                                                                Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                                                According to the limit state method

                                                                                                                                                Characteristic undrained strength at base level cuk frac14 220

                                                                                                                                                150kN=m2

                                                                                                                                                Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                                                150frac14 1320 kN=m2

                                                                                                                                                Characteristic shaft resistance qsk frac14 00150

                                                                                                                                                frac14 86

                                                                                                                                                150frac14 57 kN=m2

                                                                                                                                                Characteristic base and shaft resistances

                                                                                                                                                Rbk frac14

                                                                                                                                                4 32 1320 frac14 9330 kN

                                                                                                                                                Rsk frac14 105 139 86

                                                                                                                                                150frac14 2629 kN

                                                                                                                                                For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                                                Design bearing resistance Rcd frac14 9330

                                                                                                                                                160thorn 2629

                                                                                                                                                130

                                                                                                                                                frac14 5831thorn 2022

                                                                                                                                                frac14 7850 kN

                                                                                                                                                Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                                                812

                                                                                                                                                ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                                                qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                                                For a single pile

                                                                                                                                                Qf frac14 Abqb thorn Asqs

                                                                                                                                                frac14

                                                                                                                                                4 062 1305

                                                                                                                                                thorn eth 06 15 42THORN

                                                                                                                                                frac14 369thorn 1187 frac14 1556 kN

                                                                                                                                                Bearing capacity 69

                                                                                                                                                Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                                                eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                                                (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                                                (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                                                qbkfrac14 9cuk frac14 9 220

                                                                                                                                                150frac14 1320 kN=m2

                                                                                                                                                qskfrac14cuk frac14 040 105

                                                                                                                                                150frac14 28 kN=m2

                                                                                                                                                Rbkfrac14

                                                                                                                                                4 0602 1320 frac14 373 kN

                                                                                                                                                Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                                                Rcdfrac14 373

                                                                                                                                                160thorn 791

                                                                                                                                                130frac14 233thorn 608 frac14 841 kN

                                                                                                                                                Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                                                Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                                                (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                                                q frac14 21 000

                                                                                                                                                1762frac14 68 kN=m2

                                                                                                                                                Immediate settlement

                                                                                                                                                H

                                                                                                                                                Bfrac14 15

                                                                                                                                                176frac14 085

                                                                                                                                                D

                                                                                                                                                Bfrac14 13

                                                                                                                                                176frac14 074

                                                                                                                                                L

                                                                                                                                                Bfrac14 1

                                                                                                                                                Hence from Figure 515

                                                                                                                                                130 frac14 078 and 131 frac14 041

                                                                                                                                                70 Bearing capacity

                                                                                                                                                Thus using Equation 528

                                                                                                                                                si frac14 078 041 68 176

                                                                                                                                                65frac14 6mm

                                                                                                                                                Consolidation settlement

                                                                                                                                                Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                                                1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                                                434 (sod)

                                                                                                                                                Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                                                sc frac14 056 434 frac14 24mm

                                                                                                                                                The total settlement is (6thorn 24) frac14 30mm

                                                                                                                                                813

                                                                                                                                                At base level N frac14 26 Then using Equation 830

                                                                                                                                                qb frac14 40NDb

                                                                                                                                                Bfrac14 40 26 2

                                                                                                                                                025frac14 8320 kN=m2

                                                                                                                                                ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                                                Figure Q812

                                                                                                                                                Bearing capacity 71

                                                                                                                                                Over the length embedded in sand

                                                                                                                                                N frac14 21 ie18thorn 24

                                                                                                                                                2

                                                                                                                                                Using Equation 831

                                                                                                                                                qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                                For a single pile

                                                                                                                                                Qf frac14 Abqb thorn Asqs

                                                                                                                                                frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                                For the pile group assuming a group efficiency of 12

                                                                                                                                                XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                                Then the load factor is

                                                                                                                                                F frac14 6523

                                                                                                                                                2000thorn 1000frac14 21

                                                                                                                                                (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                                Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                                150frac14 5547 kNm2

                                                                                                                                                Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                                150frac14 28 kNm2

                                                                                                                                                Characteristic base and shaft resistances for a single pile

                                                                                                                                                Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                                Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                                For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                                Design bearing resistance Rcd frac14 347

                                                                                                                                                130thorn 56

                                                                                                                                                130frac14 310 kN

                                                                                                                                                For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                                Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                                (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                                From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                                72 Bearing capacity

                                                                                                                                                N frac14 24thorn 26thorn 34

                                                                                                                                                3frac14 28

                                                                                                                                                Ic frac14 171

                                                                                                                                                2814frac14 0016 ethEquation 818THORN

                                                                                                                                                s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                                The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                                814

                                                                                                                                                Using Equation 841

                                                                                                                                                Tf frac14 DLcu thorn

                                                                                                                                                4ethD2 d2THORNcuNc

                                                                                                                                                frac14 eth 02 5 06 110THORN thorn

                                                                                                                                                4eth022 012THORN110 9

                                                                                                                                                frac14 207thorn 23 frac14 230 kN

                                                                                                                                                Figure Q813

                                                                                                                                                Bearing capacity 73

                                                                                                                                                Chapter 9

                                                                                                                                                Stability of slopes

                                                                                                                                                91

                                                                                                                                                Referring to Figure Q91

                                                                                                                                                W frac14 417 19 frac14 792 kN=m

                                                                                                                                                Q frac14 20 28 frac14 56 kN=m

                                                                                                                                                Arc lengthAB frac14

                                                                                                                                                180 73 90 frac14 115m

                                                                                                                                                Arc length BC frac14

                                                                                                                                                180 28 90 frac14 44m

                                                                                                                                                The factor of safety is given by

                                                                                                                                                F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                                90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                                Depth of tension crack z0 frac14 2cu

                                                                                                                                                frac14 2 20

                                                                                                                                                19frac14 21m

                                                                                                                                                Arc length BD frac14

                                                                                                                                                180 13

                                                                                                                                                1

                                                                                                                                                2 90 frac14 21m

                                                                                                                                                F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                                The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                                Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                                14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                                Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                                The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                                92

                                                                                                                                                u frac14 0

                                                                                                                                                Depth factor D frac14 11

                                                                                                                                                9frac14 122

                                                                                                                                                Using Equation 92 with F frac14 10

                                                                                                                                                Ns frac14 cu

                                                                                                                                                FHfrac14 30

                                                                                                                                                10 19 9frac14 0175

                                                                                                                                                Hence from Figure 93

                                                                                                                                                frac14 50

                                                                                                                                                For F frac14 12

                                                                                                                                                Ns frac14 30

                                                                                                                                                12 19 9frac14 0146

                                                                                                                                                frac14 27

                                                                                                                                                93

                                                                                                                                                Refer to Figure Q93

                                                                                                                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                                74 m

                                                                                                                                                214 1deg

                                                                                                                                                213 1deg

                                                                                                                                                39 m

                                                                                                                                                WB

                                                                                                                                                D

                                                                                                                                                C

                                                                                                                                                28 m

                                                                                                                                                21 m

                                                                                                                                                A

                                                                                                                                                Q

                                                                                                                                                Soil (1)Soil (2)

                                                                                                                                                73deg

                                                                                                                                                Figure Q91

                                                                                                                                                Stability of slopes 75

                                                                                                                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                                599 256 328 1372

                                                                                                                                                Figure Q93

                                                                                                                                                76 Stability of slopes

                                                                                                                                                XW cos frac14 b

                                                                                                                                                Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                                W sin frac14 bX

                                                                                                                                                h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                                Arc length La frac14

                                                                                                                                                180 57

                                                                                                                                                1

                                                                                                                                                2 326 frac14 327m

                                                                                                                                                The factor of safety is given by

                                                                                                                                                F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                                W sin

                                                                                                                                                frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                                frac14 091

                                                                                                                                                According to the limit state method

                                                                                                                                                0d frac14 tan1tan 32

                                                                                                                                                125

                                                                                                                                                frac14 265

                                                                                                                                                c0 frac14 8

                                                                                                                                                160frac14 5 kN=m2

                                                                                                                                                Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                                Design disturbing moment frac14 1075 kN=m

                                                                                                                                                The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                                94

                                                                                                                                                F frac14 1

                                                                                                                                                W sin

                                                                                                                                                Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                                1thorn ethtan tan0=FTHORN

                                                                                                                                                c0 frac14 8 kN=m2

                                                                                                                                                0 frac14 32

                                                                                                                                                c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                                W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                                Try F frac14 100

                                                                                                                                                tan0

                                                                                                                                                Ffrac14 0625

                                                                                                                                                Stability of slopes 77

                                                                                                                                                Values of u are as obtained in Figure Q93

                                                                                                                                                SliceNo

                                                                                                                                                h(m)

                                                                                                                                                W frac14 bh(kNm)

                                                                                                                                                W sin(kNm)

                                                                                                                                                ub(kNm)

                                                                                                                                                c0bthorn (W ub) tan0(kNm)

                                                                                                                                                sec

                                                                                                                                                1thorn (tan tan0)FProduct(kNm)

                                                                                                                                                1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                                23 33 30 1042 31

                                                                                                                                                3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                                224 92 72 0931 67

                                                                                                                                                6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                                12 58 112 90 0889 80

                                                                                                                                                8 60 252 1812

                                                                                                                                                80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                                10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                                2154 88 116 0853 99

                                                                                                                                                14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                                1074 1091

                                                                                                                                                F frac14 1091

                                                                                                                                                1074frac14 102 (assumed value 100)

                                                                                                                                                Thus

                                                                                                                                                F frac14 101

                                                                                                                                                95

                                                                                                                                                F frac14 1

                                                                                                                                                W sin

                                                                                                                                                XfWeth1 ruTHORN tan0g sec

                                                                                                                                                1thorn ethtan tan0THORN=F

                                                                                                                                                0 frac14 33

                                                                                                                                                ru frac14 020

                                                                                                                                                W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                                eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                                Try F frac14 110

                                                                                                                                                tan 0

                                                                                                                                                Ffrac14 tan 33

                                                                                                                                                110frac14 0590

                                                                                                                                                78 Stability of slopes

                                                                                                                                                Referring to Figure Q95

                                                                                                                                                SliceNo

                                                                                                                                                h(m)

                                                                                                                                                W frac14 bh(kNm)

                                                                                                                                                W sin(kNm)

                                                                                                                                                W(1 ru) tan0(kNm)

                                                                                                                                                sec

                                                                                                                                                1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                                1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                                2120 234 0892 209

                                                                                                                                                4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                                1185 1271

                                                                                                                                                Figure Q95

                                                                                                                                                Stability of slopes 79

                                                                                                                                                F frac14 1271

                                                                                                                                                1185frac14 107

                                                                                                                                                The trial value was 110 therefore take F to be 108

                                                                                                                                                96

                                                                                                                                                (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                F frac14 0

                                                                                                                                                sat

                                                                                                                                                tan0

                                                                                                                                                tan

                                                                                                                                                ptie 15 frac14 92

                                                                                                                                                19

                                                                                                                                                tan 36

                                                                                                                                                tan

                                                                                                                                                tan frac14 0234

                                                                                                                                                frac14 13

                                                                                                                                                Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                F frac14 tan0

                                                                                                                                                tan

                                                                                                                                                frac14 tan 36

                                                                                                                                                tan 13

                                                                                                                                                frac14 31

                                                                                                                                                (b) 0d frac14 tan1tan 36

                                                                                                                                                125

                                                                                                                                                frac14 30

                                                                                                                                                Depth of potential failure surface frac14 z

                                                                                                                                                Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                frac14 504z kN

                                                                                                                                                Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                frac14 19 z sin 13 cos 13

                                                                                                                                                frac14 416z kN

                                                                                                                                                Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                80 Stability of slopes

                                                                                                                                                • Book Cover
                                                                                                                                                • Title
                                                                                                                                                • Contents
                                                                                                                                                • Basic characteristics of soils
                                                                                                                                                • Seepage
                                                                                                                                                • Effective stress
                                                                                                                                                • Shear strength
                                                                                                                                                • Stresses and displacements
                                                                                                                                                • Lateral earth pressure
                                                                                                                                                • Consolidation theory
                                                                                                                                                • Bearing capacity
                                                                                                                                                • Stability of slopes

                                                                                                                                                  Thus

                                                                                                                                                  qa frac14 150 082 frac14 120 kN=m2

                                                                                                                                                  (b)Meyerhof Use uncorrected N values between depths of 12 and 47m the averagevalue being 10 For B frac14 35m and N frac14 10 the provisional value of bearing capacityusing Figure 810 is 90 kNm2 This value is increased by 50

                                                                                                                                                  Thus

                                                                                                                                                  qa frac14 90 15 frac14 135 kN=m2

                                                                                                                                                  (c) Burland and Burbidge Using Figure 812 for B frac14 35m z1 frac14 25m Use N valuesbetween depths of 12 and 37m the average value being 10 From Equation 818

                                                                                                                                                  Ic frac14 171

                                                                                                                                                  1014frac14 0068

                                                                                                                                                  From Equation 819(a) with s frac14 25mm

                                                                                                                                                  q frac14 25

                                                                                                                                                  3507 0068frac14 150 kN=m2

                                                                                                                                                  810

                                                                                                                                                  Peak value of strain influence factor occurs at a depth of 27m and is given by

                                                                                                                                                  Izp frac14 05thorn 01130

                                                                                                                                                  16 27

                                                                                                                                                  05

                                                                                                                                                  frac14 067

                                                                                                                                                  Refer to Figure Q810

                                                                                                                                                  E frac14 25qc

                                                                                                                                                  Layer z (m) qc (MNm2) E frac14 25qc (MNm2) IzIz

                                                                                                                                                  Ez (mm3MN)

                                                                                                                                                  1 12 26 65 033 00612 08 50 125 065 00423 16 40 100 048 00774 16 72 180 024 00215 08 124 310 007 0002

                                                                                                                                                  0203

                                                                                                                                                  C1 frac14 1 0500qnfrac14 1 05 12 16

                                                                                                                                                  130frac14 093

                                                                                                                                                  C2 frac14 1 ethsayTHORN

                                                                                                                                                  s frac14 C1C2qnX Iz

                                                                                                                                                  Ez frac14 093 1 130 0203 frac14 25mm

                                                                                                                                                  Bearing capacity 67

                                                                                                                                                  811

                                                                                                                                                  At pile base level

                                                                                                                                                  cu frac14 220 kN=m2

                                                                                                                                                  qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                                                  Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                                                  00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                                                  qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                                                  Then

                                                                                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                                                                                  frac14

                                                                                                                                                  4 32 1980

                                                                                                                                                  thorn eth 105 139 86THORN

                                                                                                                                                  frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                                                  0 01 02 03 04 05 06 07

                                                                                                                                                  0 2 4 6 8 10 12 14

                                                                                                                                                  1

                                                                                                                                                  2

                                                                                                                                                  3

                                                                                                                                                  4

                                                                                                                                                  5

                                                                                                                                                  6

                                                                                                                                                  7

                                                                                                                                                  8

                                                                                                                                                  (1)

                                                                                                                                                  (2)

                                                                                                                                                  (3)

                                                                                                                                                  (4)

                                                                                                                                                  (5)

                                                                                                                                                  qc

                                                                                                                                                  qc

                                                                                                                                                  Iz

                                                                                                                                                  Iz

                                                                                                                                                  (MNm2)

                                                                                                                                                  z (m)

                                                                                                                                                  Figure Q810

                                                                                                                                                  68 Bearing capacity

                                                                                                                                                  Allowable load

                                                                                                                                                  ethaTHORN Qf

                                                                                                                                                  2frac14 17 937

                                                                                                                                                  2frac14 8968 kN

                                                                                                                                                  ethbTHORN Abqb

                                                                                                                                                  3thorn Asqs frac14 13 996

                                                                                                                                                  3thorn 3941 frac14 8606 kN

                                                                                                                                                  ie allowable load frac14 8600 kN

                                                                                                                                                  Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                                                  According to the limit state method

                                                                                                                                                  Characteristic undrained strength at base level cuk frac14 220

                                                                                                                                                  150kN=m2

                                                                                                                                                  Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                                                  150frac14 1320 kN=m2

                                                                                                                                                  Characteristic shaft resistance qsk frac14 00150

                                                                                                                                                  frac14 86

                                                                                                                                                  150frac14 57 kN=m2

                                                                                                                                                  Characteristic base and shaft resistances

                                                                                                                                                  Rbk frac14

                                                                                                                                                  4 32 1320 frac14 9330 kN

                                                                                                                                                  Rsk frac14 105 139 86

                                                                                                                                                  150frac14 2629 kN

                                                                                                                                                  For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                                                  Design bearing resistance Rcd frac14 9330

                                                                                                                                                  160thorn 2629

                                                                                                                                                  130

                                                                                                                                                  frac14 5831thorn 2022

                                                                                                                                                  frac14 7850 kN

                                                                                                                                                  Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                                                  812

                                                                                                                                                  ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                                                  qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                                                  For a single pile

                                                                                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                                                                                  frac14

                                                                                                                                                  4 062 1305

                                                                                                                                                  thorn eth 06 15 42THORN

                                                                                                                                                  frac14 369thorn 1187 frac14 1556 kN

                                                                                                                                                  Bearing capacity 69

                                                                                                                                                  Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                                                  eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                                                  (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                                                  (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                                                  qbkfrac14 9cuk frac14 9 220

                                                                                                                                                  150frac14 1320 kN=m2

                                                                                                                                                  qskfrac14cuk frac14 040 105

                                                                                                                                                  150frac14 28 kN=m2

                                                                                                                                                  Rbkfrac14

                                                                                                                                                  4 0602 1320 frac14 373 kN

                                                                                                                                                  Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                                                  Rcdfrac14 373

                                                                                                                                                  160thorn 791

                                                                                                                                                  130frac14 233thorn 608 frac14 841 kN

                                                                                                                                                  Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                                                  Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                                                  (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                                                  q frac14 21 000

                                                                                                                                                  1762frac14 68 kN=m2

                                                                                                                                                  Immediate settlement

                                                                                                                                                  H

                                                                                                                                                  Bfrac14 15

                                                                                                                                                  176frac14 085

                                                                                                                                                  D

                                                                                                                                                  Bfrac14 13

                                                                                                                                                  176frac14 074

                                                                                                                                                  L

                                                                                                                                                  Bfrac14 1

                                                                                                                                                  Hence from Figure 515

                                                                                                                                                  130 frac14 078 and 131 frac14 041

                                                                                                                                                  70 Bearing capacity

                                                                                                                                                  Thus using Equation 528

                                                                                                                                                  si frac14 078 041 68 176

                                                                                                                                                  65frac14 6mm

                                                                                                                                                  Consolidation settlement

                                                                                                                                                  Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                                                  1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                                                  434 (sod)

                                                                                                                                                  Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                                                  sc frac14 056 434 frac14 24mm

                                                                                                                                                  The total settlement is (6thorn 24) frac14 30mm

                                                                                                                                                  813

                                                                                                                                                  At base level N frac14 26 Then using Equation 830

                                                                                                                                                  qb frac14 40NDb

                                                                                                                                                  Bfrac14 40 26 2

                                                                                                                                                  025frac14 8320 kN=m2

                                                                                                                                                  ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                                                  Figure Q812

                                                                                                                                                  Bearing capacity 71

                                                                                                                                                  Over the length embedded in sand

                                                                                                                                                  N frac14 21 ie18thorn 24

                                                                                                                                                  2

                                                                                                                                                  Using Equation 831

                                                                                                                                                  qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                                  For a single pile

                                                                                                                                                  Qf frac14 Abqb thorn Asqs

                                                                                                                                                  frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                                  For the pile group assuming a group efficiency of 12

                                                                                                                                                  XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                                  Then the load factor is

                                                                                                                                                  F frac14 6523

                                                                                                                                                  2000thorn 1000frac14 21

                                                                                                                                                  (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                                  Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                                  150frac14 5547 kNm2

                                                                                                                                                  Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                                  150frac14 28 kNm2

                                                                                                                                                  Characteristic base and shaft resistances for a single pile

                                                                                                                                                  Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                                  Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                                  For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                                  Design bearing resistance Rcd frac14 347

                                                                                                                                                  130thorn 56

                                                                                                                                                  130frac14 310 kN

                                                                                                                                                  For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                                  Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                                  (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                                  From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                                  72 Bearing capacity

                                                                                                                                                  N frac14 24thorn 26thorn 34

                                                                                                                                                  3frac14 28

                                                                                                                                                  Ic frac14 171

                                                                                                                                                  2814frac14 0016 ethEquation 818THORN

                                                                                                                                                  s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                                  The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                                  814

                                                                                                                                                  Using Equation 841

                                                                                                                                                  Tf frac14 DLcu thorn

                                                                                                                                                  4ethD2 d2THORNcuNc

                                                                                                                                                  frac14 eth 02 5 06 110THORN thorn

                                                                                                                                                  4eth022 012THORN110 9

                                                                                                                                                  frac14 207thorn 23 frac14 230 kN

                                                                                                                                                  Figure Q813

                                                                                                                                                  Bearing capacity 73

                                                                                                                                                  Chapter 9

                                                                                                                                                  Stability of slopes

                                                                                                                                                  91

                                                                                                                                                  Referring to Figure Q91

                                                                                                                                                  W frac14 417 19 frac14 792 kN=m

                                                                                                                                                  Q frac14 20 28 frac14 56 kN=m

                                                                                                                                                  Arc lengthAB frac14

                                                                                                                                                  180 73 90 frac14 115m

                                                                                                                                                  Arc length BC frac14

                                                                                                                                                  180 28 90 frac14 44m

                                                                                                                                                  The factor of safety is given by

                                                                                                                                                  F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                                  90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                                  Depth of tension crack z0 frac14 2cu

                                                                                                                                                  frac14 2 20

                                                                                                                                                  19frac14 21m

                                                                                                                                                  Arc length BD frac14

                                                                                                                                                  180 13

                                                                                                                                                  1

                                                                                                                                                  2 90 frac14 21m

                                                                                                                                                  F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                                  The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                                  Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                                  14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                                  Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                                  The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                                  92

                                                                                                                                                  u frac14 0

                                                                                                                                                  Depth factor D frac14 11

                                                                                                                                                  9frac14 122

                                                                                                                                                  Using Equation 92 with F frac14 10

                                                                                                                                                  Ns frac14 cu

                                                                                                                                                  FHfrac14 30

                                                                                                                                                  10 19 9frac14 0175

                                                                                                                                                  Hence from Figure 93

                                                                                                                                                  frac14 50

                                                                                                                                                  For F frac14 12

                                                                                                                                                  Ns frac14 30

                                                                                                                                                  12 19 9frac14 0146

                                                                                                                                                  frac14 27

                                                                                                                                                  93

                                                                                                                                                  Refer to Figure Q93

                                                                                                                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                  1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                                  74 m

                                                                                                                                                  214 1deg

                                                                                                                                                  213 1deg

                                                                                                                                                  39 m

                                                                                                                                                  WB

                                                                                                                                                  D

                                                                                                                                                  C

                                                                                                                                                  28 m

                                                                                                                                                  21 m

                                                                                                                                                  A

                                                                                                                                                  Q

                                                                                                                                                  Soil (1)Soil (2)

                                                                                                                                                  73deg

                                                                                                                                                  Figure Q91

                                                                                                                                                  Stability of slopes 75

                                                                                                                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                  9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                                  599 256 328 1372

                                                                                                                                                  Figure Q93

                                                                                                                                                  76 Stability of slopes

                                                                                                                                                  XW cos frac14 b

                                                                                                                                                  Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                                  W sin frac14 bX

                                                                                                                                                  h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                                  Arc length La frac14

                                                                                                                                                  180 57

                                                                                                                                                  1

                                                                                                                                                  2 326 frac14 327m

                                                                                                                                                  The factor of safety is given by

                                                                                                                                                  F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                                  W sin

                                                                                                                                                  frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                                  frac14 091

                                                                                                                                                  According to the limit state method

                                                                                                                                                  0d frac14 tan1tan 32

                                                                                                                                                  125

                                                                                                                                                  frac14 265

                                                                                                                                                  c0 frac14 8

                                                                                                                                                  160frac14 5 kN=m2

                                                                                                                                                  Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                                  Design disturbing moment frac14 1075 kN=m

                                                                                                                                                  The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                                  94

                                                                                                                                                  F frac14 1

                                                                                                                                                  W sin

                                                                                                                                                  Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                                  1thorn ethtan tan0=FTHORN

                                                                                                                                                  c0 frac14 8 kN=m2

                                                                                                                                                  0 frac14 32

                                                                                                                                                  c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                                  W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                                  Try F frac14 100

                                                                                                                                                  tan0

                                                                                                                                                  Ffrac14 0625

                                                                                                                                                  Stability of slopes 77

                                                                                                                                                  Values of u are as obtained in Figure Q93

                                                                                                                                                  SliceNo

                                                                                                                                                  h(m)

                                                                                                                                                  W frac14 bh(kNm)

                                                                                                                                                  W sin(kNm)

                                                                                                                                                  ub(kNm)

                                                                                                                                                  c0bthorn (W ub) tan0(kNm)

                                                                                                                                                  sec

                                                                                                                                                  1thorn (tan tan0)FProduct(kNm)

                                                                                                                                                  1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                                  23 33 30 1042 31

                                                                                                                                                  3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                                  224 92 72 0931 67

                                                                                                                                                  6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                                  12 58 112 90 0889 80

                                                                                                                                                  8 60 252 1812

                                                                                                                                                  80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                                  10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                                  2154 88 116 0853 99

                                                                                                                                                  14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                                  1074 1091

                                                                                                                                                  F frac14 1091

                                                                                                                                                  1074frac14 102 (assumed value 100)

                                                                                                                                                  Thus

                                                                                                                                                  F frac14 101

                                                                                                                                                  95

                                                                                                                                                  F frac14 1

                                                                                                                                                  W sin

                                                                                                                                                  XfWeth1 ruTHORN tan0g sec

                                                                                                                                                  1thorn ethtan tan0THORN=F

                                                                                                                                                  0 frac14 33

                                                                                                                                                  ru frac14 020

                                                                                                                                                  W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                                  eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                                  Try F frac14 110

                                                                                                                                                  tan 0

                                                                                                                                                  Ffrac14 tan 33

                                                                                                                                                  110frac14 0590

                                                                                                                                                  78 Stability of slopes

                                                                                                                                                  Referring to Figure Q95

                                                                                                                                                  SliceNo

                                                                                                                                                  h(m)

                                                                                                                                                  W frac14 bh(kNm)

                                                                                                                                                  W sin(kNm)

                                                                                                                                                  W(1 ru) tan0(kNm)

                                                                                                                                                  sec

                                                                                                                                                  1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                                  1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                                  2120 234 0892 209

                                                                                                                                                  4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                                  1185 1271

                                                                                                                                                  Figure Q95

                                                                                                                                                  Stability of slopes 79

                                                                                                                                                  F frac14 1271

                                                                                                                                                  1185frac14 107

                                                                                                                                                  The trial value was 110 therefore take F to be 108

                                                                                                                                                  96

                                                                                                                                                  (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                  F frac14 0

                                                                                                                                                  sat

                                                                                                                                                  tan0

                                                                                                                                                  tan

                                                                                                                                                  ptie 15 frac14 92

                                                                                                                                                  19

                                                                                                                                                  tan 36

                                                                                                                                                  tan

                                                                                                                                                  tan frac14 0234

                                                                                                                                                  frac14 13

                                                                                                                                                  Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                  F frac14 tan0

                                                                                                                                                  tan

                                                                                                                                                  frac14 tan 36

                                                                                                                                                  tan 13

                                                                                                                                                  frac14 31

                                                                                                                                                  (b) 0d frac14 tan1tan 36

                                                                                                                                                  125

                                                                                                                                                  frac14 30

                                                                                                                                                  Depth of potential failure surface frac14 z

                                                                                                                                                  Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                  frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                  frac14 504z kN

                                                                                                                                                  Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                  frac14 19 z sin 13 cos 13

                                                                                                                                                  frac14 416z kN

                                                                                                                                                  Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                  80 Stability of slopes

                                                                                                                                                  • Book Cover
                                                                                                                                                  • Title
                                                                                                                                                  • Contents
                                                                                                                                                  • Basic characteristics of soils
                                                                                                                                                  • Seepage
                                                                                                                                                  • Effective stress
                                                                                                                                                  • Shear strength
                                                                                                                                                  • Stresses and displacements
                                                                                                                                                  • Lateral earth pressure
                                                                                                                                                  • Consolidation theory
                                                                                                                                                  • Bearing capacity
                                                                                                                                                  • Stability of slopes

                                                                                                                                                    811

                                                                                                                                                    At pile base level

                                                                                                                                                    cu frac14 220 kN=m2

                                                                                                                                                    qb frac14 cuNc frac14 220 9 frac14 1980 kN=m2

                                                                                                                                                    Disregard skin friction over a length of 2B above the under-ream Between 4 and 179m

                                                                                                                                                    00 frac14 10950 frac14 1095 112 frac14 1226 kN=m2

                                                                                                                                                    qs frac14 00 frac14 07 1226 frac14 86 kN=m2

                                                                                                                                                    Then

                                                                                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                                                                                    frac14

                                                                                                                                                    4 32 1980

                                                                                                                                                    thorn eth 105 139 86THORN

                                                                                                                                                    frac14 13 996thorn 3941 frac14 17 937 kN

                                                                                                                                                    0 01 02 03 04 05 06 07

                                                                                                                                                    0 2 4 6 8 10 12 14

                                                                                                                                                    1

                                                                                                                                                    2

                                                                                                                                                    3

                                                                                                                                                    4

                                                                                                                                                    5

                                                                                                                                                    6

                                                                                                                                                    7

                                                                                                                                                    8

                                                                                                                                                    (1)

                                                                                                                                                    (2)

                                                                                                                                                    (3)

                                                                                                                                                    (4)

                                                                                                                                                    (5)

                                                                                                                                                    qc

                                                                                                                                                    qc

                                                                                                                                                    Iz

                                                                                                                                                    Iz

                                                                                                                                                    (MNm2)

                                                                                                                                                    z (m)

                                                                                                                                                    Figure Q810

                                                                                                                                                    68 Bearing capacity

                                                                                                                                                    Allowable load

                                                                                                                                                    ethaTHORN Qf

                                                                                                                                                    2frac14 17 937

                                                                                                                                                    2frac14 8968 kN

                                                                                                                                                    ethbTHORN Abqb

                                                                                                                                                    3thorn Asqs frac14 13 996

                                                                                                                                                    3thorn 3941 frac14 8606 kN

                                                                                                                                                    ie allowable load frac14 8600 kN

                                                                                                                                                    Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                                                    According to the limit state method

                                                                                                                                                    Characteristic undrained strength at base level cuk frac14 220

                                                                                                                                                    150kN=m2

                                                                                                                                                    Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                                                    150frac14 1320 kN=m2

                                                                                                                                                    Characteristic shaft resistance qsk frac14 00150

                                                                                                                                                    frac14 86

                                                                                                                                                    150frac14 57 kN=m2

                                                                                                                                                    Characteristic base and shaft resistances

                                                                                                                                                    Rbk frac14

                                                                                                                                                    4 32 1320 frac14 9330 kN

                                                                                                                                                    Rsk frac14 105 139 86

                                                                                                                                                    150frac14 2629 kN

                                                                                                                                                    For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                                                    Design bearing resistance Rcd frac14 9330

                                                                                                                                                    160thorn 2629

                                                                                                                                                    130

                                                                                                                                                    frac14 5831thorn 2022

                                                                                                                                                    frac14 7850 kN

                                                                                                                                                    Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                                                    812

                                                                                                                                                    ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                                                    qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                                                    For a single pile

                                                                                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                                                                                    frac14

                                                                                                                                                    4 062 1305

                                                                                                                                                    thorn eth 06 15 42THORN

                                                                                                                                                    frac14 369thorn 1187 frac14 1556 kN

                                                                                                                                                    Bearing capacity 69

                                                                                                                                                    Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                                                    eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                                                    (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                                                    (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                                                    qbkfrac14 9cuk frac14 9 220

                                                                                                                                                    150frac14 1320 kN=m2

                                                                                                                                                    qskfrac14cuk frac14 040 105

                                                                                                                                                    150frac14 28 kN=m2

                                                                                                                                                    Rbkfrac14

                                                                                                                                                    4 0602 1320 frac14 373 kN

                                                                                                                                                    Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                                                    Rcdfrac14 373

                                                                                                                                                    160thorn 791

                                                                                                                                                    130frac14 233thorn 608 frac14 841 kN

                                                                                                                                                    Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                                                    Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                                                    (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                                                    q frac14 21 000

                                                                                                                                                    1762frac14 68 kN=m2

                                                                                                                                                    Immediate settlement

                                                                                                                                                    H

                                                                                                                                                    Bfrac14 15

                                                                                                                                                    176frac14 085

                                                                                                                                                    D

                                                                                                                                                    Bfrac14 13

                                                                                                                                                    176frac14 074

                                                                                                                                                    L

                                                                                                                                                    Bfrac14 1

                                                                                                                                                    Hence from Figure 515

                                                                                                                                                    130 frac14 078 and 131 frac14 041

                                                                                                                                                    70 Bearing capacity

                                                                                                                                                    Thus using Equation 528

                                                                                                                                                    si frac14 078 041 68 176

                                                                                                                                                    65frac14 6mm

                                                                                                                                                    Consolidation settlement

                                                                                                                                                    Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                                                    1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                                                    434 (sod)

                                                                                                                                                    Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                                                    sc frac14 056 434 frac14 24mm

                                                                                                                                                    The total settlement is (6thorn 24) frac14 30mm

                                                                                                                                                    813

                                                                                                                                                    At base level N frac14 26 Then using Equation 830

                                                                                                                                                    qb frac14 40NDb

                                                                                                                                                    Bfrac14 40 26 2

                                                                                                                                                    025frac14 8320 kN=m2

                                                                                                                                                    ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                                                    Figure Q812

                                                                                                                                                    Bearing capacity 71

                                                                                                                                                    Over the length embedded in sand

                                                                                                                                                    N frac14 21 ie18thorn 24

                                                                                                                                                    2

                                                                                                                                                    Using Equation 831

                                                                                                                                                    qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                                    For a single pile

                                                                                                                                                    Qf frac14 Abqb thorn Asqs

                                                                                                                                                    frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                                    For the pile group assuming a group efficiency of 12

                                                                                                                                                    XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                                    Then the load factor is

                                                                                                                                                    F frac14 6523

                                                                                                                                                    2000thorn 1000frac14 21

                                                                                                                                                    (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                                    Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                                    150frac14 5547 kNm2

                                                                                                                                                    Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                                    150frac14 28 kNm2

                                                                                                                                                    Characteristic base and shaft resistances for a single pile

                                                                                                                                                    Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                                    Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                                    For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                                    Design bearing resistance Rcd frac14 347

                                                                                                                                                    130thorn 56

                                                                                                                                                    130frac14 310 kN

                                                                                                                                                    For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                                    Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                                    (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                                    From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                                    72 Bearing capacity

                                                                                                                                                    N frac14 24thorn 26thorn 34

                                                                                                                                                    3frac14 28

                                                                                                                                                    Ic frac14 171

                                                                                                                                                    2814frac14 0016 ethEquation 818THORN

                                                                                                                                                    s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                                    The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                                    814

                                                                                                                                                    Using Equation 841

                                                                                                                                                    Tf frac14 DLcu thorn

                                                                                                                                                    4ethD2 d2THORNcuNc

                                                                                                                                                    frac14 eth 02 5 06 110THORN thorn

                                                                                                                                                    4eth022 012THORN110 9

                                                                                                                                                    frac14 207thorn 23 frac14 230 kN

                                                                                                                                                    Figure Q813

                                                                                                                                                    Bearing capacity 73

                                                                                                                                                    Chapter 9

                                                                                                                                                    Stability of slopes

                                                                                                                                                    91

                                                                                                                                                    Referring to Figure Q91

                                                                                                                                                    W frac14 417 19 frac14 792 kN=m

                                                                                                                                                    Q frac14 20 28 frac14 56 kN=m

                                                                                                                                                    Arc lengthAB frac14

                                                                                                                                                    180 73 90 frac14 115m

                                                                                                                                                    Arc length BC frac14

                                                                                                                                                    180 28 90 frac14 44m

                                                                                                                                                    The factor of safety is given by

                                                                                                                                                    F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                                    90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                                    Depth of tension crack z0 frac14 2cu

                                                                                                                                                    frac14 2 20

                                                                                                                                                    19frac14 21m

                                                                                                                                                    Arc length BD frac14

                                                                                                                                                    180 13

                                                                                                                                                    1

                                                                                                                                                    2 90 frac14 21m

                                                                                                                                                    F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                                    The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                                    Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                                    14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                                    Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                                    The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                                    92

                                                                                                                                                    u frac14 0

                                                                                                                                                    Depth factor D frac14 11

                                                                                                                                                    9frac14 122

                                                                                                                                                    Using Equation 92 with F frac14 10

                                                                                                                                                    Ns frac14 cu

                                                                                                                                                    FHfrac14 30

                                                                                                                                                    10 19 9frac14 0175

                                                                                                                                                    Hence from Figure 93

                                                                                                                                                    frac14 50

                                                                                                                                                    For F frac14 12

                                                                                                                                                    Ns frac14 30

                                                                                                                                                    12 19 9frac14 0146

                                                                                                                                                    frac14 27

                                                                                                                                                    93

                                                                                                                                                    Refer to Figure Q93

                                                                                                                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                    1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                                    74 m

                                                                                                                                                    214 1deg

                                                                                                                                                    213 1deg

                                                                                                                                                    39 m

                                                                                                                                                    WB

                                                                                                                                                    D

                                                                                                                                                    C

                                                                                                                                                    28 m

                                                                                                                                                    21 m

                                                                                                                                                    A

                                                                                                                                                    Q

                                                                                                                                                    Soil (1)Soil (2)

                                                                                                                                                    73deg

                                                                                                                                                    Figure Q91

                                                                                                                                                    Stability of slopes 75

                                                                                                                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                    9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                                    599 256 328 1372

                                                                                                                                                    Figure Q93

                                                                                                                                                    76 Stability of slopes

                                                                                                                                                    XW cos frac14 b

                                                                                                                                                    Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                                    W sin frac14 bX

                                                                                                                                                    h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                                    Arc length La frac14

                                                                                                                                                    180 57

                                                                                                                                                    1

                                                                                                                                                    2 326 frac14 327m

                                                                                                                                                    The factor of safety is given by

                                                                                                                                                    F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                                    W sin

                                                                                                                                                    frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                                    frac14 091

                                                                                                                                                    According to the limit state method

                                                                                                                                                    0d frac14 tan1tan 32

                                                                                                                                                    125

                                                                                                                                                    frac14 265

                                                                                                                                                    c0 frac14 8

                                                                                                                                                    160frac14 5 kN=m2

                                                                                                                                                    Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                                    Design disturbing moment frac14 1075 kN=m

                                                                                                                                                    The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                                    94

                                                                                                                                                    F frac14 1

                                                                                                                                                    W sin

                                                                                                                                                    Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                                    1thorn ethtan tan0=FTHORN

                                                                                                                                                    c0 frac14 8 kN=m2

                                                                                                                                                    0 frac14 32

                                                                                                                                                    c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                                    W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                                    Try F frac14 100

                                                                                                                                                    tan0

                                                                                                                                                    Ffrac14 0625

                                                                                                                                                    Stability of slopes 77

                                                                                                                                                    Values of u are as obtained in Figure Q93

                                                                                                                                                    SliceNo

                                                                                                                                                    h(m)

                                                                                                                                                    W frac14 bh(kNm)

                                                                                                                                                    W sin(kNm)

                                                                                                                                                    ub(kNm)

                                                                                                                                                    c0bthorn (W ub) tan0(kNm)

                                                                                                                                                    sec

                                                                                                                                                    1thorn (tan tan0)FProduct(kNm)

                                                                                                                                                    1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                                    23 33 30 1042 31

                                                                                                                                                    3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                                    224 92 72 0931 67

                                                                                                                                                    6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                                    12 58 112 90 0889 80

                                                                                                                                                    8 60 252 1812

                                                                                                                                                    80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                                    10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                                    2154 88 116 0853 99

                                                                                                                                                    14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                                    1074 1091

                                                                                                                                                    F frac14 1091

                                                                                                                                                    1074frac14 102 (assumed value 100)

                                                                                                                                                    Thus

                                                                                                                                                    F frac14 101

                                                                                                                                                    95

                                                                                                                                                    F frac14 1

                                                                                                                                                    W sin

                                                                                                                                                    XfWeth1 ruTHORN tan0g sec

                                                                                                                                                    1thorn ethtan tan0THORN=F

                                                                                                                                                    0 frac14 33

                                                                                                                                                    ru frac14 020

                                                                                                                                                    W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                                    eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                                    Try F frac14 110

                                                                                                                                                    tan 0

                                                                                                                                                    Ffrac14 tan 33

                                                                                                                                                    110frac14 0590

                                                                                                                                                    78 Stability of slopes

                                                                                                                                                    Referring to Figure Q95

                                                                                                                                                    SliceNo

                                                                                                                                                    h(m)

                                                                                                                                                    W frac14 bh(kNm)

                                                                                                                                                    W sin(kNm)

                                                                                                                                                    W(1 ru) tan0(kNm)

                                                                                                                                                    sec

                                                                                                                                                    1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                                    1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                                    2120 234 0892 209

                                                                                                                                                    4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                                    1185 1271

                                                                                                                                                    Figure Q95

                                                                                                                                                    Stability of slopes 79

                                                                                                                                                    F frac14 1271

                                                                                                                                                    1185frac14 107

                                                                                                                                                    The trial value was 110 therefore take F to be 108

                                                                                                                                                    96

                                                                                                                                                    (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                    F frac14 0

                                                                                                                                                    sat

                                                                                                                                                    tan0

                                                                                                                                                    tan

                                                                                                                                                    ptie 15 frac14 92

                                                                                                                                                    19

                                                                                                                                                    tan 36

                                                                                                                                                    tan

                                                                                                                                                    tan frac14 0234

                                                                                                                                                    frac14 13

                                                                                                                                                    Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                    F frac14 tan0

                                                                                                                                                    tan

                                                                                                                                                    frac14 tan 36

                                                                                                                                                    tan 13

                                                                                                                                                    frac14 31

                                                                                                                                                    (b) 0d frac14 tan1tan 36

                                                                                                                                                    125

                                                                                                                                                    frac14 30

                                                                                                                                                    Depth of potential failure surface frac14 z

                                                                                                                                                    Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                    frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                    frac14 504z kN

                                                                                                                                                    Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                    frac14 19 z sin 13 cos 13

                                                                                                                                                    frac14 416z kN

                                                                                                                                                    Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                    80 Stability of slopes

                                                                                                                                                    • Book Cover
                                                                                                                                                    • Title
                                                                                                                                                    • Contents
                                                                                                                                                    • Basic characteristics of soils
                                                                                                                                                    • Seepage
                                                                                                                                                    • Effective stress
                                                                                                                                                    • Shear strength
                                                                                                                                                    • Stresses and displacements
                                                                                                                                                    • Lateral earth pressure
                                                                                                                                                    • Consolidation theory
                                                                                                                                                    • Bearing capacity
                                                                                                                                                    • Stability of slopes

                                                                                                                                                      Allowable load

                                                                                                                                                      ethaTHORN Qf

                                                                                                                                                      2frac14 17 937

                                                                                                                                                      2frac14 8968 kN

                                                                                                                                                      ethbTHORN Abqb

                                                                                                                                                      3thorn Asqs frac14 13 996

                                                                                                                                                      3thorn 3941 frac14 8606 kN

                                                                                                                                                      ie allowable load frac14 8600 kN

                                                                                                                                                      Adding 1frasl3(DAb W) the allowable load becomes 9200 kN

                                                                                                                                                      According to the limit state method

                                                                                                                                                      Characteristic undrained strength at base level cuk frac14 220

                                                                                                                                                      150kN=m2

                                                                                                                                                      Characteristic base resistance qbk frac14 9cuk frac14 9 220

                                                                                                                                                      150frac14 1320 kN=m2

                                                                                                                                                      Characteristic shaft resistance qsk frac14 00150

                                                                                                                                                      frac14 86

                                                                                                                                                      150frac14 57 kN=m2

                                                                                                                                                      Characteristic base and shaft resistances

                                                                                                                                                      Rbk frac14

                                                                                                                                                      4 32 1320 frac14 9330 kN

                                                                                                                                                      Rsk frac14 105 139 86

                                                                                                                                                      150frac14 2629 kN

                                                                                                                                                      For a bored pile the partial factors are b frac14 160 and s frac14 130

                                                                                                                                                      Design bearing resistance Rcd frac14 9330

                                                                                                                                                      160thorn 2629

                                                                                                                                                      130

                                                                                                                                                      frac14 5831thorn 2022

                                                                                                                                                      frac14 7850 kN

                                                                                                                                                      Adding ethDAb W) the design bearing resistance becomes 9650 kN

                                                                                                                                                      812

                                                                                                                                                      ethaTHORN qb frac14 9cu frac14 9 145 frac14 1305 kN=m2

                                                                                                                                                      qs frac14 cu frac14 040 105 frac14 42 kN=m2

                                                                                                                                                      For a single pile

                                                                                                                                                      Qf frac14 Abqb thorn Asqs

                                                                                                                                                      frac14

                                                                                                                                                      4 062 1305

                                                                                                                                                      thorn eth 06 15 42THORN

                                                                                                                                                      frac14 369thorn 1187 frac14 1556 kN

                                                                                                                                                      Bearing capacity 69

                                                                                                                                                      Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                                                      eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                                                      (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                                                      (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                                                      qbkfrac14 9cuk frac14 9 220

                                                                                                                                                      150frac14 1320 kN=m2

                                                                                                                                                      qskfrac14cuk frac14 040 105

                                                                                                                                                      150frac14 28 kN=m2

                                                                                                                                                      Rbkfrac14

                                                                                                                                                      4 0602 1320 frac14 373 kN

                                                                                                                                                      Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                                                      Rcdfrac14 373

                                                                                                                                                      160thorn 791

                                                                                                                                                      130frac14 233thorn 608 frac14 841 kN

                                                                                                                                                      Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                                                      Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                                                      (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                                                      q frac14 21 000

                                                                                                                                                      1762frac14 68 kN=m2

                                                                                                                                                      Immediate settlement

                                                                                                                                                      H

                                                                                                                                                      Bfrac14 15

                                                                                                                                                      176frac14 085

                                                                                                                                                      D

                                                                                                                                                      Bfrac14 13

                                                                                                                                                      176frac14 074

                                                                                                                                                      L

                                                                                                                                                      Bfrac14 1

                                                                                                                                                      Hence from Figure 515

                                                                                                                                                      130 frac14 078 and 131 frac14 041

                                                                                                                                                      70 Bearing capacity

                                                                                                                                                      Thus using Equation 528

                                                                                                                                                      si frac14 078 041 68 176

                                                                                                                                                      65frac14 6mm

                                                                                                                                                      Consolidation settlement

                                                                                                                                                      Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                                                      1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                                                      434 (sod)

                                                                                                                                                      Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                                                      sc frac14 056 434 frac14 24mm

                                                                                                                                                      The total settlement is (6thorn 24) frac14 30mm

                                                                                                                                                      813

                                                                                                                                                      At base level N frac14 26 Then using Equation 830

                                                                                                                                                      qb frac14 40NDb

                                                                                                                                                      Bfrac14 40 26 2

                                                                                                                                                      025frac14 8320 kN=m2

                                                                                                                                                      ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                                                      Figure Q812

                                                                                                                                                      Bearing capacity 71

                                                                                                                                                      Over the length embedded in sand

                                                                                                                                                      N frac14 21 ie18thorn 24

                                                                                                                                                      2

                                                                                                                                                      Using Equation 831

                                                                                                                                                      qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                                      For a single pile

                                                                                                                                                      Qf frac14 Abqb thorn Asqs

                                                                                                                                                      frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                                      For the pile group assuming a group efficiency of 12

                                                                                                                                                      XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                                      Then the load factor is

                                                                                                                                                      F frac14 6523

                                                                                                                                                      2000thorn 1000frac14 21

                                                                                                                                                      (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                                      Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                                      150frac14 5547 kNm2

                                                                                                                                                      Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                                      150frac14 28 kNm2

                                                                                                                                                      Characteristic base and shaft resistances for a single pile

                                                                                                                                                      Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                                      Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                                      For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                                      Design bearing resistance Rcd frac14 347

                                                                                                                                                      130thorn 56

                                                                                                                                                      130frac14 310 kN

                                                                                                                                                      For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                                      Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                                      (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                                      From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                                      72 Bearing capacity

                                                                                                                                                      N frac14 24thorn 26thorn 34

                                                                                                                                                      3frac14 28

                                                                                                                                                      Ic frac14 171

                                                                                                                                                      2814frac14 0016 ethEquation 818THORN

                                                                                                                                                      s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                                      The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                                      814

                                                                                                                                                      Using Equation 841

                                                                                                                                                      Tf frac14 DLcu thorn

                                                                                                                                                      4ethD2 d2THORNcuNc

                                                                                                                                                      frac14 eth 02 5 06 110THORN thorn

                                                                                                                                                      4eth022 012THORN110 9

                                                                                                                                                      frac14 207thorn 23 frac14 230 kN

                                                                                                                                                      Figure Q813

                                                                                                                                                      Bearing capacity 73

                                                                                                                                                      Chapter 9

                                                                                                                                                      Stability of slopes

                                                                                                                                                      91

                                                                                                                                                      Referring to Figure Q91

                                                                                                                                                      W frac14 417 19 frac14 792 kN=m

                                                                                                                                                      Q frac14 20 28 frac14 56 kN=m

                                                                                                                                                      Arc lengthAB frac14

                                                                                                                                                      180 73 90 frac14 115m

                                                                                                                                                      Arc length BC frac14

                                                                                                                                                      180 28 90 frac14 44m

                                                                                                                                                      The factor of safety is given by

                                                                                                                                                      F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                                      90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                                      Depth of tension crack z0 frac14 2cu

                                                                                                                                                      frac14 2 20

                                                                                                                                                      19frac14 21m

                                                                                                                                                      Arc length BD frac14

                                                                                                                                                      180 13

                                                                                                                                                      1

                                                                                                                                                      2 90 frac14 21m

                                                                                                                                                      F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                                      The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                                      Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                                      14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                                      Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                                      The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                                      92

                                                                                                                                                      u frac14 0

                                                                                                                                                      Depth factor D frac14 11

                                                                                                                                                      9frac14 122

                                                                                                                                                      Using Equation 92 with F frac14 10

                                                                                                                                                      Ns frac14 cu

                                                                                                                                                      FHfrac14 30

                                                                                                                                                      10 19 9frac14 0175

                                                                                                                                                      Hence from Figure 93

                                                                                                                                                      frac14 50

                                                                                                                                                      For F frac14 12

                                                                                                                                                      Ns frac14 30

                                                                                                                                                      12 19 9frac14 0146

                                                                                                                                                      frac14 27

                                                                                                                                                      93

                                                                                                                                                      Refer to Figure Q93

                                                                                                                                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                      1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                                      74 m

                                                                                                                                                      214 1deg

                                                                                                                                                      213 1deg

                                                                                                                                                      39 m

                                                                                                                                                      WB

                                                                                                                                                      D

                                                                                                                                                      C

                                                                                                                                                      28 m

                                                                                                                                                      21 m

                                                                                                                                                      A

                                                                                                                                                      Q

                                                                                                                                                      Soil (1)Soil (2)

                                                                                                                                                      73deg

                                                                                                                                                      Figure Q91

                                                                                                                                                      Stability of slopes 75

                                                                                                                                                      Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                      9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                                      599 256 328 1372

                                                                                                                                                      Figure Q93

                                                                                                                                                      76 Stability of slopes

                                                                                                                                                      XW cos frac14 b

                                                                                                                                                      Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                                      W sin frac14 bX

                                                                                                                                                      h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                                      Arc length La frac14

                                                                                                                                                      180 57

                                                                                                                                                      1

                                                                                                                                                      2 326 frac14 327m

                                                                                                                                                      The factor of safety is given by

                                                                                                                                                      F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                                      W sin

                                                                                                                                                      frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                                      frac14 091

                                                                                                                                                      According to the limit state method

                                                                                                                                                      0d frac14 tan1tan 32

                                                                                                                                                      125

                                                                                                                                                      frac14 265

                                                                                                                                                      c0 frac14 8

                                                                                                                                                      160frac14 5 kN=m2

                                                                                                                                                      Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                                      Design disturbing moment frac14 1075 kN=m

                                                                                                                                                      The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                                      94

                                                                                                                                                      F frac14 1

                                                                                                                                                      W sin

                                                                                                                                                      Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                                      1thorn ethtan tan0=FTHORN

                                                                                                                                                      c0 frac14 8 kN=m2

                                                                                                                                                      0 frac14 32

                                                                                                                                                      c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                                      W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                                      Try F frac14 100

                                                                                                                                                      tan0

                                                                                                                                                      Ffrac14 0625

                                                                                                                                                      Stability of slopes 77

                                                                                                                                                      Values of u are as obtained in Figure Q93

                                                                                                                                                      SliceNo

                                                                                                                                                      h(m)

                                                                                                                                                      W frac14 bh(kNm)

                                                                                                                                                      W sin(kNm)

                                                                                                                                                      ub(kNm)

                                                                                                                                                      c0bthorn (W ub) tan0(kNm)

                                                                                                                                                      sec

                                                                                                                                                      1thorn (tan tan0)FProduct(kNm)

                                                                                                                                                      1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                                      23 33 30 1042 31

                                                                                                                                                      3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                                      224 92 72 0931 67

                                                                                                                                                      6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                                      12 58 112 90 0889 80

                                                                                                                                                      8 60 252 1812

                                                                                                                                                      80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                                      10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                                      2154 88 116 0853 99

                                                                                                                                                      14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                                      1074 1091

                                                                                                                                                      F frac14 1091

                                                                                                                                                      1074frac14 102 (assumed value 100)

                                                                                                                                                      Thus

                                                                                                                                                      F frac14 101

                                                                                                                                                      95

                                                                                                                                                      F frac14 1

                                                                                                                                                      W sin

                                                                                                                                                      XfWeth1 ruTHORN tan0g sec

                                                                                                                                                      1thorn ethtan tan0THORN=F

                                                                                                                                                      0 frac14 33

                                                                                                                                                      ru frac14 020

                                                                                                                                                      W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                                      eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                                      Try F frac14 110

                                                                                                                                                      tan 0

                                                                                                                                                      Ffrac14 tan 33

                                                                                                                                                      110frac14 0590

                                                                                                                                                      78 Stability of slopes

                                                                                                                                                      Referring to Figure Q95

                                                                                                                                                      SliceNo

                                                                                                                                                      h(m)

                                                                                                                                                      W frac14 bh(kNm)

                                                                                                                                                      W sin(kNm)

                                                                                                                                                      W(1 ru) tan0(kNm)

                                                                                                                                                      sec

                                                                                                                                                      1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                                      1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                                      2120 234 0892 209

                                                                                                                                                      4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                                      1185 1271

                                                                                                                                                      Figure Q95

                                                                                                                                                      Stability of slopes 79

                                                                                                                                                      F frac14 1271

                                                                                                                                                      1185frac14 107

                                                                                                                                                      The trial value was 110 therefore take F to be 108

                                                                                                                                                      96

                                                                                                                                                      (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                      F frac14 0

                                                                                                                                                      sat

                                                                                                                                                      tan0

                                                                                                                                                      tan

                                                                                                                                                      ptie 15 frac14 92

                                                                                                                                                      19

                                                                                                                                                      tan 36

                                                                                                                                                      tan

                                                                                                                                                      tan frac14 0234

                                                                                                                                                      frac14 13

                                                                                                                                                      Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                      F frac14 tan0

                                                                                                                                                      tan

                                                                                                                                                      frac14 tan 36

                                                                                                                                                      tan 13

                                                                                                                                                      frac14 31

                                                                                                                                                      (b) 0d frac14 tan1tan 36

                                                                                                                                                      125

                                                                                                                                                      frac14 30

                                                                                                                                                      Depth of potential failure surface frac14 z

                                                                                                                                                      Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                      frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                      frac14 504z kN

                                                                                                                                                      Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                      frac14 19 z sin 13 cos 13

                                                                                                                                                      frac14 416z kN

                                                                                                                                                      Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                      80 Stability of slopes

                                                                                                                                                      • Book Cover
                                                                                                                                                      • Title
                                                                                                                                                      • Contents
                                                                                                                                                      • Basic characteristics of soils
                                                                                                                                                      • Seepage
                                                                                                                                                      • Effective stress
                                                                                                                                                      • Shear strength
                                                                                                                                                      • Stresses and displacements
                                                                                                                                                      • Lateral earth pressure
                                                                                                                                                      • Consolidation theory
                                                                                                                                                      • Bearing capacity
                                                                                                                                                      • Stability of slopes

                                                                                                                                                        Assuming single pile failure and a group efficiency of 1 the ultimate load on the pilegroup is (1556 36) frac14 56 016 kN The width of the group is 126m and hence theultimate load assuming block failure and taking the full undrained strength on theperimeter is given by

                                                                                                                                                        eth1262 1305THORN thorn eth4 126 15 105THORNfrac14 207 180thorn 79 380 frac14 286 560 kN

                                                                                                                                                        (Even if the remoulded strength were used there would be no likelihood of blockfailure) Thus the load factor is (56 01621 000) frac14 27

                                                                                                                                                        (b) Design load Fcd frac14 15thorn (6 130) frac14 228MN

                                                                                                                                                        qbkfrac14 9cuk frac14 9 220

                                                                                                                                                        150frac14 1320 kN=m2

                                                                                                                                                        qskfrac14cuk frac14 040 105

                                                                                                                                                        150frac14 28 kN=m2

                                                                                                                                                        Rbkfrac14

                                                                                                                                                        4 0602 1320 frac14 373 kN

                                                                                                                                                        Rskfrac14 060 15 28 frac14 791 kN

                                                                                                                                                        Rcdfrac14 373

                                                                                                                                                        160thorn 791

                                                                                                                                                        130frac14 233thorn 608 frac14 841 kN

                                                                                                                                                        Design bearing resistance of pile group Rcd frac14 841 36 10frac14 30 276kNfrac14 3027MN

                                                                                                                                                        Rcd gt Fcd therefore the bearing resistance limit state is satisfied

                                                                                                                                                        (c) Settlement is estimated using the equivalent raft concept The equivalent raft islocated 10m ( 2frasl3 15m) below the top of the piles and is 176m wide (see FigureQ812) Assume that the load on the equivalent raft is spread at 21 to the underlyingclay Thus the pressure on the equivalent raft is

                                                                                                                                                        q frac14 21 000

                                                                                                                                                        1762frac14 68 kN=m2

                                                                                                                                                        Immediate settlement

                                                                                                                                                        H

                                                                                                                                                        Bfrac14 15

                                                                                                                                                        176frac14 085

                                                                                                                                                        D

                                                                                                                                                        Bfrac14 13

                                                                                                                                                        176frac14 074

                                                                                                                                                        L

                                                                                                                                                        Bfrac14 1

                                                                                                                                                        Hence from Figure 515

                                                                                                                                                        130 frac14 078 and 131 frac14 041

                                                                                                                                                        70 Bearing capacity

                                                                                                                                                        Thus using Equation 528

                                                                                                                                                        si frac14 078 041 68 176

                                                                                                                                                        65frac14 6mm

                                                                                                                                                        Consolidation settlement

                                                                                                                                                        Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                                                        1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                                                        434 (sod)

                                                                                                                                                        Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                                                        sc frac14 056 434 frac14 24mm

                                                                                                                                                        The total settlement is (6thorn 24) frac14 30mm

                                                                                                                                                        813

                                                                                                                                                        At base level N frac14 26 Then using Equation 830

                                                                                                                                                        qb frac14 40NDb

                                                                                                                                                        Bfrac14 40 26 2

                                                                                                                                                        025frac14 8320 kN=m2

                                                                                                                                                        ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                                                        Figure Q812

                                                                                                                                                        Bearing capacity 71

                                                                                                                                                        Over the length embedded in sand

                                                                                                                                                        N frac14 21 ie18thorn 24

                                                                                                                                                        2

                                                                                                                                                        Using Equation 831

                                                                                                                                                        qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                                        For a single pile

                                                                                                                                                        Qf frac14 Abqb thorn Asqs

                                                                                                                                                        frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                                        For the pile group assuming a group efficiency of 12

                                                                                                                                                        XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                                        Then the load factor is

                                                                                                                                                        F frac14 6523

                                                                                                                                                        2000thorn 1000frac14 21

                                                                                                                                                        (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                                        Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                                        150frac14 5547 kNm2

                                                                                                                                                        Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                                        150frac14 28 kNm2

                                                                                                                                                        Characteristic base and shaft resistances for a single pile

                                                                                                                                                        Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                                        Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                                        For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                                        Design bearing resistance Rcd frac14 347

                                                                                                                                                        130thorn 56

                                                                                                                                                        130frac14 310 kN

                                                                                                                                                        For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                                        Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                                        (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                                        From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                                        72 Bearing capacity

                                                                                                                                                        N frac14 24thorn 26thorn 34

                                                                                                                                                        3frac14 28

                                                                                                                                                        Ic frac14 171

                                                                                                                                                        2814frac14 0016 ethEquation 818THORN

                                                                                                                                                        s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                                        The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                                        814

                                                                                                                                                        Using Equation 841

                                                                                                                                                        Tf frac14 DLcu thorn

                                                                                                                                                        4ethD2 d2THORNcuNc

                                                                                                                                                        frac14 eth 02 5 06 110THORN thorn

                                                                                                                                                        4eth022 012THORN110 9

                                                                                                                                                        frac14 207thorn 23 frac14 230 kN

                                                                                                                                                        Figure Q813

                                                                                                                                                        Bearing capacity 73

                                                                                                                                                        Chapter 9

                                                                                                                                                        Stability of slopes

                                                                                                                                                        91

                                                                                                                                                        Referring to Figure Q91

                                                                                                                                                        W frac14 417 19 frac14 792 kN=m

                                                                                                                                                        Q frac14 20 28 frac14 56 kN=m

                                                                                                                                                        Arc lengthAB frac14

                                                                                                                                                        180 73 90 frac14 115m

                                                                                                                                                        Arc length BC frac14

                                                                                                                                                        180 28 90 frac14 44m

                                                                                                                                                        The factor of safety is given by

                                                                                                                                                        F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                                        90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                                        Depth of tension crack z0 frac14 2cu

                                                                                                                                                        frac14 2 20

                                                                                                                                                        19frac14 21m

                                                                                                                                                        Arc length BD frac14

                                                                                                                                                        180 13

                                                                                                                                                        1

                                                                                                                                                        2 90 frac14 21m

                                                                                                                                                        F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                                        The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                                        Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                                        14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                                        Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                                        The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                                        92

                                                                                                                                                        u frac14 0

                                                                                                                                                        Depth factor D frac14 11

                                                                                                                                                        9frac14 122

                                                                                                                                                        Using Equation 92 with F frac14 10

                                                                                                                                                        Ns frac14 cu

                                                                                                                                                        FHfrac14 30

                                                                                                                                                        10 19 9frac14 0175

                                                                                                                                                        Hence from Figure 93

                                                                                                                                                        frac14 50

                                                                                                                                                        For F frac14 12

                                                                                                                                                        Ns frac14 30

                                                                                                                                                        12 19 9frac14 0146

                                                                                                                                                        frac14 27

                                                                                                                                                        93

                                                                                                                                                        Refer to Figure Q93

                                                                                                                                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                        1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                                        74 m

                                                                                                                                                        214 1deg

                                                                                                                                                        213 1deg

                                                                                                                                                        39 m

                                                                                                                                                        WB

                                                                                                                                                        D

                                                                                                                                                        C

                                                                                                                                                        28 m

                                                                                                                                                        21 m

                                                                                                                                                        A

                                                                                                                                                        Q

                                                                                                                                                        Soil (1)Soil (2)

                                                                                                                                                        73deg

                                                                                                                                                        Figure Q91

                                                                                                                                                        Stability of slopes 75

                                                                                                                                                        Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                        9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                                        599 256 328 1372

                                                                                                                                                        Figure Q93

                                                                                                                                                        76 Stability of slopes

                                                                                                                                                        XW cos frac14 b

                                                                                                                                                        Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                                        W sin frac14 bX

                                                                                                                                                        h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                                        Arc length La frac14

                                                                                                                                                        180 57

                                                                                                                                                        1

                                                                                                                                                        2 326 frac14 327m

                                                                                                                                                        The factor of safety is given by

                                                                                                                                                        F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                                        W sin

                                                                                                                                                        frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                                        frac14 091

                                                                                                                                                        According to the limit state method

                                                                                                                                                        0d frac14 tan1tan 32

                                                                                                                                                        125

                                                                                                                                                        frac14 265

                                                                                                                                                        c0 frac14 8

                                                                                                                                                        160frac14 5 kN=m2

                                                                                                                                                        Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                                        Design disturbing moment frac14 1075 kN=m

                                                                                                                                                        The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                                        94

                                                                                                                                                        F frac14 1

                                                                                                                                                        W sin

                                                                                                                                                        Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                                        1thorn ethtan tan0=FTHORN

                                                                                                                                                        c0 frac14 8 kN=m2

                                                                                                                                                        0 frac14 32

                                                                                                                                                        c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                                        W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                                        Try F frac14 100

                                                                                                                                                        tan0

                                                                                                                                                        Ffrac14 0625

                                                                                                                                                        Stability of slopes 77

                                                                                                                                                        Values of u are as obtained in Figure Q93

                                                                                                                                                        SliceNo

                                                                                                                                                        h(m)

                                                                                                                                                        W frac14 bh(kNm)

                                                                                                                                                        W sin(kNm)

                                                                                                                                                        ub(kNm)

                                                                                                                                                        c0bthorn (W ub) tan0(kNm)

                                                                                                                                                        sec

                                                                                                                                                        1thorn (tan tan0)FProduct(kNm)

                                                                                                                                                        1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                                        23 33 30 1042 31

                                                                                                                                                        3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                                        224 92 72 0931 67

                                                                                                                                                        6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                                        12 58 112 90 0889 80

                                                                                                                                                        8 60 252 1812

                                                                                                                                                        80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                                        10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                                        2154 88 116 0853 99

                                                                                                                                                        14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                                        1074 1091

                                                                                                                                                        F frac14 1091

                                                                                                                                                        1074frac14 102 (assumed value 100)

                                                                                                                                                        Thus

                                                                                                                                                        F frac14 101

                                                                                                                                                        95

                                                                                                                                                        F frac14 1

                                                                                                                                                        W sin

                                                                                                                                                        XfWeth1 ruTHORN tan0g sec

                                                                                                                                                        1thorn ethtan tan0THORN=F

                                                                                                                                                        0 frac14 33

                                                                                                                                                        ru frac14 020

                                                                                                                                                        W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                                        eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                                        Try F frac14 110

                                                                                                                                                        tan 0

                                                                                                                                                        Ffrac14 tan 33

                                                                                                                                                        110frac14 0590

                                                                                                                                                        78 Stability of slopes

                                                                                                                                                        Referring to Figure Q95

                                                                                                                                                        SliceNo

                                                                                                                                                        h(m)

                                                                                                                                                        W frac14 bh(kNm)

                                                                                                                                                        W sin(kNm)

                                                                                                                                                        W(1 ru) tan0(kNm)

                                                                                                                                                        sec

                                                                                                                                                        1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                                        1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                                        2120 234 0892 209

                                                                                                                                                        4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                                        1185 1271

                                                                                                                                                        Figure Q95

                                                                                                                                                        Stability of slopes 79

                                                                                                                                                        F frac14 1271

                                                                                                                                                        1185frac14 107

                                                                                                                                                        The trial value was 110 therefore take F to be 108

                                                                                                                                                        96

                                                                                                                                                        (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                        F frac14 0

                                                                                                                                                        sat

                                                                                                                                                        tan0

                                                                                                                                                        tan

                                                                                                                                                        ptie 15 frac14 92

                                                                                                                                                        19

                                                                                                                                                        tan 36

                                                                                                                                                        tan

                                                                                                                                                        tan frac14 0234

                                                                                                                                                        frac14 13

                                                                                                                                                        Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                        F frac14 tan0

                                                                                                                                                        tan

                                                                                                                                                        frac14 tan 36

                                                                                                                                                        tan 13

                                                                                                                                                        frac14 31

                                                                                                                                                        (b) 0d frac14 tan1tan 36

                                                                                                                                                        125

                                                                                                                                                        frac14 30

                                                                                                                                                        Depth of potential failure surface frac14 z

                                                                                                                                                        Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                        frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                        frac14 504z kN

                                                                                                                                                        Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                        frac14 19 z sin 13 cos 13

                                                                                                                                                        frac14 416z kN

                                                                                                                                                        Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                        80 Stability of slopes

                                                                                                                                                        • Book Cover
                                                                                                                                                        • Title
                                                                                                                                                        • Contents
                                                                                                                                                        • Basic characteristics of soils
                                                                                                                                                        • Seepage
                                                                                                                                                        • Effective stress
                                                                                                                                                        • Shear strength
                                                                                                                                                        • Stresses and displacements
                                                                                                                                                        • Lateral earth pressure
                                                                                                                                                        • Consolidation theory
                                                                                                                                                        • Bearing capacity
                                                                                                                                                        • Stability of slopes

                                                                                                                                                          Thus using Equation 528

                                                                                                                                                          si frac14 078 041 68 176

                                                                                                                                                          65frac14 6mm

                                                                                                                                                          Consolidation settlement

                                                                                                                                                          Layer z (m) Area (m2) (kNm2) mvH (mm)

                                                                                                                                                          1 25 2012 520 2082 75 2512 333 1333 125 3012 232 93

                                                                                                                                                          434 (sod)

                                                                                                                                                          Equivalent diameter frac14 1986m thus HB frac14 151986 frac14 076 Now A frac14 028 hencefrom Figure 712 13 frac14 056 Then from Equation 710

                                                                                                                                                          sc frac14 056 434 frac14 24mm

                                                                                                                                                          The total settlement is (6thorn 24) frac14 30mm

                                                                                                                                                          813

                                                                                                                                                          At base level N frac14 26 Then using Equation 830

                                                                                                                                                          qb frac14 40NDb

                                                                                                                                                          Bfrac14 40 26 2

                                                                                                                                                          025frac14 8320 kN=m2

                                                                                                                                                          ethCheck lt400N ie 400 26 frac14 10 400 kN=m2THORN

                                                                                                                                                          Figure Q812

                                                                                                                                                          Bearing capacity 71

                                                                                                                                                          Over the length embedded in sand

                                                                                                                                                          N frac14 21 ie18thorn 24

                                                                                                                                                          2

                                                                                                                                                          Using Equation 831

                                                                                                                                                          qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                                          For a single pile

                                                                                                                                                          Qf frac14 Abqb thorn Asqs

                                                                                                                                                          frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                                          For the pile group assuming a group efficiency of 12

                                                                                                                                                          XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                                          Then the load factor is

                                                                                                                                                          F frac14 6523

                                                                                                                                                          2000thorn 1000frac14 21

                                                                                                                                                          (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                                          Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                                          150frac14 5547 kNm2

                                                                                                                                                          Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                                          150frac14 28 kNm2

                                                                                                                                                          Characteristic base and shaft resistances for a single pile

                                                                                                                                                          Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                                          Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                                          For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                                          Design bearing resistance Rcd frac14 347

                                                                                                                                                          130thorn 56

                                                                                                                                                          130frac14 310 kN

                                                                                                                                                          For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                                          Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                                          (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                                          From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                                          72 Bearing capacity

                                                                                                                                                          N frac14 24thorn 26thorn 34

                                                                                                                                                          3frac14 28

                                                                                                                                                          Ic frac14 171

                                                                                                                                                          2814frac14 0016 ethEquation 818THORN

                                                                                                                                                          s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                                          The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                                          814

                                                                                                                                                          Using Equation 841

                                                                                                                                                          Tf frac14 DLcu thorn

                                                                                                                                                          4ethD2 d2THORNcuNc

                                                                                                                                                          frac14 eth 02 5 06 110THORN thorn

                                                                                                                                                          4eth022 012THORN110 9

                                                                                                                                                          frac14 207thorn 23 frac14 230 kN

                                                                                                                                                          Figure Q813

                                                                                                                                                          Bearing capacity 73

                                                                                                                                                          Chapter 9

                                                                                                                                                          Stability of slopes

                                                                                                                                                          91

                                                                                                                                                          Referring to Figure Q91

                                                                                                                                                          W frac14 417 19 frac14 792 kN=m

                                                                                                                                                          Q frac14 20 28 frac14 56 kN=m

                                                                                                                                                          Arc lengthAB frac14

                                                                                                                                                          180 73 90 frac14 115m

                                                                                                                                                          Arc length BC frac14

                                                                                                                                                          180 28 90 frac14 44m

                                                                                                                                                          The factor of safety is given by

                                                                                                                                                          F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                                          90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                                          Depth of tension crack z0 frac14 2cu

                                                                                                                                                          frac14 2 20

                                                                                                                                                          19frac14 21m

                                                                                                                                                          Arc length BD frac14

                                                                                                                                                          180 13

                                                                                                                                                          1

                                                                                                                                                          2 90 frac14 21m

                                                                                                                                                          F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                                          The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                                          Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                                          14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                                          Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                                          The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                                          92

                                                                                                                                                          u frac14 0

                                                                                                                                                          Depth factor D frac14 11

                                                                                                                                                          9frac14 122

                                                                                                                                                          Using Equation 92 with F frac14 10

                                                                                                                                                          Ns frac14 cu

                                                                                                                                                          FHfrac14 30

                                                                                                                                                          10 19 9frac14 0175

                                                                                                                                                          Hence from Figure 93

                                                                                                                                                          frac14 50

                                                                                                                                                          For F frac14 12

                                                                                                                                                          Ns frac14 30

                                                                                                                                                          12 19 9frac14 0146

                                                                                                                                                          frac14 27

                                                                                                                                                          93

                                                                                                                                                          Refer to Figure Q93

                                                                                                                                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                          1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                                          74 m

                                                                                                                                                          214 1deg

                                                                                                                                                          213 1deg

                                                                                                                                                          39 m

                                                                                                                                                          WB

                                                                                                                                                          D

                                                                                                                                                          C

                                                                                                                                                          28 m

                                                                                                                                                          21 m

                                                                                                                                                          A

                                                                                                                                                          Q

                                                                                                                                                          Soil (1)Soil (2)

                                                                                                                                                          73deg

                                                                                                                                                          Figure Q91

                                                                                                                                                          Stability of slopes 75

                                                                                                                                                          Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                          9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                                          599 256 328 1372

                                                                                                                                                          Figure Q93

                                                                                                                                                          76 Stability of slopes

                                                                                                                                                          XW cos frac14 b

                                                                                                                                                          Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                                          W sin frac14 bX

                                                                                                                                                          h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                                          Arc length La frac14

                                                                                                                                                          180 57

                                                                                                                                                          1

                                                                                                                                                          2 326 frac14 327m

                                                                                                                                                          The factor of safety is given by

                                                                                                                                                          F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                                          W sin

                                                                                                                                                          frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                                          frac14 091

                                                                                                                                                          According to the limit state method

                                                                                                                                                          0d frac14 tan1tan 32

                                                                                                                                                          125

                                                                                                                                                          frac14 265

                                                                                                                                                          c0 frac14 8

                                                                                                                                                          160frac14 5 kN=m2

                                                                                                                                                          Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                                          Design disturbing moment frac14 1075 kN=m

                                                                                                                                                          The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                                          94

                                                                                                                                                          F frac14 1

                                                                                                                                                          W sin

                                                                                                                                                          Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                                          1thorn ethtan tan0=FTHORN

                                                                                                                                                          c0 frac14 8 kN=m2

                                                                                                                                                          0 frac14 32

                                                                                                                                                          c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                                          W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                                          Try F frac14 100

                                                                                                                                                          tan0

                                                                                                                                                          Ffrac14 0625

                                                                                                                                                          Stability of slopes 77

                                                                                                                                                          Values of u are as obtained in Figure Q93

                                                                                                                                                          SliceNo

                                                                                                                                                          h(m)

                                                                                                                                                          W frac14 bh(kNm)

                                                                                                                                                          W sin(kNm)

                                                                                                                                                          ub(kNm)

                                                                                                                                                          c0bthorn (W ub) tan0(kNm)

                                                                                                                                                          sec

                                                                                                                                                          1thorn (tan tan0)FProduct(kNm)

                                                                                                                                                          1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                                          23 33 30 1042 31

                                                                                                                                                          3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                                          224 92 72 0931 67

                                                                                                                                                          6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                                          12 58 112 90 0889 80

                                                                                                                                                          8 60 252 1812

                                                                                                                                                          80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                                          10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                                          2154 88 116 0853 99

                                                                                                                                                          14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                                          1074 1091

                                                                                                                                                          F frac14 1091

                                                                                                                                                          1074frac14 102 (assumed value 100)

                                                                                                                                                          Thus

                                                                                                                                                          F frac14 101

                                                                                                                                                          95

                                                                                                                                                          F frac14 1

                                                                                                                                                          W sin

                                                                                                                                                          XfWeth1 ruTHORN tan0g sec

                                                                                                                                                          1thorn ethtan tan0THORN=F

                                                                                                                                                          0 frac14 33

                                                                                                                                                          ru frac14 020

                                                                                                                                                          W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                                          eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                                          Try F frac14 110

                                                                                                                                                          tan 0

                                                                                                                                                          Ffrac14 tan 33

                                                                                                                                                          110frac14 0590

                                                                                                                                                          78 Stability of slopes

                                                                                                                                                          Referring to Figure Q95

                                                                                                                                                          SliceNo

                                                                                                                                                          h(m)

                                                                                                                                                          W frac14 bh(kNm)

                                                                                                                                                          W sin(kNm)

                                                                                                                                                          W(1 ru) tan0(kNm)

                                                                                                                                                          sec

                                                                                                                                                          1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                                          1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                                          2120 234 0892 209

                                                                                                                                                          4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                                          1185 1271

                                                                                                                                                          Figure Q95

                                                                                                                                                          Stability of slopes 79

                                                                                                                                                          F frac14 1271

                                                                                                                                                          1185frac14 107

                                                                                                                                                          The trial value was 110 therefore take F to be 108

                                                                                                                                                          96

                                                                                                                                                          (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                          F frac14 0

                                                                                                                                                          sat

                                                                                                                                                          tan0

                                                                                                                                                          tan

                                                                                                                                                          ptie 15 frac14 92

                                                                                                                                                          19

                                                                                                                                                          tan 36

                                                                                                                                                          tan

                                                                                                                                                          tan frac14 0234

                                                                                                                                                          frac14 13

                                                                                                                                                          Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                          F frac14 tan0

                                                                                                                                                          tan

                                                                                                                                                          frac14 tan 36

                                                                                                                                                          tan 13

                                                                                                                                                          frac14 31

                                                                                                                                                          (b) 0d frac14 tan1tan 36

                                                                                                                                                          125

                                                                                                                                                          frac14 30

                                                                                                                                                          Depth of potential failure surface frac14 z

                                                                                                                                                          Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                          frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                          frac14 504z kN

                                                                                                                                                          Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                          frac14 19 z sin 13 cos 13

                                                                                                                                                          frac14 416z kN

                                                                                                                                                          Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                          80 Stability of slopes

                                                                                                                                                          • Book Cover
                                                                                                                                                          • Title
                                                                                                                                                          • Contents
                                                                                                                                                          • Basic characteristics of soils
                                                                                                                                                          • Seepage
                                                                                                                                                          • Effective stress
                                                                                                                                                          • Shear strength
                                                                                                                                                          • Stresses and displacements
                                                                                                                                                          • Lateral earth pressure
                                                                                                                                                          • Consolidation theory
                                                                                                                                                          • Bearing capacity
                                                                                                                                                          • Stability of slopes

                                                                                                                                                            Over the length embedded in sand

                                                                                                                                                            N frac14 21 ie18thorn 24

                                                                                                                                                            2

                                                                                                                                                            Using Equation 831

                                                                                                                                                            qs frac14 2N frac14 2 21 frac14 42 kN=m2

                                                                                                                                                            For a single pile

                                                                                                                                                            Qf frac14 Abqb thorn Asqs

                                                                                                                                                            frac14 eth0252 8320THORN thorn eth4 025 2 42THORNfrac14 520thorn 84 frac14 604 kN

                                                                                                                                                            For the pile group assuming a group efficiency of 12

                                                                                                                                                            XQf frac14 12 9 604 frac14 6523 kN

                                                                                                                                                            Then the load factor is

                                                                                                                                                            F frac14 6523

                                                                                                                                                            2000thorn 1000frac14 21

                                                                                                                                                            (b) Design load Fcd frac14 2000thorn (1000 130) frac14 3300 kN

                                                                                                                                                            Characteristic base resistance per unit area qbk frac14 8320

                                                                                                                                                            150frac14 5547 kNm2

                                                                                                                                                            Characteristic shaft resistance per unit area qsk frac14 42

                                                                                                                                                            150frac14 28 kNm2

                                                                                                                                                            Characteristic base and shaft resistances for a single pile

                                                                                                                                                            Rbk frac14 0252 5547 frac14 347 kN

                                                                                                                                                            Rsk frac14 4 025 2 28 frac14 56 kN

                                                                                                                                                            For a driven pile the partial factors are b frac14 s frac14 130

                                                                                                                                                            Design bearing resistance Rcd frac14 347

                                                                                                                                                            130thorn 56

                                                                                                                                                            130frac14 310 kN

                                                                                                                                                            For the pile group Rcd frac14 12 9 310 frac14 3348 kN

                                                                                                                                                            Rcd gt Fcd (3348 gt 3300) therefore the bearing resistance limit state is satisfied

                                                                                                                                                            (c) Referring to Figure Q813 the equivalent raft is 242m squareFor the serviceability limit state the design load Fcd frac14 2000thorn 1000 frac14 3000 kNThe pressure on the equivalent raft q frac14 3000=2422 frac14 512 kNm2

                                                                                                                                                            From Figure 812 for B frac14 242m the value of zI is 18m Therefore N values betweendepths of 133 and 313m should be used Thus

                                                                                                                                                            72 Bearing capacity

                                                                                                                                                            N frac14 24thorn 26thorn 34

                                                                                                                                                            3frac14 28

                                                                                                                                                            Ic frac14 171

                                                                                                                                                            2814frac14 0016 ethEquation 818THORN

                                                                                                                                                            s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                                            The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                                            814

                                                                                                                                                            Using Equation 841

                                                                                                                                                            Tf frac14 DLcu thorn

                                                                                                                                                            4ethD2 d2THORNcuNc

                                                                                                                                                            frac14 eth 02 5 06 110THORN thorn

                                                                                                                                                            4eth022 012THORN110 9

                                                                                                                                                            frac14 207thorn 23 frac14 230 kN

                                                                                                                                                            Figure Q813

                                                                                                                                                            Bearing capacity 73

                                                                                                                                                            Chapter 9

                                                                                                                                                            Stability of slopes

                                                                                                                                                            91

                                                                                                                                                            Referring to Figure Q91

                                                                                                                                                            W frac14 417 19 frac14 792 kN=m

                                                                                                                                                            Q frac14 20 28 frac14 56 kN=m

                                                                                                                                                            Arc lengthAB frac14

                                                                                                                                                            180 73 90 frac14 115m

                                                                                                                                                            Arc length BC frac14

                                                                                                                                                            180 28 90 frac14 44m

                                                                                                                                                            The factor of safety is given by

                                                                                                                                                            F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                                            90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                                            Depth of tension crack z0 frac14 2cu

                                                                                                                                                            frac14 2 20

                                                                                                                                                            19frac14 21m

                                                                                                                                                            Arc length BD frac14

                                                                                                                                                            180 13

                                                                                                                                                            1

                                                                                                                                                            2 90 frac14 21m

                                                                                                                                                            F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                                            The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                                            Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                                            14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                                            Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                                            The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                                            92

                                                                                                                                                            u frac14 0

                                                                                                                                                            Depth factor D frac14 11

                                                                                                                                                            9frac14 122

                                                                                                                                                            Using Equation 92 with F frac14 10

                                                                                                                                                            Ns frac14 cu

                                                                                                                                                            FHfrac14 30

                                                                                                                                                            10 19 9frac14 0175

                                                                                                                                                            Hence from Figure 93

                                                                                                                                                            frac14 50

                                                                                                                                                            For F frac14 12

                                                                                                                                                            Ns frac14 30

                                                                                                                                                            12 19 9frac14 0146

                                                                                                                                                            frac14 27

                                                                                                                                                            93

                                                                                                                                                            Refer to Figure Q93

                                                                                                                                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                            1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                                            74 m

                                                                                                                                                            214 1deg

                                                                                                                                                            213 1deg

                                                                                                                                                            39 m

                                                                                                                                                            WB

                                                                                                                                                            D

                                                                                                                                                            C

                                                                                                                                                            28 m

                                                                                                                                                            21 m

                                                                                                                                                            A

                                                                                                                                                            Q

                                                                                                                                                            Soil (1)Soil (2)

                                                                                                                                                            73deg

                                                                                                                                                            Figure Q91

                                                                                                                                                            Stability of slopes 75

                                                                                                                                                            Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                            9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                                            599 256 328 1372

                                                                                                                                                            Figure Q93

                                                                                                                                                            76 Stability of slopes

                                                                                                                                                            XW cos frac14 b

                                                                                                                                                            Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                                            W sin frac14 bX

                                                                                                                                                            h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                                            Arc length La frac14

                                                                                                                                                            180 57

                                                                                                                                                            1

                                                                                                                                                            2 326 frac14 327m

                                                                                                                                                            The factor of safety is given by

                                                                                                                                                            F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                                            W sin

                                                                                                                                                            frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                                            frac14 091

                                                                                                                                                            According to the limit state method

                                                                                                                                                            0d frac14 tan1tan 32

                                                                                                                                                            125

                                                                                                                                                            frac14 265

                                                                                                                                                            c0 frac14 8

                                                                                                                                                            160frac14 5 kN=m2

                                                                                                                                                            Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                                            Design disturbing moment frac14 1075 kN=m

                                                                                                                                                            The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                                            94

                                                                                                                                                            F frac14 1

                                                                                                                                                            W sin

                                                                                                                                                            Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                                            1thorn ethtan tan0=FTHORN

                                                                                                                                                            c0 frac14 8 kN=m2

                                                                                                                                                            0 frac14 32

                                                                                                                                                            c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                                            W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                                            Try F frac14 100

                                                                                                                                                            tan0

                                                                                                                                                            Ffrac14 0625

                                                                                                                                                            Stability of slopes 77

                                                                                                                                                            Values of u are as obtained in Figure Q93

                                                                                                                                                            SliceNo

                                                                                                                                                            h(m)

                                                                                                                                                            W frac14 bh(kNm)

                                                                                                                                                            W sin(kNm)

                                                                                                                                                            ub(kNm)

                                                                                                                                                            c0bthorn (W ub) tan0(kNm)

                                                                                                                                                            sec

                                                                                                                                                            1thorn (tan tan0)FProduct(kNm)

                                                                                                                                                            1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                                            23 33 30 1042 31

                                                                                                                                                            3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                                            224 92 72 0931 67

                                                                                                                                                            6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                                            12 58 112 90 0889 80

                                                                                                                                                            8 60 252 1812

                                                                                                                                                            80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                                            10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                                            2154 88 116 0853 99

                                                                                                                                                            14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                                            1074 1091

                                                                                                                                                            F frac14 1091

                                                                                                                                                            1074frac14 102 (assumed value 100)

                                                                                                                                                            Thus

                                                                                                                                                            F frac14 101

                                                                                                                                                            95

                                                                                                                                                            F frac14 1

                                                                                                                                                            W sin

                                                                                                                                                            XfWeth1 ruTHORN tan0g sec

                                                                                                                                                            1thorn ethtan tan0THORN=F

                                                                                                                                                            0 frac14 33

                                                                                                                                                            ru frac14 020

                                                                                                                                                            W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                                            eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                                            Try F frac14 110

                                                                                                                                                            tan 0

                                                                                                                                                            Ffrac14 tan 33

                                                                                                                                                            110frac14 0590

                                                                                                                                                            78 Stability of slopes

                                                                                                                                                            Referring to Figure Q95

                                                                                                                                                            SliceNo

                                                                                                                                                            h(m)

                                                                                                                                                            W frac14 bh(kNm)

                                                                                                                                                            W sin(kNm)

                                                                                                                                                            W(1 ru) tan0(kNm)

                                                                                                                                                            sec

                                                                                                                                                            1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                                            1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                                            2120 234 0892 209

                                                                                                                                                            4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                                            1185 1271

                                                                                                                                                            Figure Q95

                                                                                                                                                            Stability of slopes 79

                                                                                                                                                            F frac14 1271

                                                                                                                                                            1185frac14 107

                                                                                                                                                            The trial value was 110 therefore take F to be 108

                                                                                                                                                            96

                                                                                                                                                            (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                            F frac14 0

                                                                                                                                                            sat

                                                                                                                                                            tan0

                                                                                                                                                            tan

                                                                                                                                                            ptie 15 frac14 92

                                                                                                                                                            19

                                                                                                                                                            tan 36

                                                                                                                                                            tan

                                                                                                                                                            tan frac14 0234

                                                                                                                                                            frac14 13

                                                                                                                                                            Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                            F frac14 tan0

                                                                                                                                                            tan

                                                                                                                                                            frac14 tan 36

                                                                                                                                                            tan 13

                                                                                                                                                            frac14 31

                                                                                                                                                            (b) 0d frac14 tan1tan 36

                                                                                                                                                            125

                                                                                                                                                            frac14 30

                                                                                                                                                            Depth of potential failure surface frac14 z

                                                                                                                                                            Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                            frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                            frac14 504z kN

                                                                                                                                                            Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                            frac14 19 z sin 13 cos 13

                                                                                                                                                            frac14 416z kN

                                                                                                                                                            Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                            80 Stability of slopes

                                                                                                                                                            • Book Cover
                                                                                                                                                            • Title
                                                                                                                                                            • Contents
                                                                                                                                                            • Basic characteristics of soils
                                                                                                                                                            • Seepage
                                                                                                                                                            • Effective stress
                                                                                                                                                            • Shear strength
                                                                                                                                                            • Stresses and displacements
                                                                                                                                                            • Lateral earth pressure
                                                                                                                                                            • Consolidation theory
                                                                                                                                                            • Bearing capacity
                                                                                                                                                            • Stability of slopes

                                                                                                                                                              N frac14 24thorn 26thorn 34

                                                                                                                                                              3frac14 28

                                                                                                                                                              Ic frac14 171

                                                                                                                                                              2814frac14 0016 ethEquation 818THORN

                                                                                                                                                              s frac14 512 24207 0016 frac14 15mm ethEquation 819ethaTHORNTHORN

                                                                                                                                                              The settlement is less than 20mm therefore the serviceability limit state is satisfied

                                                                                                                                                              814

                                                                                                                                                              Using Equation 841

                                                                                                                                                              Tf frac14 DLcu thorn

                                                                                                                                                              4ethD2 d2THORNcuNc

                                                                                                                                                              frac14 eth 02 5 06 110THORN thorn

                                                                                                                                                              4eth022 012THORN110 9

                                                                                                                                                              frac14 207thorn 23 frac14 230 kN

                                                                                                                                                              Figure Q813

                                                                                                                                                              Bearing capacity 73

                                                                                                                                                              Chapter 9

                                                                                                                                                              Stability of slopes

                                                                                                                                                              91

                                                                                                                                                              Referring to Figure Q91

                                                                                                                                                              W frac14 417 19 frac14 792 kN=m

                                                                                                                                                              Q frac14 20 28 frac14 56 kN=m

                                                                                                                                                              Arc lengthAB frac14

                                                                                                                                                              180 73 90 frac14 115m

                                                                                                                                                              Arc length BC frac14

                                                                                                                                                              180 28 90 frac14 44m

                                                                                                                                                              The factor of safety is given by

                                                                                                                                                              F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                                              90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                                              Depth of tension crack z0 frac14 2cu

                                                                                                                                                              frac14 2 20

                                                                                                                                                              19frac14 21m

                                                                                                                                                              Arc length BD frac14

                                                                                                                                                              180 13

                                                                                                                                                              1

                                                                                                                                                              2 90 frac14 21m

                                                                                                                                                              F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                                              The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                                              Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                                              14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                                              Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                                              The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                                              92

                                                                                                                                                              u frac14 0

                                                                                                                                                              Depth factor D frac14 11

                                                                                                                                                              9frac14 122

                                                                                                                                                              Using Equation 92 with F frac14 10

                                                                                                                                                              Ns frac14 cu

                                                                                                                                                              FHfrac14 30

                                                                                                                                                              10 19 9frac14 0175

                                                                                                                                                              Hence from Figure 93

                                                                                                                                                              frac14 50

                                                                                                                                                              For F frac14 12

                                                                                                                                                              Ns frac14 30

                                                                                                                                                              12 19 9frac14 0146

                                                                                                                                                              frac14 27

                                                                                                                                                              93

                                                                                                                                                              Refer to Figure Q93

                                                                                                                                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                              1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                                              74 m

                                                                                                                                                              214 1deg

                                                                                                                                                              213 1deg

                                                                                                                                                              39 m

                                                                                                                                                              WB

                                                                                                                                                              D

                                                                                                                                                              C

                                                                                                                                                              28 m

                                                                                                                                                              21 m

                                                                                                                                                              A

                                                                                                                                                              Q

                                                                                                                                                              Soil (1)Soil (2)

                                                                                                                                                              73deg

                                                                                                                                                              Figure Q91

                                                                                                                                                              Stability of slopes 75

                                                                                                                                                              Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                              9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                                              599 256 328 1372

                                                                                                                                                              Figure Q93

                                                                                                                                                              76 Stability of slopes

                                                                                                                                                              XW cos frac14 b

                                                                                                                                                              Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                                              W sin frac14 bX

                                                                                                                                                              h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                                              Arc length La frac14

                                                                                                                                                              180 57

                                                                                                                                                              1

                                                                                                                                                              2 326 frac14 327m

                                                                                                                                                              The factor of safety is given by

                                                                                                                                                              F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                                              W sin

                                                                                                                                                              frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                                              frac14 091

                                                                                                                                                              According to the limit state method

                                                                                                                                                              0d frac14 tan1tan 32

                                                                                                                                                              125

                                                                                                                                                              frac14 265

                                                                                                                                                              c0 frac14 8

                                                                                                                                                              160frac14 5 kN=m2

                                                                                                                                                              Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                                              Design disturbing moment frac14 1075 kN=m

                                                                                                                                                              The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                                              94

                                                                                                                                                              F frac14 1

                                                                                                                                                              W sin

                                                                                                                                                              Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                                              1thorn ethtan tan0=FTHORN

                                                                                                                                                              c0 frac14 8 kN=m2

                                                                                                                                                              0 frac14 32

                                                                                                                                                              c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                                              W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                                              Try F frac14 100

                                                                                                                                                              tan0

                                                                                                                                                              Ffrac14 0625

                                                                                                                                                              Stability of slopes 77

                                                                                                                                                              Values of u are as obtained in Figure Q93

                                                                                                                                                              SliceNo

                                                                                                                                                              h(m)

                                                                                                                                                              W frac14 bh(kNm)

                                                                                                                                                              W sin(kNm)

                                                                                                                                                              ub(kNm)

                                                                                                                                                              c0bthorn (W ub) tan0(kNm)

                                                                                                                                                              sec

                                                                                                                                                              1thorn (tan tan0)FProduct(kNm)

                                                                                                                                                              1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                                              23 33 30 1042 31

                                                                                                                                                              3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                                              224 92 72 0931 67

                                                                                                                                                              6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                                              12 58 112 90 0889 80

                                                                                                                                                              8 60 252 1812

                                                                                                                                                              80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                                              10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                                              2154 88 116 0853 99

                                                                                                                                                              14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                                              1074 1091

                                                                                                                                                              F frac14 1091

                                                                                                                                                              1074frac14 102 (assumed value 100)

                                                                                                                                                              Thus

                                                                                                                                                              F frac14 101

                                                                                                                                                              95

                                                                                                                                                              F frac14 1

                                                                                                                                                              W sin

                                                                                                                                                              XfWeth1 ruTHORN tan0g sec

                                                                                                                                                              1thorn ethtan tan0THORN=F

                                                                                                                                                              0 frac14 33

                                                                                                                                                              ru frac14 020

                                                                                                                                                              W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                                              eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                                              Try F frac14 110

                                                                                                                                                              tan 0

                                                                                                                                                              Ffrac14 tan 33

                                                                                                                                                              110frac14 0590

                                                                                                                                                              78 Stability of slopes

                                                                                                                                                              Referring to Figure Q95

                                                                                                                                                              SliceNo

                                                                                                                                                              h(m)

                                                                                                                                                              W frac14 bh(kNm)

                                                                                                                                                              W sin(kNm)

                                                                                                                                                              W(1 ru) tan0(kNm)

                                                                                                                                                              sec

                                                                                                                                                              1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                                              1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                                              2120 234 0892 209

                                                                                                                                                              4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                                              1185 1271

                                                                                                                                                              Figure Q95

                                                                                                                                                              Stability of slopes 79

                                                                                                                                                              F frac14 1271

                                                                                                                                                              1185frac14 107

                                                                                                                                                              The trial value was 110 therefore take F to be 108

                                                                                                                                                              96

                                                                                                                                                              (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                              F frac14 0

                                                                                                                                                              sat

                                                                                                                                                              tan0

                                                                                                                                                              tan

                                                                                                                                                              ptie 15 frac14 92

                                                                                                                                                              19

                                                                                                                                                              tan 36

                                                                                                                                                              tan

                                                                                                                                                              tan frac14 0234

                                                                                                                                                              frac14 13

                                                                                                                                                              Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                              F frac14 tan0

                                                                                                                                                              tan

                                                                                                                                                              frac14 tan 36

                                                                                                                                                              tan 13

                                                                                                                                                              frac14 31

                                                                                                                                                              (b) 0d frac14 tan1tan 36

                                                                                                                                                              125

                                                                                                                                                              frac14 30

                                                                                                                                                              Depth of potential failure surface frac14 z

                                                                                                                                                              Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                              frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                              frac14 504z kN

                                                                                                                                                              Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                              frac14 19 z sin 13 cos 13

                                                                                                                                                              frac14 416z kN

                                                                                                                                                              Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                              80 Stability of slopes

                                                                                                                                                              • Book Cover
                                                                                                                                                              • Title
                                                                                                                                                              • Contents
                                                                                                                                                              • Basic characteristics of soils
                                                                                                                                                              • Seepage
                                                                                                                                                              • Effective stress
                                                                                                                                                              • Shear strength
                                                                                                                                                              • Stresses and displacements
                                                                                                                                                              • Lateral earth pressure
                                                                                                                                                              • Consolidation theory
                                                                                                                                                              • Bearing capacity
                                                                                                                                                              • Stability of slopes

                                                                                                                                                                Chapter 9

                                                                                                                                                                Stability of slopes

                                                                                                                                                                91

                                                                                                                                                                Referring to Figure Q91

                                                                                                                                                                W frac14 417 19 frac14 792 kN=m

                                                                                                                                                                Q frac14 20 28 frac14 56 kN=m

                                                                                                                                                                Arc lengthAB frac14

                                                                                                                                                                180 73 90 frac14 115m

                                                                                                                                                                Arc length BC frac14

                                                                                                                                                                180 28 90 frac14 44m

                                                                                                                                                                The factor of safety is given by

                                                                                                                                                                F frac14 rethcuLaTHORNWd1 thornQd2 frac14

                                                                                                                                                                90frac12eth30 44THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 167

                                                                                                                                                                Depth of tension crack z0 frac14 2cu

                                                                                                                                                                frac14 2 20

                                                                                                                                                                19frac14 21m

                                                                                                                                                                Arc length BD frac14

                                                                                                                                                                180 13

                                                                                                                                                                1

                                                                                                                                                                2 90 frac14 21m

                                                                                                                                                                F frac14 90frac12eth30 21THORN thorn eth45 115THORNeth792 39THORN thorn eth56 74THORN frac14 149

                                                                                                                                                                The surcharge is a variable action therefore a partial factor of 130 is applied In thelimit state method the design values of undrained strength (cud) are (30140) and (45140) kNm2

                                                                                                                                                                Design resisting moment frac14 rXethcudLaTHORN frac14 90

                                                                                                                                                                14frac12eth30 44THORN thorn eth45 115THORN frac14 4175 kN=m

                                                                                                                                                                Design disturbing moment frac14Wd1 thornQd2 frac14 eth792 39THORN thorn eth56 130 74THORNfrac14 3628 kN=m

                                                                                                                                                                The design resisting moment is greater than the design disturbing moment thereforeoverall stability is assured

                                                                                                                                                                92

                                                                                                                                                                u frac14 0

                                                                                                                                                                Depth factor D frac14 11

                                                                                                                                                                9frac14 122

                                                                                                                                                                Using Equation 92 with F frac14 10

                                                                                                                                                                Ns frac14 cu

                                                                                                                                                                FHfrac14 30

                                                                                                                                                                10 19 9frac14 0175

                                                                                                                                                                Hence from Figure 93

                                                                                                                                                                frac14 50

                                                                                                                                                                For F frac14 12

                                                                                                                                                                Ns frac14 30

                                                                                                                                                                12 19 9frac14 0146

                                                                                                                                                                frac14 27

                                                                                                                                                                93

                                                                                                                                                                Refer to Figure Q93

                                                                                                                                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                                1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                                                74 m

                                                                                                                                                                214 1deg

                                                                                                                                                                213 1deg

                                                                                                                                                                39 m

                                                                                                                                                                WB

                                                                                                                                                                D

                                                                                                                                                                C

                                                                                                                                                                28 m

                                                                                                                                                                21 m

                                                                                                                                                                A

                                                                                                                                                                Q

                                                                                                                                                                Soil (1)Soil (2)

                                                                                                                                                                73deg

                                                                                                                                                                Figure Q91

                                                                                                                                                                Stability of slopes 75

                                                                                                                                                                Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                                9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                                                599 256 328 1372

                                                                                                                                                                Figure Q93

                                                                                                                                                                76 Stability of slopes

                                                                                                                                                                XW cos frac14 b

                                                                                                                                                                Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                                                W sin frac14 bX

                                                                                                                                                                h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                                                Arc length La frac14

                                                                                                                                                                180 57

                                                                                                                                                                1

                                                                                                                                                                2 326 frac14 327m

                                                                                                                                                                The factor of safety is given by

                                                                                                                                                                F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                                                W sin

                                                                                                                                                                frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                                                frac14 091

                                                                                                                                                                According to the limit state method

                                                                                                                                                                0d frac14 tan1tan 32

                                                                                                                                                                125

                                                                                                                                                                frac14 265

                                                                                                                                                                c0 frac14 8

                                                                                                                                                                160frac14 5 kN=m2

                                                                                                                                                                Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                                                Design disturbing moment frac14 1075 kN=m

                                                                                                                                                                The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                                                94

                                                                                                                                                                F frac14 1

                                                                                                                                                                W sin

                                                                                                                                                                Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                                                1thorn ethtan tan0=FTHORN

                                                                                                                                                                c0 frac14 8 kN=m2

                                                                                                                                                                0 frac14 32

                                                                                                                                                                c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                                                W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                                                Try F frac14 100

                                                                                                                                                                tan0

                                                                                                                                                                Ffrac14 0625

                                                                                                                                                                Stability of slopes 77

                                                                                                                                                                Values of u are as obtained in Figure Q93

                                                                                                                                                                SliceNo

                                                                                                                                                                h(m)

                                                                                                                                                                W frac14 bh(kNm)

                                                                                                                                                                W sin(kNm)

                                                                                                                                                                ub(kNm)

                                                                                                                                                                c0bthorn (W ub) tan0(kNm)

                                                                                                                                                                sec

                                                                                                                                                                1thorn (tan tan0)FProduct(kNm)

                                                                                                                                                                1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                                                23 33 30 1042 31

                                                                                                                                                                3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                                                224 92 72 0931 67

                                                                                                                                                                6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                                                12 58 112 90 0889 80

                                                                                                                                                                8 60 252 1812

                                                                                                                                                                80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                                                10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                                                2154 88 116 0853 99

                                                                                                                                                                14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                                                1074 1091

                                                                                                                                                                F frac14 1091

                                                                                                                                                                1074frac14 102 (assumed value 100)

                                                                                                                                                                Thus

                                                                                                                                                                F frac14 101

                                                                                                                                                                95

                                                                                                                                                                F frac14 1

                                                                                                                                                                W sin

                                                                                                                                                                XfWeth1 ruTHORN tan0g sec

                                                                                                                                                                1thorn ethtan tan0THORN=F

                                                                                                                                                                0 frac14 33

                                                                                                                                                                ru frac14 020

                                                                                                                                                                W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                                                eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                                                Try F frac14 110

                                                                                                                                                                tan 0

                                                                                                                                                                Ffrac14 tan 33

                                                                                                                                                                110frac14 0590

                                                                                                                                                                78 Stability of slopes

                                                                                                                                                                Referring to Figure Q95

                                                                                                                                                                SliceNo

                                                                                                                                                                h(m)

                                                                                                                                                                W frac14 bh(kNm)

                                                                                                                                                                W sin(kNm)

                                                                                                                                                                W(1 ru) tan0(kNm)

                                                                                                                                                                sec

                                                                                                                                                                1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                                                1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                                                2120 234 0892 209

                                                                                                                                                                4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                                                1185 1271

                                                                                                                                                                Figure Q95

                                                                                                                                                                Stability of slopes 79

                                                                                                                                                                F frac14 1271

                                                                                                                                                                1185frac14 107

                                                                                                                                                                The trial value was 110 therefore take F to be 108

                                                                                                                                                                96

                                                                                                                                                                (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                                F frac14 0

                                                                                                                                                                sat

                                                                                                                                                                tan0

                                                                                                                                                                tan

                                                                                                                                                                ptie 15 frac14 92

                                                                                                                                                                19

                                                                                                                                                                tan 36

                                                                                                                                                                tan

                                                                                                                                                                tan frac14 0234

                                                                                                                                                                frac14 13

                                                                                                                                                                Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                                F frac14 tan0

                                                                                                                                                                tan

                                                                                                                                                                frac14 tan 36

                                                                                                                                                                tan 13

                                                                                                                                                                frac14 31

                                                                                                                                                                (b) 0d frac14 tan1tan 36

                                                                                                                                                                125

                                                                                                                                                                frac14 30

                                                                                                                                                                Depth of potential failure surface frac14 z

                                                                                                                                                                Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                                frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                                frac14 504z kN

                                                                                                                                                                Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                                frac14 19 z sin 13 cos 13

                                                                                                                                                                frac14 416z kN

                                                                                                                                                                Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                                80 Stability of slopes

                                                                                                                                                                • Book Cover
                                                                                                                                                                • Title
                                                                                                                                                                • Contents
                                                                                                                                                                • Basic characteristics of soils
                                                                                                                                                                • Seepage
                                                                                                                                                                • Effective stress
                                                                                                                                                                • Shear strength
                                                                                                                                                                • Stresses and displacements
                                                                                                                                                                • Lateral earth pressure
                                                                                                                                                                • Consolidation theory
                                                                                                                                                                • Bearing capacity
                                                                                                                                                                • Stability of slopes

                                                                                                                                                                  92

                                                                                                                                                                  u frac14 0

                                                                                                                                                                  Depth factor D frac14 11

                                                                                                                                                                  9frac14 122

                                                                                                                                                                  Using Equation 92 with F frac14 10

                                                                                                                                                                  Ns frac14 cu

                                                                                                                                                                  FHfrac14 30

                                                                                                                                                                  10 19 9frac14 0175

                                                                                                                                                                  Hence from Figure 93

                                                                                                                                                                  frac14 50

                                                                                                                                                                  For F frac14 12

                                                                                                                                                                  Ns frac14 30

                                                                                                                                                                  12 19 9frac14 0146

                                                                                                                                                                  frac14 27

                                                                                                                                                                  93

                                                                                                                                                                  Refer to Figure Q93

                                                                                                                                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                                  1 05 ndash 07 69 12 82 12 01 17 167 20 333 24 ndash 30 294 20 594 34 02 39 382 20 765 43 05 47 461 21 976 49 09 51 500 21 1057 53 14 57 559 21 1178 57 18 58 568 21 119

                                                                                                                                                                  74 m

                                                                                                                                                                  214 1deg

                                                                                                                                                                  213 1deg

                                                                                                                                                                  39 m

                                                                                                                                                                  WB

                                                                                                                                                                  D

                                                                                                                                                                  C

                                                                                                                                                                  28 m

                                                                                                                                                                  21 m

                                                                                                                                                                  A

                                                                                                                                                                  Q

                                                                                                                                                                  Soil (1)Soil (2)

                                                                                                                                                                  73deg

                                                                                                                                                                  Figure Q91

                                                                                                                                                                  Stability of slopes 75

                                                                                                                                                                  Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                                  9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                                                  599 256 328 1372

                                                                                                                                                                  Figure Q93

                                                                                                                                                                  76 Stability of slopes

                                                                                                                                                                  XW cos frac14 b

                                                                                                                                                                  Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                                                  W sin frac14 bX

                                                                                                                                                                  h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                                                  Arc length La frac14

                                                                                                                                                                  180 57

                                                                                                                                                                  1

                                                                                                                                                                  2 326 frac14 327m

                                                                                                                                                                  The factor of safety is given by

                                                                                                                                                                  F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                                                  W sin

                                                                                                                                                                  frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                                                  frac14 091

                                                                                                                                                                  According to the limit state method

                                                                                                                                                                  0d frac14 tan1tan 32

                                                                                                                                                                  125

                                                                                                                                                                  frac14 265

                                                                                                                                                                  c0 frac14 8

                                                                                                                                                                  160frac14 5 kN=m2

                                                                                                                                                                  Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                                                  Design disturbing moment frac14 1075 kN=m

                                                                                                                                                                  The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                                                  94

                                                                                                                                                                  F frac14 1

                                                                                                                                                                  W sin

                                                                                                                                                                  Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                                                  1thorn ethtan tan0=FTHORN

                                                                                                                                                                  c0 frac14 8 kN=m2

                                                                                                                                                                  0 frac14 32

                                                                                                                                                                  c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                                                  W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                                                  Try F frac14 100

                                                                                                                                                                  tan0

                                                                                                                                                                  Ffrac14 0625

                                                                                                                                                                  Stability of slopes 77

                                                                                                                                                                  Values of u are as obtained in Figure Q93

                                                                                                                                                                  SliceNo

                                                                                                                                                                  h(m)

                                                                                                                                                                  W frac14 bh(kNm)

                                                                                                                                                                  W sin(kNm)

                                                                                                                                                                  ub(kNm)

                                                                                                                                                                  c0bthorn (W ub) tan0(kNm)

                                                                                                                                                                  sec

                                                                                                                                                                  1thorn (tan tan0)FProduct(kNm)

                                                                                                                                                                  1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                                                  23 33 30 1042 31

                                                                                                                                                                  3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                                                  224 92 72 0931 67

                                                                                                                                                                  6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                                                  12 58 112 90 0889 80

                                                                                                                                                                  8 60 252 1812

                                                                                                                                                                  80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                                                  10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                                                  2154 88 116 0853 99

                                                                                                                                                                  14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                                                  1074 1091

                                                                                                                                                                  F frac14 1091

                                                                                                                                                                  1074frac14 102 (assumed value 100)

                                                                                                                                                                  Thus

                                                                                                                                                                  F frac14 101

                                                                                                                                                                  95

                                                                                                                                                                  F frac14 1

                                                                                                                                                                  W sin

                                                                                                                                                                  XfWeth1 ruTHORN tan0g sec

                                                                                                                                                                  1thorn ethtan tan0THORN=F

                                                                                                                                                                  0 frac14 33

                                                                                                                                                                  ru frac14 020

                                                                                                                                                                  W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                                                  eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                                                  Try F frac14 110

                                                                                                                                                                  tan 0

                                                                                                                                                                  Ffrac14 tan 33

                                                                                                                                                                  110frac14 0590

                                                                                                                                                                  78 Stability of slopes

                                                                                                                                                                  Referring to Figure Q95

                                                                                                                                                                  SliceNo

                                                                                                                                                                  h(m)

                                                                                                                                                                  W frac14 bh(kNm)

                                                                                                                                                                  W sin(kNm)

                                                                                                                                                                  W(1 ru) tan0(kNm)

                                                                                                                                                                  sec

                                                                                                                                                                  1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                                                  1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                                                  2120 234 0892 209

                                                                                                                                                                  4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                                                  1185 1271

                                                                                                                                                                  Figure Q95

                                                                                                                                                                  Stability of slopes 79

                                                                                                                                                                  F frac14 1271

                                                                                                                                                                  1185frac14 107

                                                                                                                                                                  The trial value was 110 therefore take F to be 108

                                                                                                                                                                  96

                                                                                                                                                                  (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                                  F frac14 0

                                                                                                                                                                  sat

                                                                                                                                                                  tan0

                                                                                                                                                                  tan

                                                                                                                                                                  ptie 15 frac14 92

                                                                                                                                                                  19

                                                                                                                                                                  tan 36

                                                                                                                                                                  tan

                                                                                                                                                                  tan frac14 0234

                                                                                                                                                                  frac14 13

                                                                                                                                                                  Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                                  F frac14 tan0

                                                                                                                                                                  tan

                                                                                                                                                                  frac14 tan 36

                                                                                                                                                                  tan 13

                                                                                                                                                                  frac14 31

                                                                                                                                                                  (b) 0d frac14 tan1tan 36

                                                                                                                                                                  125

                                                                                                                                                                  frac14 30

                                                                                                                                                                  Depth of potential failure surface frac14 z

                                                                                                                                                                  Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                                  frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                                  frac14 504z kN

                                                                                                                                                                  Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                                  frac14 19 z sin 13 cos 13

                                                                                                                                                                  frac14 416z kN

                                                                                                                                                                  Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                                  80 Stability of slopes

                                                                                                                                                                  • Book Cover
                                                                                                                                                                  • Title
                                                                                                                                                                  • Contents
                                                                                                                                                                  • Basic characteristics of soils
                                                                                                                                                                  • Seepage
                                                                                                                                                                  • Effective stress
                                                                                                                                                                  • Shear strength
                                                                                                                                                                  • Stresses and displacements
                                                                                                                                                                  • Lateral earth pressure
                                                                                                                                                                  • Consolidation theory
                                                                                                                                                                  • Bearing capacity
                                                                                                                                                                  • Stability of slopes

                                                                                                                                                                    Slice No h cos (m) h sin (m) uw (m) u (kNm2) l (m) ul (kNm)

                                                                                                                                                                    9 59 24 59 578 22 12710 59 29 60 588 22 12911 56 33 57 559 23 12912 52 35 52 510 24 12213 46 37 45 441 25 11014 34 32 34 333 27 9015 016 0190 18 176 029 0051

                                                                                                                                                                    599 256 328 1372

                                                                                                                                                                    Figure Q93

                                                                                                                                                                    76 Stability of slopes

                                                                                                                                                                    XW cos frac14 b

                                                                                                                                                                    Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                                                    W sin frac14 bX

                                                                                                                                                                    h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                                                    Arc length La frac14

                                                                                                                                                                    180 57

                                                                                                                                                                    1

                                                                                                                                                                    2 326 frac14 327m

                                                                                                                                                                    The factor of safety is given by

                                                                                                                                                                    F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                                                    W sin

                                                                                                                                                                    frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                                                    frac14 091

                                                                                                                                                                    According to the limit state method

                                                                                                                                                                    0d frac14 tan1tan 32

                                                                                                                                                                    125

                                                                                                                                                                    frac14 265

                                                                                                                                                                    c0 frac14 8

                                                                                                                                                                    160frac14 5 kN=m2

                                                                                                                                                                    Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                                                    Design disturbing moment frac14 1075 kN=m

                                                                                                                                                                    The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                                                    94

                                                                                                                                                                    F frac14 1

                                                                                                                                                                    W sin

                                                                                                                                                                    Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                                                    1thorn ethtan tan0=FTHORN

                                                                                                                                                                    c0 frac14 8 kN=m2

                                                                                                                                                                    0 frac14 32

                                                                                                                                                                    c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                                                    W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                                                    Try F frac14 100

                                                                                                                                                                    tan0

                                                                                                                                                                    Ffrac14 0625

                                                                                                                                                                    Stability of slopes 77

                                                                                                                                                                    Values of u are as obtained in Figure Q93

                                                                                                                                                                    SliceNo

                                                                                                                                                                    h(m)

                                                                                                                                                                    W frac14 bh(kNm)

                                                                                                                                                                    W sin(kNm)

                                                                                                                                                                    ub(kNm)

                                                                                                                                                                    c0bthorn (W ub) tan0(kNm)

                                                                                                                                                                    sec

                                                                                                                                                                    1thorn (tan tan0)FProduct(kNm)

                                                                                                                                                                    1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                                                    23 33 30 1042 31

                                                                                                                                                                    3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                                                    224 92 72 0931 67

                                                                                                                                                                    6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                                                    12 58 112 90 0889 80

                                                                                                                                                                    8 60 252 1812

                                                                                                                                                                    80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                                                    10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                                                    2154 88 116 0853 99

                                                                                                                                                                    14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                                                    1074 1091

                                                                                                                                                                    F frac14 1091

                                                                                                                                                                    1074frac14 102 (assumed value 100)

                                                                                                                                                                    Thus

                                                                                                                                                                    F frac14 101

                                                                                                                                                                    95

                                                                                                                                                                    F frac14 1

                                                                                                                                                                    W sin

                                                                                                                                                                    XfWeth1 ruTHORN tan0g sec

                                                                                                                                                                    1thorn ethtan tan0THORN=F

                                                                                                                                                                    0 frac14 33

                                                                                                                                                                    ru frac14 020

                                                                                                                                                                    W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                                                    eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                                                    Try F frac14 110

                                                                                                                                                                    tan 0

                                                                                                                                                                    Ffrac14 tan 33

                                                                                                                                                                    110frac14 0590

                                                                                                                                                                    78 Stability of slopes

                                                                                                                                                                    Referring to Figure Q95

                                                                                                                                                                    SliceNo

                                                                                                                                                                    h(m)

                                                                                                                                                                    W frac14 bh(kNm)

                                                                                                                                                                    W sin(kNm)

                                                                                                                                                                    W(1 ru) tan0(kNm)

                                                                                                                                                                    sec

                                                                                                                                                                    1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                                                    1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                                                    2120 234 0892 209

                                                                                                                                                                    4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                                                    1185 1271

                                                                                                                                                                    Figure Q95

                                                                                                                                                                    Stability of slopes 79

                                                                                                                                                                    F frac14 1271

                                                                                                                                                                    1185frac14 107

                                                                                                                                                                    The trial value was 110 therefore take F to be 108

                                                                                                                                                                    96

                                                                                                                                                                    (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                                    F frac14 0

                                                                                                                                                                    sat

                                                                                                                                                                    tan0

                                                                                                                                                                    tan

                                                                                                                                                                    ptie 15 frac14 92

                                                                                                                                                                    19

                                                                                                                                                                    tan 36

                                                                                                                                                                    tan

                                                                                                                                                                    tan frac14 0234

                                                                                                                                                                    frac14 13

                                                                                                                                                                    Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                                    F frac14 tan0

                                                                                                                                                                    tan

                                                                                                                                                                    frac14 tan 36

                                                                                                                                                                    tan 13

                                                                                                                                                                    frac14 31

                                                                                                                                                                    (b) 0d frac14 tan1tan 36

                                                                                                                                                                    125

                                                                                                                                                                    frac14 30

                                                                                                                                                                    Depth of potential failure surface frac14 z

                                                                                                                                                                    Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                                    frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                                    frac14 504z kN

                                                                                                                                                                    Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                                    frac14 19 z sin 13 cos 13

                                                                                                                                                                    frac14 416z kN

                                                                                                                                                                    Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                                    80 Stability of slopes

                                                                                                                                                                    • Book Cover
                                                                                                                                                                    • Title
                                                                                                                                                                    • Contents
                                                                                                                                                                    • Basic characteristics of soils
                                                                                                                                                                    • Seepage
                                                                                                                                                                    • Effective stress
                                                                                                                                                                    • Shear strength
                                                                                                                                                                    • Stresses and displacements
                                                                                                                                                                    • Lateral earth pressure
                                                                                                                                                                    • Consolidation theory
                                                                                                                                                                    • Bearing capacity
                                                                                                                                                                    • Stability of slopes

                                                                                                                                                                      XW cos frac14 b

                                                                                                                                                                      Xh cos frac14 21 2 599 frac14 2516 kN=mX

                                                                                                                                                                      W sin frac14 bX

                                                                                                                                                                      h sin frac14 21 2 256 frac14 1075 kN=mXethW cos ulTHORN frac14 2516 1372 frac14 1144 kN=m

                                                                                                                                                                      Arc length La frac14

                                                                                                                                                                      180 57

                                                                                                                                                                      1

                                                                                                                                                                      2 326 frac14 327m

                                                                                                                                                                      The factor of safety is given by

                                                                                                                                                                      F frac14 c0La thorn tan0ethW cos ulTHORN

                                                                                                                                                                      W sin

                                                                                                                                                                      frac14 eth8 327THORN thorn ethtan 32 1144THORN1075

                                                                                                                                                                      frac14 091

                                                                                                                                                                      According to the limit state method

                                                                                                                                                                      0d frac14 tan1tan 32

                                                                                                                                                                      125

                                                                                                                                                                      frac14 265

                                                                                                                                                                      c0 frac14 8

                                                                                                                                                                      160frac14 5 kN=m2

                                                                                                                                                                      Design resisting moment frac14 eth5 327THORN thorn ethtan 265 1144THORN frac14 734 kN=m

                                                                                                                                                                      Design disturbing moment frac14 1075 kN=m

                                                                                                                                                                      The design resisting moment is less than the design disturbing moment therefore a slipwill occur

                                                                                                                                                                      94

                                                                                                                                                                      F frac14 1

                                                                                                                                                                      W sin

                                                                                                                                                                      Xfc0bthorn ethW ubTHORN tan0g sec

                                                                                                                                                                      1thorn ethtan tan0=FTHORN

                                                                                                                                                                      c0 frac14 8 kN=m2

                                                                                                                                                                      0 frac14 32

                                                                                                                                                                      c0b frac14 8 2 frac14 16 kN=m

                                                                                                                                                                      W frac14 bh frac14 21 2 h frac14 42h kN=m

                                                                                                                                                                      Try F frac14 100

                                                                                                                                                                      tan0

                                                                                                                                                                      Ffrac14 0625

                                                                                                                                                                      Stability of slopes 77

                                                                                                                                                                      Values of u are as obtained in Figure Q93

                                                                                                                                                                      SliceNo

                                                                                                                                                                      h(m)

                                                                                                                                                                      W frac14 bh(kNm)

                                                                                                                                                                      W sin(kNm)

                                                                                                                                                                      ub(kNm)

                                                                                                                                                                      c0bthorn (W ub) tan0(kNm)

                                                                                                                                                                      sec

                                                                                                                                                                      1thorn (tan tan0)FProduct(kNm)

                                                                                                                                                                      1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                                                      23 33 30 1042 31

                                                                                                                                                                      3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                                                      224 92 72 0931 67

                                                                                                                                                                      6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                                                      12 58 112 90 0889 80

                                                                                                                                                                      8 60 252 1812

                                                                                                                                                                      80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                                                      10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                                                      2154 88 116 0853 99

                                                                                                                                                                      14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                                                      1074 1091

                                                                                                                                                                      F frac14 1091

                                                                                                                                                                      1074frac14 102 (assumed value 100)

                                                                                                                                                                      Thus

                                                                                                                                                                      F frac14 101

                                                                                                                                                                      95

                                                                                                                                                                      F frac14 1

                                                                                                                                                                      W sin

                                                                                                                                                                      XfWeth1 ruTHORN tan0g sec

                                                                                                                                                                      1thorn ethtan tan0THORN=F

                                                                                                                                                                      0 frac14 33

                                                                                                                                                                      ru frac14 020

                                                                                                                                                                      W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                                                      eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                                                      Try F frac14 110

                                                                                                                                                                      tan 0

                                                                                                                                                                      Ffrac14 tan 33

                                                                                                                                                                      110frac14 0590

                                                                                                                                                                      78 Stability of slopes

                                                                                                                                                                      Referring to Figure Q95

                                                                                                                                                                      SliceNo

                                                                                                                                                                      h(m)

                                                                                                                                                                      W frac14 bh(kNm)

                                                                                                                                                                      W sin(kNm)

                                                                                                                                                                      W(1 ru) tan0(kNm)

                                                                                                                                                                      sec

                                                                                                                                                                      1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                                                      1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                                                      2120 234 0892 209

                                                                                                                                                                      4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                                                      1185 1271

                                                                                                                                                                      Figure Q95

                                                                                                                                                                      Stability of slopes 79

                                                                                                                                                                      F frac14 1271

                                                                                                                                                                      1185frac14 107

                                                                                                                                                                      The trial value was 110 therefore take F to be 108

                                                                                                                                                                      96

                                                                                                                                                                      (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                                      F frac14 0

                                                                                                                                                                      sat

                                                                                                                                                                      tan0

                                                                                                                                                                      tan

                                                                                                                                                                      ptie 15 frac14 92

                                                                                                                                                                      19

                                                                                                                                                                      tan 36

                                                                                                                                                                      tan

                                                                                                                                                                      tan frac14 0234

                                                                                                                                                                      frac14 13

                                                                                                                                                                      Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                                      F frac14 tan0

                                                                                                                                                                      tan

                                                                                                                                                                      frac14 tan 36

                                                                                                                                                                      tan 13

                                                                                                                                                                      frac14 31

                                                                                                                                                                      (b) 0d frac14 tan1tan 36

                                                                                                                                                                      125

                                                                                                                                                                      frac14 30

                                                                                                                                                                      Depth of potential failure surface frac14 z

                                                                                                                                                                      Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                                      frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                                      frac14 504z kN

                                                                                                                                                                      Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                                      frac14 19 z sin 13 cos 13

                                                                                                                                                                      frac14 416z kN

                                                                                                                                                                      Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                                      80 Stability of slopes

                                                                                                                                                                      • Book Cover
                                                                                                                                                                      • Title
                                                                                                                                                                      • Contents
                                                                                                                                                                      • Basic characteristics of soils
                                                                                                                                                                      • Seepage
                                                                                                                                                                      • Effective stress
                                                                                                                                                                      • Shear strength
                                                                                                                                                                      • Stresses and displacements
                                                                                                                                                                      • Lateral earth pressure
                                                                                                                                                                      • Consolidation theory
                                                                                                                                                                      • Bearing capacity
                                                                                                                                                                      • Stability of slopes

                                                                                                                                                                        Values of u are as obtained in Figure Q93

                                                                                                                                                                        SliceNo

                                                                                                                                                                        h(m)

                                                                                                                                                                        W frac14 bh(kNm)

                                                                                                                                                                        W sin(kNm)

                                                                                                                                                                        ub(kNm)

                                                                                                                                                                        c0bthorn (W ub) tan0(kNm)

                                                                                                                                                                        sec

                                                                                                                                                                        1thorn (tan tan0)FProduct(kNm)

                                                                                                                                                                        1 05 21 6 2 8 24 1078 262 13 55 31

                                                                                                                                                                        23 33 30 1042 31

                                                                                                                                                                        3 24 101 0 0 59 42 1000 424 34 143 4 10 76 58 0960 565 43 181 71

                                                                                                                                                                        224 92 72 0931 67

                                                                                                                                                                        6 50 210 11 40 100 85 0907 777 55 231 14

                                                                                                                                                                        12 58 112 90 0889 80

                                                                                                                                                                        8 60 252 1812

                                                                                                                                                                        80 114 102 0874 899 63 265 22 99 116 109 0861 94

                                                                                                                                                                        10 65 273 26 120 118 113 0854 9711 65 273 30 136 112 117 0850 9912 63 265 34 148 102 118 0847 10013 59 248 381

                                                                                                                                                                        2154 88 116 0853 99

                                                                                                                                                                        14 46 193 43 132 67 95 0862 8215 25 105 48 0078 35 59 0882 0052

                                                                                                                                                                        1074 1091

                                                                                                                                                                        F frac14 1091

                                                                                                                                                                        1074frac14 102 (assumed value 100)

                                                                                                                                                                        Thus

                                                                                                                                                                        F frac14 101

                                                                                                                                                                        95

                                                                                                                                                                        F frac14 1

                                                                                                                                                                        W sin

                                                                                                                                                                        XfWeth1 ruTHORN tan0g sec

                                                                                                                                                                        1thorn ethtan tan0THORN=F

                                                                                                                                                                        0 frac14 33

                                                                                                                                                                        ru frac14 020

                                                                                                                                                                        W frac14 bh frac14 20 5 h frac14 100h kN=m

                                                                                                                                                                        eth1 ruTHORN tan0 frac14 080 tan 33 frac14 0520

                                                                                                                                                                        Try F frac14 110

                                                                                                                                                                        tan 0

                                                                                                                                                                        Ffrac14 tan 33

                                                                                                                                                                        110frac14 0590

                                                                                                                                                                        78 Stability of slopes

                                                                                                                                                                        Referring to Figure Q95

                                                                                                                                                                        SliceNo

                                                                                                                                                                        h(m)

                                                                                                                                                                        W frac14 bh(kNm)

                                                                                                                                                                        W sin(kNm)

                                                                                                                                                                        W(1 ru) tan0(kNm)

                                                                                                                                                                        sec

                                                                                                                                                                        1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                                                        1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                                                        2120 234 0892 209

                                                                                                                                                                        4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                                                        1185 1271

                                                                                                                                                                        Figure Q95

                                                                                                                                                                        Stability of slopes 79

                                                                                                                                                                        F frac14 1271

                                                                                                                                                                        1185frac14 107

                                                                                                                                                                        The trial value was 110 therefore take F to be 108

                                                                                                                                                                        96

                                                                                                                                                                        (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                                        F frac14 0

                                                                                                                                                                        sat

                                                                                                                                                                        tan0

                                                                                                                                                                        tan

                                                                                                                                                                        ptie 15 frac14 92

                                                                                                                                                                        19

                                                                                                                                                                        tan 36

                                                                                                                                                                        tan

                                                                                                                                                                        tan frac14 0234

                                                                                                                                                                        frac14 13

                                                                                                                                                                        Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                                        F frac14 tan0

                                                                                                                                                                        tan

                                                                                                                                                                        frac14 tan 36

                                                                                                                                                                        tan 13

                                                                                                                                                                        frac14 31

                                                                                                                                                                        (b) 0d frac14 tan1tan 36

                                                                                                                                                                        125

                                                                                                                                                                        frac14 30

                                                                                                                                                                        Depth of potential failure surface frac14 z

                                                                                                                                                                        Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                                        frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                                        frac14 504z kN

                                                                                                                                                                        Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                                        frac14 19 z sin 13 cos 13

                                                                                                                                                                        frac14 416z kN

                                                                                                                                                                        Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                                        80 Stability of slopes

                                                                                                                                                                        • Book Cover
                                                                                                                                                                        • Title
                                                                                                                                                                        • Contents
                                                                                                                                                                        • Basic characteristics of soils
                                                                                                                                                                        • Seepage
                                                                                                                                                                        • Effective stress
                                                                                                                                                                        • Shear strength
                                                                                                                                                                        • Stresses and displacements
                                                                                                                                                                        • Lateral earth pressure
                                                                                                                                                                        • Consolidation theory
                                                                                                                                                                        • Bearing capacity
                                                                                                                                                                        • Stability of slopes

                                                                                                                                                                          Referring to Figure Q95

                                                                                                                                                                          SliceNo

                                                                                                                                                                          h(m)

                                                                                                                                                                          W frac14 bh(kNm)

                                                                                                                                                                          W sin(kNm)

                                                                                                                                                                          W(1 ru) tan0(kNm)

                                                                                                                                                                          sec

                                                                                                                                                                          1thorn ( tan tan0)FProduct(kNm)

                                                                                                                                                                          1 15 75 4 5 20 0963 192 31 310 9 48 161 0926 1493 45 450 151

                                                                                                                                                                          2120 234 0892 209

                                                                                                                                                                          4 53 530 21 190 276 0873 2415 60 600 28 282 312 0862 2696 50 500 35 287 260 0864 2257 34 340 43 232 177 0882 1568 14 28 49 21 3 0908 3

                                                                                                                                                                          1185 1271

                                                                                                                                                                          Figure Q95

                                                                                                                                                                          Stability of slopes 79

                                                                                                                                                                          F frac14 1271

                                                                                                                                                                          1185frac14 107

                                                                                                                                                                          The trial value was 110 therefore take F to be 108

                                                                                                                                                                          96

                                                                                                                                                                          (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                                          F frac14 0

                                                                                                                                                                          sat

                                                                                                                                                                          tan0

                                                                                                                                                                          tan

                                                                                                                                                                          ptie 15 frac14 92

                                                                                                                                                                          19

                                                                                                                                                                          tan 36

                                                                                                                                                                          tan

                                                                                                                                                                          tan frac14 0234

                                                                                                                                                                          frac14 13

                                                                                                                                                                          Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                                          F frac14 tan0

                                                                                                                                                                          tan

                                                                                                                                                                          frac14 tan 36

                                                                                                                                                                          tan 13

                                                                                                                                                                          frac14 31

                                                                                                                                                                          (b) 0d frac14 tan1tan 36

                                                                                                                                                                          125

                                                                                                                                                                          frac14 30

                                                                                                                                                                          Depth of potential failure surface frac14 z

                                                                                                                                                                          Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                                          frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                                          frac14 504z kN

                                                                                                                                                                          Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                                          frac14 19 z sin 13 cos 13

                                                                                                                                                                          frac14 416z kN

                                                                                                                                                                          Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                                          80 Stability of slopes

                                                                                                                                                                          • Book Cover
                                                                                                                                                                          • Title
                                                                                                                                                                          • Contents
                                                                                                                                                                          • Basic characteristics of soils
                                                                                                                                                                          • Seepage
                                                                                                                                                                          • Effective stress
                                                                                                                                                                          • Shear strength
                                                                                                                                                                          • Stresses and displacements
                                                                                                                                                                          • Lateral earth pressure
                                                                                                                                                                          • Consolidation theory
                                                                                                                                                                          • Bearing capacity
                                                                                                                                                                          • Stability of slopes

                                                                                                                                                                            F frac14 1271

                                                                                                                                                                            1185frac14 107

                                                                                                                                                                            The trial value was 110 therefore take F to be 108

                                                                                                                                                                            96

                                                                                                                                                                            (a) Water table at surface the factor of safety is given by Equation 912

                                                                                                                                                                            F frac14 0

                                                                                                                                                                            sat

                                                                                                                                                                            tan0

                                                                                                                                                                            tan

                                                                                                                                                                            ptie 15 frac14 92

                                                                                                                                                                            19

                                                                                                                                                                            tan 36

                                                                                                                                                                            tan

                                                                                                                                                                            tan frac14 0234

                                                                                                                                                                            frac14 13

                                                                                                                                                                            Water table well below surface the factor of safety is given by Equation 911

                                                                                                                                                                            F frac14 tan0

                                                                                                                                                                            tan

                                                                                                                                                                            frac14 tan 36

                                                                                                                                                                            tan 13

                                                                                                                                                                            frac14 31

                                                                                                                                                                            (b) 0d frac14 tan1tan 36

                                                                                                                                                                            125

                                                                                                                                                                            frac14 30

                                                                                                                                                                            Depth of potential failure surface frac14 z

                                                                                                                                                                            Design resisting moment per unit area Rd frac14 eth uTHORN tan 0

                                                                                                                                                                            frac14 0z cos2 tan 0dfrac14 92 z cos2 13 tan 30

                                                                                                                                                                            frac14 504z kN

                                                                                                                                                                            Design disturbing moment per unit area Sd frac14 sat sin cos

                                                                                                                                                                            frac14 19 z sin 13 cos 13

                                                                                                                                                                            frac14 416z kN

                                                                                                                                                                            Rd gtSd therefore the limit state for overall stability is satisfied

                                                                                                                                                                            80 Stability of slopes

                                                                                                                                                                            • Book Cover
                                                                                                                                                                            • Title
                                                                                                                                                                            • Contents
                                                                                                                                                                            • Basic characteristics of soils
                                                                                                                                                                            • Seepage
                                                                                                                                                                            • Effective stress
                                                                                                                                                                            • Shear strength
                                                                                                                                                                            • Stresses and displacements
                                                                                                                                                                            • Lateral earth pressure
                                                                                                                                                                            • Consolidation theory
                                                                                                                                                                            • Bearing capacity
                                                                                                                                                                            • Stability of slopes

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